Generalized Continued Logarithms and Related Continued Fractions

Jonathan M. Borwein^{1}

CARMA University of Newcastle

Building V University Drive

Callaghan NSW 2308 Australia

Kevin G. Hare^{2} and Jason G. Lynch^{1} Department of Pure Mathematics

University of Waterloo 200 University Ave West Waterloo, ON N2L 3G1

Canada

kghare@uwaterloo.ca
j4lynch@uwaterloo.ca

In memory of Jonathan Borwein (1951–2016)

Abstract

We study continued logarithms, as introduced by Gosper and studied by Borwein et al. After providing an overview of the type I and type II generalizations of binary continued logarithms introduced by Borwein et al., we focus on a new generalization to an arbitrary integer base b. We show that all of our so-called type III continued logarithms converge and all rational numbers have finite type III continued logarithms. As with simple continued fractions, we show that the continued logarithm terms, for almost every real number, follow a specific distribution. We also generalize Khinchin’s constant from simple continued fractions to continued logarithms, and show that these logarithmic Khinchin constants have an elementary closed form. Finally, we show that simple continued fractions are the limiting case of our continued logarithms, and briefly consider how we could generalize beyond continued logarithms.

1Research of J. M. Borwein and J. G. Lynch was supported by CARMA, University of Newcastle.

2Research of K. G. Hare was supported by NSERC Grant RGPIN-2014-03154.

1 Introduction

Continued fractions, especially simple continued fractions, have been well studied throughout history. Continued binary logarithms, however, appear to have first been introduced by Gosper in his appendix on continued fraction arithmetic [4]. More recently, Borwein et al. proved some basic results about binary continued logarithms and applied experimental methods to determine the term distribution of binary continued logarithms [2]. They conjectured and indicated a proof that, like in the case of continued fractions, almost every real number has continued logarithm terms that follow a specific distribution. They then introduced two different generalizations of binary continued logarithms to arbitrary bases.

1.1 The structure of this paper

Section 1 introduces some basic definitions and results for continued fractions, briefly describes binary continued logarithms as introduced by Gosper, and provides an overview of results relating to the Khinchin constant for continued fractions. Sections 2 and 3 then provide an overview of the type I and type II continued logarithms introduced by Borwein et al. Further details on these can be found in the work of Borwein et al. [2].

Section 4 comprises the main body of the paper. In Section 4.1 we define type III continued logarithms and extend to them the standard continued fraction recurrences. Section 4.2 then proves that type III continued logarithms are guaranteed to converge to the correct value, and that every rational number has a finite type III continued logarithm. These are two desirable properties of continued fractions and binary continued logarithms that a complete generalization should have. In Section 4.3 we describe how measure theory can be used to investigate the distribution of continued logarithm terms. This is then applied in Section 4.4 to determine the distribution, and Section 4.5 to determine the logarithmic Khinchin constant. The main proofs of these sections are quite technical, and are separated out into Appendices A and B, respectively. Finally, Section 4.6 derives some relationships between simple continued fractions and the limiting case of type III continued logarithms.

Finally, we close the paper in Section 5 by briefly introducing one way to generalize past continued logarithms.

1.2 Continued fractions

The material in this section can be found in many places, including Borwein et al. [3] and Khinchin [5].

Definition 1. A continued fraction is an expression of the form

y_{1} = α_{0} + \frac{β_{1}}{α_{1} + \frac{β_{2}}{α_{2} + \frac{β_{3}}{...}}}

or

y_{2} = α_{0} + \frac{β_{1}}{α_{1} + \frac{β_{2}}{α_{2} + \frac{β_{3}}{⋯ + \frac{β_{n}}{α_{n}}}}}.

For the sake of simplicity, we will sometimes denote the above as

y_{1} = α_{0} + \frac{β_{1}}{α_{1}} + \frac{β_{2}}{α_{2}} + ⋯

or

y_{2} = α_{0} + \frac{β_{1}}{α_{1}} + \frac{β_{2}}{α_{2}} + ⋯ + \frac{β_{n}}{α_{n}} ,

respectively. The terms α_{0}, α_{1}, … are called denominator terms and the terms β_{1}, β_{2}, … are called numerator terms.

Definition 2. Two continued fractions

y = α_{0} + \frac{β_{1}}{α_{1}} + \frac{β_{2}}{α_{2}} + ⋯  and  y′ = α′_{0} + \frac{β′_{1}}{α′_{1}} + \frac{β′_{2}}{α′_{2}} + ⋯

are called equivalent if there is a sequence (d_{n})_{n=0}^{∞} with d_{0} = 1 such that α′_{n} = d_{n}α_{n} for all n ≥ 0 and β′_{n} = d_{n}d_{n−1}β_{n} for all n ≥ 1.

The c_{n} terms can be thought of as constants that are multiplied by both numerators and denominators of successive terms.

Definition 3. The nth convergent of the continued fraction

y = α_{0} + \frac{β_{1}}{α_{1}} + \frac{β_{2}}{α_{2}} + ⋯

is given by

x_{n} = α_{0} + \frac{β_{1}}{α_{1}} + \frac{β_{2}}{α_{2}} + ⋯ + \frac{β_{n}}{α_{n}} .

Definition 4. The nth remainder term of the continued fraction

y = α_{0} + \frac{β_{1}}{α_{1}} + \frac{β_{2}}{α_{2}} + ⋯

is given by

r_{n} = α_{n} + \frac{β_{n+1}}{α_{n+1}} + \frac{β_{n+2}}{α_{n+2}} + · · · .

The following results will be useful for generalizing to continued logarithms.

Fact 5. Suppose x = α_{0}+ \frac{β_{1}}{α_{1}} + \frac{β_{2}}{α_{2}} +⋯, where α_{n}, β_{n} > 0 for all n. Then the convergents

are given by

x_{n} = \frac{p_{n}}{q_{n}}

where

p_{−1} = 1, q_{−1} = 0, p_{0} = α_{0}, q_{0} = 1,
p_{n} = α_{n}p_{n−1} + β_{n}p_{n−2}  n ≥ 1,
q_{n} = α_{n}q_{n−1} + β_{n}q_{n−2}  n ≥ 1.

Fact 6. Suppose x = α_{0} + \frac{β_{1}}{α_{1}} + \frac{β_{2}}{α_{2}} + ⋯ where α_{n}, β_{n} > 0 for all n. Then the continued

fraction for x converges to x if ∑_{n=1}^{∞} \frac{α_{n}α_{n+1}}{β_{n+1}} = ∞.

Remark 7. Throughout this paper, we will use λ(A) or just λA to denote the Lebesgue measure of a set A ⊆ ℝ.

1.3 Binary continued logarithms

Let 1 ≤ α ∈ ℝ. Let y_{0} = α and recursively define a_{n} = ⌊log_{2} y_{n}⌋. If y_{n} − 2^{a_{n}} = 0, then terminate. Otherwise, set

y_{n+1} = \frac{2^{a_{n}}}{y_{n} − 2^{a_{n}}}

and recurse. This produces the binary (base 2) continued logarithm for y_{0}:

y_{0} = 2^{a_{0}} + \frac{2^{a_{0}}}{2^{a_{1}}} + \frac{2^{a_{1}}}{2^{a_{2}}} + \frac{2^{a_{2}}}{2^{a_{3}}} + ⋯

These binary continued logarithms were introduced explicitly by Gosper in his appendix on continued fraction arithmetic [4]. Borwein et al. [2] further studied binary continued logarithms further, extending classical continued fraction recurrences for binary continued logs and investigating the distribution of aperiodic binary continued logarithm terms for quadratic irrationalities – such as can not occur for simple continued fractions.

Remark 8. Shallit [8] proved some limits on the length of a finite binary continued logarithm. Specifically, the binary continued logarithm for a rational number p/q ≥ 1 has at most 2 log_{2} p + O(1) terms. Furthermore, this bound is tight, as can be seen by considering the continued fraction for 2^{n} − 1. Moreover, the sum of the terms of the continued logarithm of p/q ≥ 1 is bounded by (log_{2} p)(2 log_{2} p + 2).

1.4 Khinchin’s constant

Khinchin [5] proved that for almost every α ∈ (0, 1), where

α = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + ⋯,

the denominator terms a_{1}, a_{2}, a_{3}, … follow a specific limiting distribution. That is, let P_{α}(k) = lim_{N→∞} \frac{1}{N} |\{n ≤ N : a_{n} = k\}|. This is the limiting ratio of the denominator terms that equal k, if this limit exists. Then for almost every α ∈ (0, 1),

P_{α}(k) = \frac{log (1 + \frac{1}{k(k+2)})}{log 2}

for every k ∈ ℕ. It then follows for almost every α ∈ (0, 1) that the limiting geometric mean of the denominator terms is given by

lim_{n→∞} \sqrt[n]{a_{1}a_{2} ⋯ a_{n}} = ∏_{k=1}^{∞} (1 + \frac{1}{k(k + 2)})^{log_{2} k} ≈ 2.685452.

This constant is now known as Khinchin’s constant, K.

2 Type I continued logarithms

2.1 Type I definition and preliminaries

Fix an integer base b ≥ 2. We define type I continued logarithms as follows.

Definition 9. Let α ∈ (1, ∞). The base b continued logarithm of type I for α is

b^{a_{0}} + \frac{(b − 1)b^{a_{0}}}{b^{a_{1}}} + \frac{(b − 1)b^{a_{1}}}{b^{a_{2}}} + \frac{(b − 1)b^{a_{2}}}{b^{a_{3}}} + ⋯ = [b^{a_{0}}, b^{a_{1}}, b^{a_{2}}, …]_{cl_{1}(b)},

where the terms a_{0}, a_{1}, a_{2}, … are determined by the recursive process below, terminating at the term b^{a_{n}} if at any point y_{n} = b^{a_{n}}.

y_{0} = α
a_{n} = ⌊log_{b} y_{n}⌋  n ≥ 0
y_{n+1} = \frac{(b − 1)b^{a_{n}}}{y_{n} − b^{a_{n}}}  n ≥ 0.

The numerator terms (b − 1)b^{a_{n}} are defined as such to ensure that y_{n} ∈ (1, ∞) for all n. Indeed, notice that for each n, we must have b^{a_{n}} ≤ y_{n} < b^{a_{n}+1}. Thus 0 ≤ y_{n} − b^{a_{n}} < (b − 1)b^{a_{n}}. If y_{n} − b^{a_{n}} = 0, then we terminate, otherwise we get 0 < y_{n} − b^{a_{n}} < (b − 1)b^{a_{n}}, so y_{n+1} = \frac{(b−1)b^{an}}{y_{n}−b^{an}} ∈ (1, ∞).

Borwein et al. proved that the type I continued fraction of α ∈ (1, ∞) will converge to α [2, Thm. 15]. Additionally, numbers with finite type I continued logarithms must be rational. However for b ≥ 3, rationals need not have finite continued logarithms. For example, the type I ternary continued logarithm for 2 is [3^{0}, 3^{0}, 3^{0}, …]_{cl_{1}(3)}.

2.2 Distribution of type I continued logarithm terms and type I logarithmic Khinchin constant

We now look at the limiting distribution of the type I continued logarithm terms. Consider α = [b^{a_{0}}, b^{a_{1}}, b^{a_{2}}, …]_{cl_{1}(b)}. Assume that the continued logarithm for α is infinite. Furthermore, assume (without loss of generality) that a_{0} = 0, so that α ∈ (1, b). These results can also be found in Lascu [7].

Definition 10. For n ∈ ℕ, let

D_{n}(k) = {α ∈ (1, b) : a_{n} = k}

denote the set of α ∈ (1, b) for which the nth continued logarithm term is b^{k}.

Definition 11. Let x = [1, b^{a_{1}}, b^{a_{2}}, …]_{cl_{1}(b)} ∈ (1, b). The nth remainder term of x is r_{n} = r_{n}(x) = [b^{a_{n}}, b^{a_{n+1}}, …]_{cl_{1}(b)}, as in Definition 4. Define

z_{n} = z_{n}(x) = \frac{r_{n}}{b^{a_{n}}} = [1, b^{a_{n+1}}, b^{a_{n+2}}, …]_{cl_{1}(b)} ∈ (1, b),

M_{n}(x) = {α ∈ (1, b) : z_{n}(α) < x} ⊆ (1, b),

m_{n}(x) = \frac{1}{b − 1}λ(M_{n}(x)) ∈ (0, 1),

and

m(x) = lim_{n→∞} m_{n}(x),

wherever this limit exists.

Notice that since 1 < z_{n}(α) < b for all n ∈ ℕ and α ∈ (1, b), we must have m_{n}(1) = 0 and m_{n}(b) = 1 for all n ∈ ℕ. We can now derive a recursion for the functions m_{n}.

Theorem 12. The sequence of functions m_{n} is given by the recursive relationship

m_{0}(x) = \frac{x − 1}{b − 1}  (1)
m_{n}(x) = ∑_{k=0}^{∞} m_{n−1}(1 + (b − 1)b^{−k}) − m_{n−1}(1 + x^{−1}(b − 1)b^{−k})  n ≥ 1  (2)

for 1 ≤ x ≤ b.

The proof of this is similar to that of Theorem 40.

We next derive a formula for D_{n}(k) in terms of the function m_{n}.

Theorem 13.

\frac{1}{b − 1}λ(D_{n+1}(k)) = m_{n}(1 + (b − 1)b^{−k}) − m_{n}(1 + (b − 1)b^{−(k+1)}).

The proof of this theorem is similar to that of Theorem 43.

Thus, if the limiting distribution m(x) exists, it immediately follows that

lim_{n→∞} \frac{1}{b − 1}λ(D_{n}(k)) = m(1 + (b − 1)b^{−k}) − m(1 + (b − 1)b^{−(k+1)}).  (3)

2.3 Experimentally determining the type I distribution

Now suppose b > 1 is an arbitrary integer. Let μ_{b} denote the limiting distribution function m for the base b, assuming it exists.

We may investigate the form of μ_{b}(x) by iterating the recurrence relation of Theorem 12 at points evenly spaced over the interval [1, b], starting with m_{0}(x) = \frac{x−1}{b−1}. At each iteration, we fit a spline to these points, evaluating each “infinite” sum to 100 terms, and breaking the interval [1, b] into 100 pieces. This is practicable since the continued logarithm converges much more rapidly than the simple continued fraction.

We find good convergence of μ_{b}(x) after around 10 iterations. We use the 101 data points from this process to seek the best fit to a function of the form

μ_{b}(x) = C log_{b} \frac{αx + β}{γx + δ} .

We set γ = 1 to eliminate any common factor between the numerator and denominator. To meet the boundary condition μ_{b}(1) = 0, we must have δ = α + β − 1, and to meet the boundary condition μ_{b}(b) = 1, we must have C = _{\frac{1}{log_{b} \frac{bα+β}{α+β+b−1}}} , leaving the functional form to

be fit as

μ_{b}(x) = \begin{matrix}log & b & \frac{αx+β}{x+α+β−1} \\ log & b & \frac{bα+β}{α+β+b−1} \end{matrix} .  (4)

We sought this superposition form when the simpler structure for simple continued fractions failed.

Fitting our data to the model suggests candidate values of α = \frac{1}{b} and β = \frac{b−1}{b}, from which we get

μ_{b}(x) = \begin{matrix}log & \frac{bx}{x+b−1} \\ log & \frac{b^{2}}{2b−1} \end{matrix} .  (5)

When we then apply (3), we get

lim_{n→∞} \frac{1}{b − 1}λD_{n}(k) = \begin{matrix}log & (1 + \frac{b^{k}(b−1)^{3}}{(b^{k+1}+b−1)^{2}} ) \\ log & \frac{b^{2}}{2b−1} \end{matrix}.

A proof of this distribution and of the type I Khinchin constant for each integer base b, using ergodic theory, was given by Lascu [6, 7]. Additionally, it is likely that the proofs in Appendices A and B for the type III continued logarithm distribution and logarithmic Khinchin constant could be appropriately adjusted to prove these results.

If a type I base b Khinchin constant KL^{I}_{b} exists (i.e., almost every α ∈ (1, ∞) has the same limiting geometric mean of denominator terms), and if a limiting distribution D(k) = lim_{n→∞} D_{n}(k) of denominator terms exists, then

KL^{I}_{b} = ∏_{k=0}^{∞} b^{k} \frac{λD(k)}{b − 1} = b^{∑_{k=0}^{∞} k \frac{MD(k)}{b−1}} .

This is because the limiting distribution of denominator terms (if it exists) is essentially the “average” distribution over all numbers α ∈ (1, b). If we then assume that almost every α ∈ (1, b) has the same limiting geometric mean of denominator terms, then this limiting geometric mean (the logarithmic type I Khinchin constant) must equal the limiting geometric mean of the “average” distribution.

Thus, if we assume KL^{I}_{b} exists and that the distribution in (3) is correct, then we must have KL^{I}_{b} = b^{A}, where

A = ∑_{k=0}^{∞} k \frac{λD(k)}{b − 1} = ∑_{k=0}^{∞} k[μ_{b}(1 + (b − 1)b^{−k}) − μ_{b}(1 + (b − 1)b^{−(k+1)})] = \frac{log b}{log \frac{b^{2}}{2b−1}} − 1,

by Theorem 13 and a lengthy but straightforward algebraic manipulation. These conjectured type I logarithmic Khinchin constants for 2 ≤ b ≤ 10 are given in Table 1.

\begin{matrix}b & KL^{I}_{b} \\ 2 & 2.656305058 \\ 3 & 2.598065150 \\ 4 & 2.556003239 \\ 5 & 2.524285360 \\ 6 & 2.499311827 \\ 7 & 2.478977440 \\ 8 & 2.461986788 \\ 9 & 2.447498976 \\ 10 & 2.434942582 \end{matrix}

Table 1: Type I logarithmic Khinchin constants for 2 ≤ b ≤ 10

These conjectured values of the type I logarithmic Khinchin constants were supported by empirical evidence, as the numerically computed limiting geometric means of denominator terms for various irrational constants give the expected values.

Notice that the type I logarithmic Khinchin constants have a simple closed form, which is noteworthy as no simple closed form has been found for the Khinchin constant for simple continued fractions.

3 Type II continued logarithms

3.1 Type II definition and preliminaries

Fix an integer base b ≥ 2. We define type II continued logarithms as follows.

Definition 14. Let α ∈ ℝ_{≥1}. The base b continued logarithm of type II for α is

c_{0}b^{a_{0}} + \frac{c_{0}b^{a_{0}}}{c_{1}b^{a_{1}}} + \frac{c_{1}b^{a_{1}}}{c_{2}b^{a_{2}}} + \frac{c_{2}b^{a_{2}}}{c_{3}b^{a_{3}}} + ⋯ = [c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{2}(b)},

where the terms a_{0}, a_{1}, a_{2}, … and c_{0}, c_{1}, c_{2}, … are determined by the recursive process below, terminating at the term c_{n}b^{a_{n}} if at any point y_{n} = c_{n}b^{a_{n}}.

y_{0} = α
a_{n} = ⌊log_{b} y_{n}⌋  n ≥ 0
c_{n} = ⌊\frac{y_{n}}{b^{a_{n}}} ⌋  n ≥ 0
y_{n+1} = \frac{c_{n}b^{a_{n}}}{y_{n} − c_{n}b^{a_{n}}}  n ≥ 0.

Remark 15. The numerator terms c_{n}b^{a_{n}} are defined to match the corresponding denominator terms. Recall that in the type I case, the term y_{n+1} could take any value in (1, ∞), regardless of the value of a_{n}. This is no longer true, since y_{n} − c_{n}b^{a_{n}} ∈ (0, b^{a_{n}}), so y_{n+1} ∈ (c_{n}, ∞). We will see later that this results in type II continued logarithms having a more complicated distribution for which we could not find a closed form. This issue was the inspiration for the definition of type III continued logarithms, where the numerator terms are b^{a_{n}} instead of c_{n}b^{a_{n}}.

Borwein et al. proved that the type II continued fraction of α ∈ (1, ∞) will converge to α, and that α ∈ (1, ∞) has a finite continued logarithm if and only if α ∈ ℚ [2, Thms. 19 & 20] – unlike the situation for type I.

3.2 Distribution of type II continued logarithm terms and type II logarithmic Khinchin constant

We now look at the limiting distribution of the type II continued logarithm terms. Consider α = [c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{2}(b)}. Assume that α ∉ ℚ, so that the continued logarithm for α is infinite. Furthermore, assume (without loss of generality) a_{0} = 0 and c_{0} = 1, so that α ∈ (1, 2).

Definition 16. Let n ∈ ℕ. Let

D_{n}(k, ℓ) = {α ∈ (1, 2) : a_{n} = k, c_{n} = ℓ}

denote the α ∈ (1, 2) for which the nth continued logarithm term is ℓb^{k}. Definition 17. Let x = [1, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{2}(b)} ∈ (1, 2) with nth remainder term r_{n} = r_{n}(x) = [c_{n}b^{a_{n}}, c_{n+1}b^{a_{n+1}}, …]_{cl_{2}(b)}, as in Definition 4. Define

z_{n} = z_{n}(x) = \frac{r_{n}}{c_{n}b^{a_{n}}} = [1, c_{n+1}b^{a_{n+1}}, c_{n+2}b^{a_{n+2}}, …]_{cl_{2}(b)} ∈ (1, 2),

M_{n}(x) = {α ∈ (1, 2) : z_{n}(α) < x} ⊆ (1, 2),
m_{n}(x) = λ(M_{n}(x)) ∈ (0, 1),

and

m(x) = lim_{n→∞} m_{n}(x),

wherever this limit exists.

Notice that since 1 ≤ z_{n}(α) ≤ 2 for all n ∈ ℕ and α ∈ (1, 2), we must have m_{n}(1) = 0 and m_{n}(2) = 1 for all n ∈ ℕ.

We may now derive a recursion relation for the functions m_{n}. Theorem 18. The sequence of functions m_{n} is given by the recursive relationship

m_{0}(x) = x − 1  (6)
m_{n}(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} m_{n−1}(1 + ℓ^{−1}b^{−k}) − m_{n−1}(max{1 + ℓ^{−1}b^{−k}x^{−1}, 1 + (ℓ + 1)^{−1}b^{−k}}) n ≥ 1 (7)

for 1 ≤ x ≤ 2.

We can now derive a formula for D_{n}(k, ℓ) in terms of the function m_{n}. Theorem 19.

λ(D_{n+1}(k, ℓ)) = m_{n}(1 + ℓ^{−1}b^{−k}) − m_{n}(1 + (ℓ + 1)^{−1}b^{−k}).

Thus, if the limiting distribution m(x) exists, it immediately follows that

lim_{n→∞} λ(D_{n}(k, ℓ)) = m(1 + ℓ^{−1}b^{−k}) − m(1 + (ℓ + 1)^{−1}b^{−k}).  (8)

3.3 Experimentally determining the type II distribution

Again, suppose b is arbitrary. Let μ_{b} denote the limiting distribution function m for the base b, assuming it exists.

We may again investigate the form of μ_{b}(x) by iterating the recurrence relation of Theorem 18 at points evenly spaced over the interval [1, 2], starting with m_{0}(x) = x − 1. At each iteration, we fit a spline to these points, evaluating each “infinite” sum to 100 terms, and breaking the interval [1, 2] into 100 pieces. We find good convergence of μ_{b}(x) after around 10 iterations. However, we have been unable to find a closed form for μ_{b} for b > 2. It appears that μ_{b} is a continuous non-monotonic function that is smooth on (1, 2) except at x = \frac{j+1}{j} for j = 2, …, b − 1.

If a logarithmic Khinchin constant KL^{II}_{b} exists (i.e., almost every α ∈ (1, ∞) has the same limiting geometric mean of denominator terms), and if a limiting distribution D(k, ℓ) = lim_{n→∞} D_{n}(k, ℓ) of denominator terms exists, then

KL^{II}_{b} = ∏_{k=0}^{∞} ∏_{ℓ=1}^{b−1} ℓb^{k}λD(k, ℓ).

This is because the limiting distribution of denominator terms (if it exists) is essentially the “average” distribution over all numbers α ∈ (1, 2). If we then assume that almost every α ∈ (1, 2) has the same limiting geometric mean of denominator terms, then this limiting geometric mean (the logarithmic Khinchin constant) must equal the limiting geometric mean of the “average” distribution.

However, since we do not know the limiting distribution, we can only approximate the logarithmic Khinchin constants.

\begin{matrix}b & KL^{II}_{b} \\ 2 & 2.656305048 \\ 3 & 3.415974174 \\ 4 & 4.064209949 \\ 5 & 4.636437895 \\ 6 & 5.152343739 \\ 7 & 5.624290253 \\ 8 & 6.060673548 \\ 9 & 6.467518102 \\ 10 & 6.849326402 \end{matrix}

Table 2: Experimental type II logarithmic Khinchin constants for 2 ≤ b ≤ 10

This conjectured values of the type II logarithmic Khinchin constants are supported by empirical evidence, as the limiting geometric means of denominator terms for various irrational constants give the conjectured values.

4 Type III continued logarithms

Fix an integer base b ≥ 2. In this section, we will introduce our third generalization of base 2 continued logarithms. This appears to be the best of the three generalizations, as we will show that type III continued logarithms have guaranteed convergence, rational finiteness, and closed forms for the limiting distribution and logarithmic Khinchin constant. Additionally, type III continued logarithms ‘converge’ to simple continued fractions if one looks at limiting behavior as b → ∞.

4.1 Type III definitions and recurrences

We start with some definitions, notation, and lemmas related to continued logarithm recurrences.

Definition 20. Let α ∈ ℝ_{≥1}. The base b continued logarithm of type III for α is

c_{0}b^{a_{0}} + \frac{b^{a_{0}}}{c_{1}b^{a_{1}}} + \frac{b^{a_{1}}}{c_{2}b^{a_{2}}} + \frac{b^{a_{2}}}{c_{3}b^{a_{3}}} + ⋯ = [c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{3}(b)}.

where the terms a_{0}, a_{1}, a_{2}, … and c_{0}, c_{1}, c_{2}, … are determined by the recursive process below, terminating at the term c_{n}b^{a_{n}} if at any point y_{n} = c_{n}b^{a_{n}}.

y_{0} = α
a_{n} = ⌊log_{b} y_{n}⌋  n ≥ 0
c_{n} = ⌊\frac{y_{n}}{b^{a_{n}}} ⌋  n ≥ 0
y_{n+1} = \frac{b^{a_{n}}}{y_{n} − c_{n}b^{a_{n}}}  n ≥ 0

Remark 21. We can (and often will) think of the a_{n} and c_{n} as functions a_{0}, a_{1}, a_{2}, …: (1, ∞) → ℤ_{≥0} and c_{0}, c_{1}, c_{2}, …: (1, ∞) → {1, 2, …, b−1}, since the terms a_{0}, c_{0}, a_{1}, c_{1}, a_{2}, c_{2}, … are uniquely determined by α. Conversely, given the complete sequences a_{0}, a_{1}, a_{2}, … and c_{0}, c_{1}, c_{2}, …, one can recover the value of α. Remark 22. Let α = [c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{3}(b)} ∈ (1, ∞). Based on Definitions 3 and 4, the nth convergent and nth remainder term of α are given by

x_{n}(α) = c_{0}b^{a_{0}} + \frac{b^{a_{0}}}{c_{1}b^{a_{1}}} + \frac{b^{a_{1}}}{c_{2}b^{a_{2}}} + \frac{b^{a_{2}}}{c_{3}b^{a_{3}}} + ⋯ + \frac{b^{a_{n−1}}}{c_{n}b^{a_{n}}} .

and

r_{n}(α) = c_{n}b^{a_{n}} + \frac{b^{a_{n}}}{c_{n+1}b^{a_{n+1}}} + \frac{b^{a_{n+1}}}{c_{n+2}b^{a_{n+2}}} + ⋯,

respectively.

Note that the terms r_{n} are the same as the terms y_{n} from Definition 20.

Lemma 23. The nth convergent of α = [c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{3}(b)} is given by

x_{n} = \frac{p_{n}}{q_{n}}

where

p_{−1} = 1, q_{−1} = 0, p_{0} = c_{0}b^{a_{0}}, q_{0} = 1,

and for n ≥ 1,

p_{n} = c_{n}b^{a_{n}}p_{n−1} + b^{a_{n−1}}p_{n−2},
q_{n} = c_{n}b^{a_{n}}q_{n−1} + b^{a_{n−1}}q_{n−2}.

Proof. This follows from Fact 5, where for continued logarithms we have α_{n} = c_{n}b^{a_{n}} and β_{n} = b^{a_{n−1}}. Lemma 24. We have the following lower bounds on the denominators q_{n}:

• q_{n} ≥ 2^{(n−1)/2} > \frac{1}{2}2^{n/2} for n ≥ 0,

• q_{n} ≥ b^{a_{1}+···+a_{n}} for n ≥ 0. Proof. For the first bound, note that

q_{n} = c_{n}b^{a_{n}}q_{n−1} + b^{a_{n−1}}q_{n−2} ≥ (c_{n}b^{a_{n}} + b^{a_{n−1}})q_{n−2} ≥ 2q_{n−2}.

A simple inductive argument then gives q_{n} ≥ 2^{n/2}q_{0} = 2^{n/2} > 2^{(n−1)/2} for even n and q_{n} ≥ 2^{(n−1)/2}q_{1} ≥ 2^{(n−1)/2} for odd n.

For the second bound, note that q_{n} = c_{n}b^{a_{n}}q_{n−1}+b^{a_{n−1}}q_{n−2} ≥ b^{a_{n}}q_{n−1} from which another simple inductive argument gives q_{n} ≥ b^{a_{n}+a_{n−1}+···+a_{1}}q_{0} = b^{a_{1}+···+a_{n}}. Lemma 25. For n ≥ 0,

p_{n}q_{n−1} − q_{n}p_{n−1} = (−1)^{n−1}b^{a_{0}+···+a_{n−1}}.

Proof. For n = 0, we have

p_{0}q_{−1} − q_{0}p_{−1} = c_{0}b^{a_{0}}(0) − 1(1) = −1 = (−1)^{−1}b^{0}.

Now suppose that the statement is true for some n ≥ 0. Then by Lemma 23,

p_{n+1}q_{n} − q_{n+1}p_{n} = (c_{n+1}b^{a_{n+1}}p_{n} + b^{a_{n}}p_{n−1})q_{n} − (c_{n+1}b^{a_{n+1}}q_{n} + b^{a_{n}}q_{n−1})p_{n}
= −b^{a_{n}}(p_{n}q_{n−1} − q_{n}p_{n−1}) = −b^{a_{n}}(−1)^{n−1}b^{a_{0}+···+a_{n−1}}
= (−1)^{n}b^{a_{0}+···+a_{n}}.

so the result follows by induction.

The following lemma is equivalent to Lemma 23, and will be used to prove Theorem 27. Lemma 26. Let a_{−1} = 0. Then for all n ≥ 0,

(\begin{matrix}p_{n} & p_{n−1} \\ q_{n} & q_{n−1} \end{matrix}) = ∏_{j=0}^{n} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) .

Proof. For n = 0, we have

∏_{j=0}^{0} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) = (\begin{matrix}c_{0}b^{a_{0}} & 1 \\ b^{a_{−1}} & 0 \end{matrix}) = (\begin{matrix}c_{0}b^{a_{0}} & 1 \\ 1 & 0 \end{matrix}) = (\begin{matrix}p_{0} & p_{−1} \\ q_{0} & q_{−1} \end{matrix}) .

Now suppose for induction that

∏_{j=0}^{n−1} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) = (\begin{matrix}p_{n−1} & p_{n−2} \\ q_{n−1} & q_{n−2} \end{matrix}) .

Then by Lemma 23,

∏_{j=0}^{n} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) = (\begin{matrix}p_{n−1} & p_{n−2} \\ q_{n−1} & q_{n−2} \end{matrix}) (\begin{matrix}c_{n}b^{a_{n}} & 1 \\ b^{a_{n−1}} & 0 \end{matrix}) = (\begin{matrix}c_{n}b^{a_{n}}p_{n−1} + b^{a_{n−1}}p_{n−2} & p_{n−1} \\ c_{n}b^{a_{n}}q_{n−1} + b^{a_{n−1}}q_{n−2} & q_{n−1} \end{matrix}) = (\begin{matrix}p_{n} & p_{n−1} \\ q_{n} & q_{n−1} \end{matrix}) ,

as asserted. Theorem 27. For arbitrary 1 ≤ k ≤ n,

[c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, …, c_{n}b^{a_{n}}]_{cl_{3}(b)} = \frac{p_{k−1}r_{k} + p_{k−2}b^{a_{k−1}}}{q_{k−1}r_{k} + q_{k−2}b^{a_{k−1}}} .

Proof. First notice that r_{k} = [c_{k}b^{a_{k}}, …, c_{n}b^{a_{n}}]_{cl_{3}(b)} = \frac{p′_{k}}{q′_{k}}, where

(\begin{matrix}p′_{k} \\ q′_{k} \end{matrix}) = (\begin{matrix}c_{k}b^{a_{k}} & 1 \\ 1 & 0 \end{matrix}) ∏_{j=k+1}^{n} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) (\begin{matrix}1 \\ 0 \end{matrix}) .

Also note that

(\begin{matrix}c_{k}b^{a_{k}} & 1 \\ b^{a_{k−1}} & 0 \end{matrix}) = (\begin{matrix}1 & 0 \\ 0 & b^{a_{k−1}} \end{matrix}) (\begin{matrix}c_{k}b^{a_{k}} & 1 \\ 1 & 0 \end{matrix}) .

Then

(\begin{matrix}p_{n} \\ q_{n} \end{matrix}) = ∏_{j=0}^{n} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) (\begin{matrix}1 \\ 0 \end{matrix}) = ∏_{j=0}^{k−1} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) (\begin{matrix}c_{k}b^{a_{k}} & 1 \\ b^{a_{k−1}} & 0 \end{matrix}) ∏_{j=k+1}^{n} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) (\begin{matrix}1 \\ 0 \end{matrix})

= (\begin{matrix}p_{k−1} & p_{k−2} \\ q_{k−1} & q_{k−2} \end{matrix}) (\begin{matrix}1 & 0 \\ 0 & b^{a_{k−1}} \end{matrix}) (\begin{matrix}c_{k}b^{a_{k}} & 1 \\ 1 & 0 \end{matrix}) ∏_{j=k+1}^{n} (\begin{matrix}c_{j}b^{a_{j}} & 1 \\ b^{a_{j−1}} & 0 \end{matrix}) (\begin{matrix}1 \\ 0 \end{matrix})

= (\begin{matrix}p_{k−1} & p_{k−2}b^{a_{k−1}} \\ q_{k−1} & q_{k−2}b^{a_{k−1}} \end{matrix}) (\begin{matrix}p′_{k} \\ q′_{k} \end{matrix}) = (\begin{matrix}p_{k−1}p′_{k} + p_{k−2}b^{a_{k−1}}q′_{k} \\ q_{k−1}p′_{k} + q_{k−2}b^{a_{k−1}}q′_{k} \end{matrix}) .

Thus

[c_{0}b^{a_{0}}, …, c_{n}b^{a_{n}}]_{cl_{3}(b)} = \frac{p_{n}}{q_{n}} = \frac{p_{k−1}p′_{k} + p_{k−2}b^{a_{k−1}}q′_{k}}{q_{k−1}p′_{k} + q_{k−2}b^{a_{k−1}}q′_{k}} = \frac{p_{k−1} \frac{p′_{k}}{q′_{k}} + p_{k−2}b^{a_{k−1}}}{q_{k−1} \frac{p′_{k}}{q′_{k}} + q_{k−2}b^{a_{k−1}}} = \frac{p_{k−1}r_{k} + p_{k−2}b^{a_{k−1}}}{q_{k−1}r_{k} + q_{k−2}b^{a_{k−1}}} ,

as required.

4.2 Convergence and rational finiteness of type III continued logarithms

Theorem 28. The type III continued logarithm for a number x ≥ 1 converges to x.

Proof. Suppose that the continued logarithm for x = [c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, …, c_{n}b^{a_{n}}]_{cl_{3}(b)} is finite. From the construction, we have x = y_{0} where

y_{k} = c_{k}b^{a_{k}} + \frac{b^{a_{k}}}{y_{k+1}}

for 0 ≤ k ≤ n − 1. From Definition 20, since the continued logarithm terminates, we have y_{n} = c_{n}b^{a_{n}}, at which point we simply have

x = c_{0}b^{a_{0}} + \frac{b^{a_{0}}}{c_{1}b^{a_{1}}} + \frac{b^{a_{1}}}{c_{2}b^{a_{2}}} + \frac{b^{a_{2}}}{c_{3}b^{a_{3}}} + ⋯ + \frac{b^{a_{n−1}}}{c_{n}b^{a_{n}}}.

This shows convergence in the case of finite termination. If the continued logarithm for x does not terminate, then convergence follows from Fact 6, since

∑_{n=1}^{∞} \frac{α_{n}α_{n+1}}{β_{n+1}} = ∑_{n=1}^{∞} \frac{c_{n}b^{a_{n}}c_{n+1}b^{a_{n+1}}}{b^{a_{n}}} = ∑_{n=1}^{∞} c_{n}c_{n+1}b^{a_{n+1}} = ∞,

while all terms are positive as required.

Lemma 29. If

y = c_{0}b^{a_{0}} + \frac{b^{a_{0}}}{c_{1}b^{a_{1}}} + \frac{b^{a_{1}}}{c_{2}b^{a_{2}}} + \frac{b^{a_{2}}}{c_{3}b^{a_{3}}} + ⋯,
y_{1} = c_{0}b^{a_{0}} + \frac{c_{1}^{−1}b^{a_{0}−a_{1}}}{1} + \frac{c_{1}^{−1}c_{2}^{−1}b^{−a_{2}}}{1} + \frac{c_{2}^{−1}c_{3}^{−1}b^{−a_{3}}}{1} + ⋯,

then y and y_{1} are equivalent. (The form y_{1} is called the denominator-reduced continued logarithm for y.)

Proof. Take d_{0} = 1 and d_{n} = c_{n}^{−1}b^{−a_{n}} for n ≥ 1 to satisfy the conditions of Definition 2.

Theorem 30. The type III continued logarithm for a number x ≥ 1 will terminate finitely if and only if x ∈ ℚ.

Proof. Clearly, if the continued logarithm for x terminates finitely, then x ∈ ℚ. Conversely, suppose

x = c_{0}b^{a_{0}} + \frac{b^{a_{0}}}{c_{1}b^{a_{1}}} + \frac{b^{a_{1}}}{c_{2}b^{a_{2}}} + ⋯

is rational. By Lemma 29, we can write

x = c_{0}b^{a_{0}} (1 + \frac{c_{0}^{−1}c_{1}^{−1}b^{−a_{1}}}{1} + \frac{c_{1}^{−1}c_{2}^{−1}b^{−a_{2}}}{1} + ⋯) .

Let y_{n} denote the nth tail of the continued logarithm, that is,

y_{n} = 1 + \frac{c_{n}^{−1}c_{n+1}^{−1}b^{−a_{n+1}}}{1} + \frac{c_{n+1}^{−1}c_{n+2}^{−1}b^{−a_{n+2}}}{1} + · · · .

Notice that

y_{n} = 1 + \frac{c_{n}^{−1}c_{n+1}^{−1}b^{−a_{n+1}}}{y_{n+1}} ,

so

y_{n+1} = \frac{c_{n}^{−1}c_{n+1}^{−1}b^{−a_{n+1}}}{y_{n} − 1} .

Since each y_{n} is rational, write y_{n} = \frac{u_{n}}{v_{n}} for positive relatively prime integers u_{n} and v_{n}. Hence

\frac{u_{n+1}}{v_{n+1}} = y_{n+1} = \frac{c_{n}^{−1}c_{n+1}^{−1}b^{−a_{n+1}}}{\frac{u_{n}−v_{n}}{v_{n}}} = \frac{v_{n}}{c_{n}c_{n+1}b^{a_{n+1}}(u_{n} − v_{n})},

or equivalently,

c_{n}c_{n+1}b^{a_{n+1}}(u_{n} − v_{n})u_{n+1} = v_{n}v_{n+1}.

Notice that since y_{n} ≥ 1 for all n, u_{n} − v_{n} ≥ 0, so each multiplicative term in the above equation is a nonnegative integer. Since u_{n+1} and v_{n+1} are relatively prime, we must have u_{n+1} | v_{n}, so u_{n+1} ≤ v_{n} ≤ u_{n}. If at any point we have u_{n+1} = v_{n} = u_{n}, then y_{n} = \frac{u_{n}}{v_{n}} = 1 and the continued logarithm terminates. Otherwise, u_{n+1} < u_{n}, so (u_{n}) is a strictly decreasing sequence of nonnegative integers, so the process must terminate, again giving a finite continued logarithm.

4.3 Using measure theory to study the type III continued logarithm terms

We now look at the relative frequency of the continued logarithm terms. Specifically, the main theorem of this section places bounds on the measure of the set

\{x ∈ (1, 2) : \begin{matrix}a_{1} = k_{1}, & a_{2} = k_{2}, & …, & a_{n} = k_{n}, & a_{n+1} = k \\ c_{1} = ℓ_{1}, & c_{2} = ℓ_{2}, & …, & c_{n} = ℓ_{n}, & c_{n+1} = ℓ \end{matrix} \}

in terms of the measure of the set

\{x ∈ (1, 2) : \begin{matrix}a_{1} = k_{1}, & a_{2} = k_{2}, & …, & a_{n} = k_{n} \\ c_{1} = ℓ_{1}, & c_{2} = ℓ_{2}, & …, & c_{n} = ℓ_{n} \end{matrix} \}

and the value of k and ℓ. From that, we can get preliminary bounds on the measure of {x ∈ (1, 2) : a_{n} = k, c_{n} = ℓ} in terms of k and ℓ.

Consider α = [c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{3}(b)}. Assume that α ∉ ℚ, so that the continued logarithm for α is infinite. Furthermore, assume a_{0} = 1 and c_{0} = 1, so that α ∈ (1, 2). Notice that in order to have a_{1} = k_{1} and c_{1} = ℓ_{1}, we must have 1+(ℓ_{1} +1)b^{−k_{1}} < α ≤ 1+ℓ_{1}b^{−k_{1}}. Thus we

can partition (1, 2) into countably many intervals J_{1} (\begin{matrix}0 \\ 1 \end{matrix}) , J_{1} (\begin{matrix}0 \\ 2 \end{matrix}) , …, J_{1} (\begin{matrix}0 \\ b − 1 \end{matrix}) , J_{1} (\begin{matrix}1 \\ 1 \end{matrix}) , J_{1} (\begin{matrix}1 \\ 2 \end{matrix}) , …

such that a_{1} = k_{1} and c_{1} = ℓ_{1} for all α ∈ J_{1} (\begin{matrix}k_{1} \\ ℓ_{1} \end{matrix}). This gives, in general,

J_{1} (\begin{matrix}k_{1} \\ ℓ_{1} \end{matrix}) = (1 + \frac{1}{(ℓ_{1} + 1)b^{k_{1}}} , 1 + \frac{1}{ℓ_{1}b^{k_{1}}} ) .

We call these intervals the intervals of first rank.

Now fix some interval of first rank, J_{1} (\begin{matrix}k_{1} \\ ℓ_{1} \end{matrix}), and consider the values of a_{2} and c_{2} for

α ∈ J_{1} (\begin{matrix}k_{1} \\ ℓ_{1} \end{matrix}). One can show that we have a_{1} = k_{1}, c_{1} = ℓ_{1}, a_{2} = k_{2}, and c_{2} = ℓ_{2} on the

interval

J_{2} (\begin{matrix}k_{1}, & k_{2} \\ ℓ_{1}, & ℓ_{2} \end{matrix}) = (1 + \frac{1}{ℓ_{1}b^{k_{1}} + \frac{b^{k_{1}}}{ℓ_{2}b^{k_{2}}}} , 1 + \frac{1}{ℓ_{1}b^{k_{1}} + \frac{b^{k_{1}}}{(ℓ_{2}+1)b^{k_{2}}}} ) .

These are the intervals of second rank. We may repeat this process indefinitely to get the intervals of nth rank, noting that each interval of rank n is just a subinterval of an interval of rank n − 1.

Definition 31. Let n ∈ ℕ. The intervals of nth rank are the intervals of the form

J_{n} (\begin{matrix}k_{1}, & k_{2}, & …, & k_{n} \\ ℓ_{1}, & ℓ_{2}, & …, & ℓ_{n} \end{matrix}) = \{α ∈ (1, 2) : \begin{matrix}a_{1} = k_{1}, & a_{2} = k_{2}, & …, & a_{n} = k_{n} \\ c_{1} = ℓ_{1}, & c_{2} = ℓ_{2}, & …, & c_{n} = ℓ_{n} \end{matrix} \} ,

where k_{1}, k_{2}, …, k_{n} ∈ ℤ_{≥0} and ℓ_{1}, ℓ_{2}, …, ℓ_{n} ∈ {1, 2, …, b − 1}.

Remark 32. The intervals of nth rank will be half-open intervals that are open on the left if n is odd and open on the right if n is even. However, for simplicity, we will ignore what happens at the endpoints and treat these intervals as open intervals. This will not affect the main theorems of this paper, as the set of endpoints is a set of measure zero.

Definition 33. Suppose m, n ∈ ℕ with m ≥ n. Let a_{n+1}, …, a_{m} ∈ ℤ_{≥0} and c_{n+1}, …, c_{m} ∈ {1, …, b − 1}. Let f be a function that maps intervals of rank m to real numbers. Then we define

∑^{(n)} f (J_{m} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{m} \\ c_{1}, & c_{2}, & …, & c_{m} \end{matrix})) = ∑_{a_{1}=1}^{∞} ∑_{c_{1}=1}^{b−1} ⋯ ∑_{a_{n}=1}^{∞} ∑_{c_{n}=1}^{b−1} f (J_{m} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{m} \\ c_{1}, & c_{2}, & …, & c_{m} \end{matrix})) ,

and similarly we define

⋃^{(n)} J_{m} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{m} \\ c_{1}, & c_{2}, & …, & c_{m} \end{matrix}) = ⋃_{a_{1}=1}^{∞} ⋃_{c_{1}=1}^{b−1} ⋯ ⋃_{a_{n}=1}^{∞} ⋃_{c_{n}=1}^{b−1} J_{m} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{m} \\ c_{1}, & c_{2}, & …, & c_{m} \end{matrix}) .

Definition 34. Let n ∈ ℕ. Let

D_{n}(k, ℓ) = {α ∈ (1, 2) : a_{n} = k, c_{n} = ℓ}

denote the set of points where the nth continued logarithm term is ℓb^{k}.

Remark 35. The set D_{n}(k, ℓ) is a countable union of intervals of rank n, specifically,

D_{n}(k, ℓ) = ⋃^{(n−1)} J_{n} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{n−1}, & k \\ c_{1}, & c_{2}, & …, & c_{n−1}, & ℓ \end{matrix}) .

Lemma 36. Let J_{n} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{n} \\ c_{1}, & c_{2}, & …, & c_{n} \end{matrix}) be an interval of rank n. The endpoints of J_{n} are

\frac{p_{n}}{q_{n}}  and  \frac{p_{n} + p_{n−1}b^{a_{n}}}{q_{n} + q_{n−1}b^{a_{n}}} .

Proof. Let α ∈ J_{n} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{n} \\ c_{1}, & c_{2}, & …, & c_{n} \end{matrix}) be arbitrary. Note that α = [1, c_{1}b^{a_{1}}, …, c_{n}b^{a_{n}}, r_{n+1}]_{cl_{3}(b)},

where r_{n+1} can take any real value in [1, ∞). From Theorem 27, we have

α = \frac{p_{n}r_{n+1} + p_{n−1}b^{a_{n}}}{q_{n}r_{n+1} + q_{n−1}b^{a_{n}}} .

Notice that

α − \frac{p_{n}}{q_{n}} = \frac{p_{n}r_{n+1} + p_{n−1}b^{a_{n}}}{q_{n}r_{n+1} + q_{n−1}b^{a_{n}}} = \frac{(q_{n}p_{n−1} − p_{n}q_{n−1})b^{a_{n}}}{q_{n}(q_{n}r_{n+1} + q_{n−1}b^{a_{n}})} ,

and on J_{n} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{n} \\ c_{1}, & c_{2}, & …, & c_{n} \end{matrix}), all of p_{n}, q_{n}, p_{n−1}, q_{n−1}, a_{n} are fixed. Thus α is a monotonic

function of r_{n+1}, so the extreme values of α on J_{n} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{n} \\ c_{1}, & c_{2}, & …, & c_{n} \end{matrix}) will occur at the extreme values of r_{n+1}. Taking r_{n+1} = 1 gives α = \frac{p_{n}+p_{n−1}b^{an}}{q_{n}+q_{n−1}b^{an}}, and letting r_{n+1} → ∞ gives

α = \frac{p_{n}}{q_{n}}. Thus the endpoints of J_{n} (\begin{matrix}a_{1}, & a_{2}, & …, & a_{n} \\ c_{1}, & c_{2}, & …, & c_{n} \end{matrix}) are

\frac{p_{n}}{q_{n}}  and  \frac{p_{n} + p_{n−1}b^{a_{n}}}{q_{n} + q_{n−1}b^{a_{n}}} ,

as claimed. Theorem 37. Suppose n ∈ ℕ, a_{1}, a_{2}, …, a_{n}, k ∈ ℤ_{≥0}, and c_{1}, c_{2}, …, c_{n}, ℓ ∈ {1, …, b − 1}. Let a = (a_{1}, …, a_{n}) and c = (c_{1}, …, c_{n}). Then

\frac{1}{4ℓ(ℓ + 1)b^{k}} λJ_{n} (\begin{matrix}a \\ c \end{matrix}) ≤ λJ_{n+1} (\begin{matrix}a, & k \\ c, & ℓ \end{matrix}) ≤ \frac{2}{ℓ(ℓ + 1)b^{k}} λJ_{n} (\begin{matrix}a \\ c \end{matrix}) .

Proof. From Lemma 36, we know that the endpoints of J_{n} (\begin{matrix}a \\ c \end{matrix}) are

\frac{p_{n}}{q_{n}}  and  \frac{p_{n} + p_{n−1}b^{a_{n}}}{q_{n} + q_{n−1}b^{a_{n}}} ,

Now in order to be in J_{n+1} (\begin{matrix}a, & k \\ c, & ℓ \end{matrix}), we must have a_{n+1} = k and c_{n+1} = ℓ, so ℓb^{k} ≤ r_{n+1} ≤ (ℓ + 1)b^{k}. Thus the endpoints of J_{n+1} (\begin{matrix}a, & k \\ c, & ℓ \end{matrix}) will be

\frac{p_{n}ℓb^{k} + p_{n−1}b^{a_{n}}}{q_{n}ℓb^{k} + q_{n−1}b^{a_{n}}}  and  \frac{p_{n}(ℓ + 1)b^{k} + p_{n−1}b^{a_{n}}}{q_{n}(ℓ + 1)b^{k} + q_{n−1}b^{a_{n}}} .

Thus

λJ_{n} (\begin{matrix}a \\ c \end{matrix}) = |\frac{p_{n}}{q_{n}} − \frac{p_{n} + p_{n−1}b^{a_{n}}}{q_{n} + q_{n−1}b^{a_{n}}} | = |\frac{p_{n}q_{n−1}b^{a_{n}} − p_{n−1}q_{n}b^{a_{n}}}{q_{n}(q_{n} + q_{n−1}b^{a_{n}})} |

= \frac{b^{a_{1}+···+a_{n}}}{q_{n}(q_{n} + q_{n−1}b^{a_{n}})} = \frac{b^{a_{1}+···+a_{n}}}{q_{n}^{2} (1 + \frac{q_{n−1}b^{an}}{q_{n}} )},

and

λJ_{n+1} (\begin{matrix}a, & k \\ c, & ℓ \end{matrix}) = |\frac{p_{n}ℓb^{k} + p_{n−1}b^{a_{n}}}{q_{n}ℓb^{k} + q_{n−1}b^{a_{n}}} − \frac{p_{n}(ℓ + 1)b^{k} + p_{n−1}b^{a_{n}}}{q_{n}(ℓ + 1)b^{k} + q_{n−1}b^{a_{n}}} |

= |\frac{p_{n}q_{n−1}ℓb^{a_{n}+k} + p_{n−1}q_{n}(ℓ + 1)b^{a_{n}+k} − p_{n}q_{n−1}(ℓ + 1)b^{a_{n}+k} − p_{n−1}q_{n}ℓb^{a_{n}+k}}{(q_{n}ℓb^{k} + q_{n−1}b^{a_{n}})(q_{n}(ℓ + 1)b^{k} + q_{n−1}b^{a_{n}}} |

= |\frac{b^{a_{1}+···+a_{n}+k}}{ℓ(ℓ + 1)b^{2k}q_{n}^{2} (1 + \frac{q_{n−1}b^{an}}{q_{n}ℓb^{k}} ) (1 + \frac{q_{n−1}b^{an}}{q_{n}(ℓ+1)b^{k}} )}|

= \frac{b^{a_{1}+···+a_{n}}}{ℓ(ℓ + 1)b^{k}q_{n}^{2} (1 + \frac{q_{n−1}b^{an}}{q_{n}ℓb^{k}} ) (1 + \frac{q_{n−1}b^{an}}{q_{n}(ℓ+1)b^{k}} )},

so

\frac{λJ_{n+1} (\begin{matrix}a, & k \\ c, & ℓ \end{matrix})}{λJ_{n} (\begin{matrix}a \\ c \end{matrix})} = \frac{1 + \frac{q_{n−1}b^{an}}{q_{n}}}{ℓ(ℓ + 1)b^{k} (1 + \frac{q_{n−1}b^{an}}{q_{n}ℓb^{k}} ) (1 + \frac{q_{n−1}b^{an}}{q_{n}(ℓ+1)b^{k}} )}.

Now notice that q_{n} = c_{n}b^{a_{n}}q_{n−1} + b^{a_{n−1}}q_{n−2} ≥ b^{a_{n}}q_{n−1}, so 0 ≤ \frac{q_{n−1}b^{an}}{q_{n}} ≤ 1, 0 ≤ \frac{q_{n−1}b^{an}}{q_{n}ℓb^{k}} ≤ 1, and 0 ≤ \frac{q_{n−1}b^{an}}{q_{n}(ℓ+1)b^{k}} ≤ 1, and thus

\frac{1}{4} ≤ \frac{1 + \frac{q_{n−1}b^{an}}{q_{n}}}{(1 + \frac{q_{n−1}b^{an}}{q_{n}ℓb^{s}} ) (1 + \frac{q_{n−1}b^{an}}{q_{n}(ℓ+1)b^{k}} )} ≤ 2.

Therefore

\frac{1}{4ℓ(ℓ + 1)b^{k}} λJ_{n} (\begin{matrix}a \\ c \end{matrix}) ≤ λJ_{n+1} (\begin{matrix}a, & k \\ c, & ℓ \end{matrix}) ≤ \frac{2}{ℓ(ℓ + 1)b^{k}} λJ_{n} (\begin{matrix}a \\ c \end{matrix}) ,

and we are done.

Corollary 38. Let n ∈ ℕ, k ∈ ℤ_{≥0}, and ℓ ∈ {1, …, b − 1}. Then

\frac{1}{4ℓ(ℓ + 1)b^{k}} ≤ λ(D_{n+1}(k, ℓ)) ≤ \frac{2}{ℓ(ℓ + 1)b^{k}} .

Proof. Note that any two distinct intervals of rank n are disjoint. Thus we can add up the above inequality over all intervals of rank n, noting that

⋃^{(n)} J_{n} (\begin{matrix}a_{1}, & …, & a_{n} \\ c_{1}, & …, & c_{n} \end{matrix}) = (1, 2),

so

∑^{(n)} λJ_{n} (\begin{matrix}a_{1}, & …, & a_{n} \\ c_{1}, & …, & c_{n} \end{matrix}) = λ(1, 2) = 1,

and that

⋃^{(n)} J_{n+1} (\begin{matrix}a_{1}, & …, & a_{n}, & k \\ c_{1}, & …, & c_{n}, & ℓ \end{matrix}) = D_{n+1}(k, ℓ),

so

∑^{(n)} λJ_{n+1} (\begin{matrix}a_{1}, & …, & a_{n}, & k \\ c_{1}, & …, & c_{n}, & ℓ \end{matrix}) = λD_{n+1}(k, ℓ).

This gives

\frac{1}{4ℓ(ℓ + 1)b^{k}} ≤ λD_{n+1} (\begin{matrix}k \\ ℓ \end{matrix}) ≤ \frac{2}{ℓ(ℓ + 1)b^{k}} ,

as needed.

4.4 Distribution of type III continued logarithm terms

Definition 39. Let x = [1, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{3}(b)} ∈ (1, 2) and r_{n} = r_{n}(x) = [c_{n}b^{a_{n}}, c_{n+1}b^{a_{n+1}}, …]_{cl_{3}(b)}, as per Remark 22. Define

z_{n} = z_{n}(x) = \frac{r_{n}}{b^{a_{n}}} − c_{n} + 1 = [1, c_{n+1}b^{a_{n+1}}, c_{n+2}b^{a_{n+2}}, …]_{cl_{3}(b)} ∈ (1, 2),

M_{n}(x) = {α ∈ (1, 2) : z_{n}(α) < x} ⊆ (1, 2),

m_{n}(x) = λM_{n}(x) ∈ (0, 1),

and

m(x) = lim_{n→∞} m_{n}(x),

wherever this limit exists.

We now get a recursion relation for the sequence of functions m_{n}. Theorem 40. The sequence of functions m_{n} is given by the recursive relationship

m_{0}(x) = x − 1  (9)
m_{n}(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} m_{n−1}(1 + ℓ^{−1}b^{−k}) − m_{n−1}(1 + (x + ℓ − 1)^{−1}b^{−k})  n ≥ 1  (10)

for 1 ≤ x ≤ 2. Proof. Notice that r_{0}(α) = α, a_{0} = 0, and c_{0} = 1, so z_{0}(α) = \frac{r_{0}}{b^{a_{0}}} − c_{0} + 1 = α and thus

M_{0}(x) = {α ∈ (1, 2) : z_{0}(α) < x} = {α ∈ (1, 2) : α < x} = (1, x),

so m_{0}(x) = x − 1. Now fix n ≥ 1. Since a_{n} ∈ ℤ_{≥0} and c_{n} ∈ {1, …, b − 1}, we have

m_{n}(x) = λ{α ∈ (1, 2) : z_{n} < x} = λ ⋃_{k=0}^{∞} ⋃_{ℓ=1}^{b−1}{α ∈ (1, 2) : z_{n} < x, a_{n} = k, c_{n} = ℓ}.

Fix x ∈ (1, 2) and let

A_{k,ℓ} = {α ∈ (1, 2) : z_{n} < x, a_{n} = k, c_{n} = ℓ}

for k ∈ ℤ_{≥0} and ℓ ∈ {1, …, b − 1}. By Definition 39, z_{n} < x if and only if

\frac{r_{n}}{b^{a_{n}}} − c_{n} + 1 < x.

Notice that

z_{n−1} = [1, c_{n}b^{a_{n}}, c_{n+1}b^{a_{n+1}}, …]_{cl_{3}(b)} = [1, r_{n}]_{cl_{3}(b)} = 1 + \frac{1}{r_{n}} ,

so z_{n} < x if and only if

\frac{1}{b^{a_{n}}(z_{n−1} − 1)} − c_{n} + 1 < x,

or equivalently

z_{n−1} > 1 + (x + c_{n} − 1)^{−1}b^{−a_{n}} = 1 + (x + ℓ − 1)^{−1}b^{−k}.  (11)

Additionally, in order to have a_{n} = k and c_{n} = ℓ, we must have ℓb^{k} ≤ r_{n} < (ℓ + 1)b^{k}, or equivalently,

1 + (ℓ + 1)^{−1}b^{−k} < z_{n−1} ≤ 1 + ℓ^{−1}b^{−k}.  (12)

Now notice that since x < 2,

1 + (ℓ + 1)^{−1}b^{−k} < 1 + (x + ℓ − 1)^{−1}b^{−k},

and thus the left hand inequality in (12) is implied by (11). Therefore z_{n} < x with a_{n} = k and c_{n} = ℓ if and only if

1 + (x + ℓ − 1)^{−1}b^{−k} < z_{n−1} ≤ 1 + ℓ^{−1}b^{−k}.  (13)

Thus

A_{k,ℓ} = {α ∈ (1, 2) : 1 + (x + ℓ − 1)^{−1}b^{−k} < z_{n−1} ≤ 1 + ℓ^{−1}b^{−k}}.  (14)

Now suppose k_{1}, k_{2} ∈ ℤ and ℓ_{1}, ℓ_{2} ∈ {1, …, b − 1} with (k_{1}, ℓ_{1}) ≠ (k_{2}, ℓ_{2}). We claim that A_{k_{1},ℓ_{1}} and A_{k_{2},ℓ_{2}} are disjoint. Consider two cases:

Case 1: k_{1} ≠ k_{2}. Suppose (without loss of generality) that k_{2} < k_{1}, so k_{2} − k_{1} ≤ −1. Also note that 1 ≤ ℓ_{2} ≤ b − 1 and x < 2 so ℓ_{2} + x − 1 < b. Then we have

1+ℓ_{1}^{−1}b^{−k_{1}} = 1+ℓ_{1}^{−1}b^{k_{2}−k_{1}}b^{−k_{2}} ≤ 1+ℓ_{1}^{−1}b^{−1}b^{−k_{2}} ≤ 1+b^{−1}b^{−k_{2}} < 1+(ℓ_{2} +x−1)^{−1}b^{−k_{2}}. (15)

Case 2: k_{1} = k_{2}, ℓ_{1} ≠ ℓ_{2}. Suppose (without loss of generality) that ℓ_{1} > ℓ_{2}, so indeed ℓ_{1} ≥ ℓ_{2} + 1. Then since x − 1 < 1,

1 + ℓ_{1}^{−1}b^{−k_{1}} = 1 + ℓ_{1}^{−1}b^{−k_{2}} ≤ 1 + (ℓ_{2} + 1)^{−1}b^{−k_{2}} < 1 + (ℓ_{2} + x − 1)^{−1}b^{−k_{2}}.  (16)

Now suppose a_{1} ∈ A_{k_{1},ℓ_{1}} and a_{2} ∈ A_{k_{2},ℓ_{2}}. By (14) and either (15) or (16),

a_{1} ≤ 1 + ℓ_{1}^{−1}b^{−k_{1}} < 1 + (ℓ_{2} + x − 1)^{−1}b^{−k_{2}} ≤ a_{2},

so a_{1} ≠ a_{2} and thus A_{k_{1},ℓ_{1}} and A_{k_{2},ℓ_{2}} must be disjoint. Therefore

m_{n}(x) = λ ⋃_{k=0}^{∞} ⋃_{ℓ=1}^{b−1} A_{k,ℓ} = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} M(A_{k,ℓ}).  (17)

Finally, since m_{n−1}(x) = M{α ∈ (1, 2) : z_{n−1} < x}, by (14) and (17) we can conclude

m_{n}(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} (m_{n−1}(1 + ℓ^{−1}b^{−k}) − m_{n−1} (1 + (x + ℓ − 1)^{−1}b^{−k})) ,

which proves the recursion (10), and completes the proof of the theorem. Theorem 41. There exist constants A, λ > 0 such that

|m_{n}(x) − \begin{matrix}log & \frac{bx}{x+b−1} \\ log & \frac{2b}{b+1} \end{matrix} | < Ae^{−λ\sqrt{n}}

for all n ≥ 0 and x ∈ (1, 2).

The proof of this theorem, which is based on the Khinchin’s proof [5, Sec. 15] for simple continued fractions, is lengthy and somewhat technical. It is provided in appendix A, and the following corollary immediately follows. Corollary 42. We have

m(x) = \begin{matrix}log & \frac{bx}{x + b − 1} \\ log & \frac{2b}{b + 1} \end{matrix}

for all x ∈ (1, 2). Theorem 43. We have

λ(D_{n+1}(k, ℓ)) = m_{n}(1 + (ℓ + 1)^{−1}b^{−k}) − m_{n}(1 + ℓ^{−1}b^{−k}).

Proof. Suppose that α ∈ D_{n+1}(k, ℓ). Then a_{n+1} = k and c_{n+1} = ℓ, so

z_{n} = [1, ℓb^{k}, r_{n+2}] = 1 + \frac{1}{ℓb^{k} + \frac{b^{k}}{r_{n+2}}},

where r_{n+2} can take any value in (1, ∞). Clearly z_{n} is a monotonic function of r_{n+2} for fixed k, ℓ, so the extreme values of z_{n} on D_{n+1}(k, ℓ) will occur at the extreme values of r_{n+2}. Letting r_{n} → 1 gives z_{n} = 1 + \frac{1}{ℓb^{k}+b^{k}} = 1 + (ℓ + 1)^{−1}b^{−k} and letting r_{n} → ∞ gives z_{n} = 1 + \frac{1}{ℓb^{k}+0} = ℓ^{−1}b^{−k}. Thus

D_{n+1}(k, ℓ) = {α ∈ (1, 2) : 1 + (ℓ + 1)^{−1}b^{−k} < z_{n}(α) ≤ 1 + ℓ^{−1}b^{−k}}
= M_{n}(1 + ℓ^{−1}b^{−k}) \ M_{n}(1 + (ℓ + 1)^{−1}b^{−k}),

so

λD_{n+1}(k, ℓ) = m_{n}(1 + ℓ^{−1}b^{−k}) − m_{n}(1 + (ℓ + 1)^{−1}b^{−k}).

Theorem 44. There exist constants A, λ > 0 such that

|λ(D_{n}(k, ℓ)) − \begin{matrix}log & \frac{(ℓb^{k}+1)((ℓ+1)b^{k+1}+1)}{(ℓb^{k+1}+1)((ℓ+1)b^{k}+1)} \\ log & \frac{2b}{b+1} \end{matrix} | < \frac{Ae^{−λ\sqrt{n−1}}}{ℓ(ℓ + 1)b^{k}}

for all k ∈ ℤ_{≥0}, ℓ ∈ {1, 2, …, b − 1} and n ∈ ℤ_{≥0}.

We then immediately get the following limiting distribution.

Notice that like with type II continued fractions, the distribution is non-monotonic. This is due to the gaps in possible denominator terms. For example, for base 4, the possible denominator terms are 1, 2, 3, 4, 8, 12, . . .. The jump from 4 to 8 causes a spike in the limiting distribution.

Corollary 45. We have

lim_{n→∞} λ(D_{n}(k, ℓ))) = \begin{matrix}log & \frac{(ℓb^{k}+1)((ℓ+1)b^{k+1}+1)}{(ℓb^{k+1}+1)((ℓ+1)b^{k}+1)} \\ log & \frac{2b}{b+1} \end{matrix}

for k ∈ ℤ_{≥0} and ℓ ∈ {1, 2, …, b − 1}.

4.5 Type III logarithmic Khinchin constant

We now extend the Khinchin constant to type III continued logarithms. Note that we only gave an overview for type I and type II, but here we will be much more rigorous.

Definition 46. Let α ∈ (1, ∞) have type III continued logarithm [c_{0}b^{a_{0}}, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{3}(b)}. Let k ∈ ℤ_{≥0} and ℓ ∈ {1, 2, …, b − 1}. We define

P_{α}(k, ℓ) = lim_{N→∞} \frac{|\{n ∈ ℕ : a_{n} = k, c_{n} = ℓ\}|}{N}

to be the limiting proportion of continued logarithm terms of α that have a_{n} = k and c_{n} = ℓ, if this limit exists.

Note that for the theorems which follow, we will restrict our study to (1, 2) instead of (1, ∞). The results can be easily extended to (1, ∞) by noting that every α ∈ (1, ∞) corresponds to an α′ ∈ (1, 2) in the sense that the continued logarithm of α′ is just the continued logarithm of α with the first term replace by 1. Since we are looking at limiting behavior over all terms, changing the first term will have no impact.

The following two theorems are proved in Appendix B. The proofs are based on the analogous proofs for simple continued fractions given by Khinchin [5, Sec. 15 & 16].

Theorem 47. For almost every α ∈ (1, 2) with continued logarithm [1, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{3}(b)} we have

P_{α}(k, ℓ) = \begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix}

for all k ∈ ℤ_{≥0} and ℓ ∈ {1, 2, …, b − 1}.

Theorem 48. For almost every α ∈ (1, 2) with continued logarithm [1, c_{1}b^{a_{1}}, c_{2}b^{a_{2}}, …]_{cl_{3}(b)} we have

lim_{N→∞} (∏_{n=1}^{N}(c_{n}b^{a_{n}}))^{1/N} = b^{A_{b}},

where

A_{b} = \frac{1}{log b log \frac{b+1}{2b}} ∑_{ℓ=2}^{b} log (1 − \frac{1}{ℓ}) log (1 + \frac{1}{ℓ}) .

The values of the Khinchin constant given by the above formula for 2 ≤ b ≤ 10 are shown in Table 3.

\begin{matrix}b & KL^{III}_{b} \\ 2 & 2.656305058 \\ 3 & 2.666666667 \\ 4 & 2.671738848 \\ 5 & 2.674705520 \\ 6 & 2.676638451 \\ 7 & 2.677992355 \\ 8 & 2.678991102 \\ 9 & 2.679757051 \\ 10 & 2.680362475 \end{matrix}

Table 3: Type III logarithmic Khinchin constants for 2 ≤ b ≤ 10

Remark 49. Notice that Theorem 47 is similar to Corollary 45. However, Corollary 45 is about the limiting proportion of numbers α ∈ (1, 2) that have a_{n} = k and c_{n} = ℓ, whereas Theorem 47 is about the limiting proportion of terms of a number α ∈ (1, 2) for which a_{n} = k and c_{n} = ℓ. The fact that these two limits are the same is not a coincidence: one can show that Corollary 45 is a consequence of Theorem 47.

Based on Theorem 48, we denote

KL^{III}_{b} = b^{A_{b}},

where A_{b} is as in Theorem 48.

4.6 Type III continued logarithms and simple continued fractions

Now suppose b is no longer fixed. Let μ_{b} denote the limiting distribution for a given base b, as shown in Corollary 42. That is,

μ_{b}(x) = \begin{matrix}log & \frac{bx}{x+b−1} \\ log & \frac{2b}{b+1} \end{matrix} .

Furthermore, let KL^{III}_{b} denote the base b logarithmic Khinchin constant, as in Remark 49, and let K denote the Khinchin constant for simple continued fractions, as in Section 1.4.

We now have an interesting relationship between these logarithmic Khinchin constants and the Khinchin constant for simple continued fractions, based on the following lemma.

Lemma 50 ([1], Lemma 1(c)).

∑_{ℓ=2}^{∞} log (1 − \frac{1}{ℓ}) log (1 + \frac{1}{ℓ}) = − log K log 2.

Theorem 51.

lim_{b→∞} KL^{III}_{b} = K.

Proof. We will show that lim_{b→∞} log KL^{III}_{b} = log K, from which the desired limit immediately follows.

lim_{b→∞} log KL^{III}_{b} = lim_{b→∞} log b^{A_{b}} = lim_{b→∞}(log b)A_{b}

= lim_{b→∞} \frac{log b}{log b log \frac{b+1}{2b}} ∑_{k=2}^{b} log (1 − \frac{1}{k}) log (1 + \frac{1}{k})

= lim_{b→∞} \frac{1}{log \frac{b+1}{2b}} ∑_{k=2}^{b} log (1 − \frac{1}{k}) log (1 + \frac{1}{k})

= \frac{1}{lim_{b→∞} log (\frac{1}{2} \frac{b+1}{b})} ∑_{k=2}^{∞} log (1 − \frac{1}{k}) log (1 + \frac{1}{k})

= −\frac{1}{log 2} ∑_{k=2}^{∞} log (1 − \frac{1}{k}) log (1 + \frac{1}{k}) = log K.

Furthermore, as b → ∞, the distribution function μ_{b} approaches the appropriately shifted continued fraction distribution μ_{cl}. The continued fraction distribution function is given by

μ_{cl}(x) = log_{2}(1 + x) x ∈ (0, 1).

(See Borwein et al. [3, Section 3.4].) Since the continued fraction for a number will be unchanged (except for the first term) when adding an integer, we can shift this distribution to the right and think of it as a distribution over (1, 2) instead of (0, 1), in order to compare it to μ_{b}. We define the shifted continued fraction distribution

μ_{cl}^{∗}(x) = μ_{cl}(x − 1) = log_{2} x x ∈ (1, 2).

We then have

lim_{b→∞} μ_{b}(x) = lim_{b→∞} \begin{matrix}log & \frac{x+b−1}{bx} \\ log & \frac{b+1}{2b} \end{matrix} = lim_{b→∞} \begin{matrix}log & (\frac{1}{b} + \frac{1}{x} − \frac{1}{bx}) \\ log & (\frac{1}{2} \frac{b+1}{b}) \end{matrix} = \begin{matrix}log & \frac{1}{x} \\ log & \frac{1}{2} \end{matrix} = \frac{− log x}{− log 2} = log_{2}(x) = μ_{cf}^{∗}(x).

This shows that, in some sense, as we let b → ∞ for type III continued logarithms, we get in the limit simple continued fractions.

5 Generalizing beyond continued logarithms

A natural question that arises is how one can define something more general than continued logarithms. Consider the following definition of generalized continued fractions.

Definition 52. Let (c_{n})_{n=0}^{∞} be an increasing sequence of natural numbers with c_{0} = 1. Let α ∈ (1, ∞). The generalized continued fraction for α determined by (c_{n})_{n=0}^{∞} is

a_{0} + \frac{b_{0}}{a_{1}} + \frac{b_{1}}{a_{2}} + \frac{b_{2}}{a_{3}} + ⋯ = [a_{0}, a_{1}, a_{2}, …]_{gcf},

where the the terms a_{0}, a_{1}, … and b_{0}, b_{1}, … are determined by the following recursive process, terminating at the term a_{n} if y_{n} = a_{n}.

y_{0} = α
j_{n} = max{j : c_{j} ≤ y_{n}}  n ≥ 0
a_{n} = c_{j_{n}}  n ≥ 0
b_{n} = c_{j_{n}+1} − c_{j_{n}}  n ≥ 0
y_{n+1} = \frac{b_{n}}{y_{n} − a_{n}} = \frac{c_{j_{n}+1} − c_{j_{n}}}{y_{n} − c_{j_{n}}}  n ≥ 0.

Remark 53. This is a generalization of simple continued fractions, and of type I and type III continued logarithms. Indeed, for simple continued fractions, the term sequence (c_{n})_{n=0}^{∞} consists of the natural numbers. For type I continued logarithms, the term sequence consists of the powers b^{0}, b^{1}, b^{2}, . . .. For type III continued logarithms, the term sequence consists of terms of the form ℓb^{k}, where k ∈ ℤ_{≥0} and ℓ ∈ {1, …, b − 1}.

Recall from Remark 15 that type II continued logarithms did not have the property that y_{n+1} could take any value in (1, ∞), regardless of the values of a_{n}, c_{n}. This is a desirable property to have, since it uniquely determines the numerator terms based on the corresponding denominator terms. We have defined generalized continued logarithms so that they have this property, and for that reason they are not a generalization of type II continued logarithms. Remark 54. As per Definitions 3 and 4, the nth convergent and nth remainder term are given by

x_{n} = [a_{0}, a_{1}, …, a_{n}]_{gcf}  and  r_{n} = [a_{n}, a_{n+1}, a_{n+2}, …]_{gcf},

respectively. Note that the remainder terms r_{n} and the terms y_{n} from Definition 52 are in fact the same.

We can derive various results for generalized continued fractions that are similar to those for continued logarithms. Most notably, we get the following sufficient criteria for guaranteed convergence and rational finiteness.

Theorem 55. Suppose there is a constant M > 0 such that c_{j+1} − c_{j} < Mc_{j} for all j. Then every infinite continued fraction with term sequence (c_{n})_{n=0}^{∞} will converge.

Theorem 56. Suppose (c_{n+1} − c_{n}) | c_{n} for all n ≥ 1. Then for every α > 1, the continued fraction of α is finite if and only if α ∈ ℚ.

We are also able to extend some of the measure-theoretic results to generalized continued fractions, though details are not provided here. We conjecture that the main results that we derived for the distribution and Khinchin constant of continued logarithms would extend (likely with some additional restrictions on the sequence (c_{n})_{n=0}^{∞}) to our generalized continued fractions.

6 Acknowledgements

We would like to thank Andrew Mattingly for his input and assistance. This research was initiated at and supported by the Priority Research Center for Computer-Assisted Research Mathematics and its Applications at the University of Newcastle.

A Proof of the type III continued logarithm distribution

This appendix is devoted to proving Theorems 41 and 44.

These proofs are based extensively on the proof presented in Section 15 of Khinchin [5], which proves similar statements for simple continued fractions.

Lemma A.1. For x > 1,

∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} \frac{1}{(1 + b^{−k}(x + ℓ − 1)^{−1})(b + b^{−k}(x + ℓ − 1)^{−1})} = \frac{1}{x(x + b − 1)}.

Proof.

∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} \frac{1}{(1 + b^{−k}(x + ℓ − 1)^{−1})(b + b^{−k}(x + ℓ − 1)^{−1})}

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{k}}{(b^{k}(x + ℓ − 1) + 1)(b^{k+1}(x + ℓ − 1) + 1)}

= \frac{1}{1 − b} ∑_{ℓ=1}^{b−1} ∑_{k=0}^{∞} \frac{b^{k}}{b^{k}(x + ℓ − 1) + 1} − \frac{b^{k+1}}{b^{k+1}(x + ℓ − 1) + 1}

= \frac{1}{1 − b} ∑_{ℓ=1}^{b−1} (\frac{1}{x + ℓ} − lim_{k→∞} \frac{b^{k}}{b^{k}(x + ℓ − 1) + 1})

= \frac{1}{1 − b} ∑_{ℓ=1}^{b−1} (\frac{1}{x + ℓ} − \frac{1}{x + ℓ − 1})

= \frac{1}{1 − b} (\frac{1}{x + b − 1} − \frac{1}{x}) = \frac{1}{1 − b} (\frac{1 − b}{x(x + b − 1)}) = \frac{1}{x(x + b − 1)}.

Theorem A.2. The sequence of functions m′_{n}(x) = \frac{d}{dx}m_{n}(x) is given by the recursive relationship

m′_{0}(x) = 1  (A.1)
m′_{n}(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} b^{−k}(x + ℓ − 1)^{−2}m′_{n−1}(1 + b^{−k}(x + ℓ − 1)^{−1})  n ≥ 1  (A.2)

for 1 ≤ x ≤ 2.

Proof. Equation (A.1) follows immediately from (9). Notice that (A.2) is the result of differentiating both sides of (10). In general, if m′_{n+1} is bounded and continuous for some n, then the series on the right hand side of (A.2) will converge uniformly on (1, 2). Thus the sum of the series will be bounded and continuous and will equal m′_{n+1}, so (A.2) follows by induction, since m′_{0} is clearly bounded and continuous.

We will now prove a number of lemmas and theorems about the following classes of sequences of functions, to which (m′_{n})_{n=0}^{∞} belongs.

Definition A.3. Let f_{0}, f_{1}, … be a sequence of functions on (1, 2). We will say (f_{n})_{n=0}^{∞} ∈ A^{∗} if for all x ∈ (1, 2) and n ≥ 0,

f_{n+1}(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} f_{n} (1 + \frac{b^{−k}}{x + ℓ − 1}) .  (A.3)

Furthermore, we say that (f_{n})_{n=0}^{∞} ∈ A^{∗∗} if (f_{n})_{n=0}^{∞} ∈ A^{∗} and there exist constants M, μ > 0 such that for all x ∈ (1, 2), we have 0 < f_{0}(x) < M and |f′_{0}(x)| < μ.

Lemma A.4.

∑^{(n)} \frac{b^{a_{0}+···+a_{n}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})} = 1.

Proof. Since the intervals of rank n are disjoint and

⋃^{(n)} J_{n} (\begin{matrix}a_{1}, & …, & a_{n} \\ c_{1}, & …, & c_{n} \end{matrix}) = (1, 2),

we have that

∑^{(n)} λJ_{n} (\begin{matrix}a_{1}, & …, & a_{n} \\ c_{1}, & …, & c_{n} \end{matrix}) = λ(1, 2) = 1.

Now notice that by Lemma 36 and Lemma 25,

λJ_{n} (\begin{matrix}a_{1}, & …, & a_{n} \\ c_{1}, & …, & c_{n} \end{matrix}) = |\frac{p_{n}}{q_{n}} − \frac{p_{n} + b^{a_{n}}p_{n−1}}{q_{n} + b^{a_{n}}q_{n−1}} | = |\frac{b^{a_{n}}(p_{n}q_{n−1} − q_{n}p_{n−1})}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})}|

= |\frac{(−1)^{n−1}b^{a_{0}+···+a_{n}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})} | = \frac{b^{a_{0}+···+a_{n}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})},

and thus

∑^{(n)} \frac{b^{a_{0}+···+a_{n}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})} = ∑^{(n)} λJ_{n} (\begin{matrix}a_{1}, & …, & a_{n} \\ c_{1}, & …, & c_{n} \end{matrix}) = 1.

Lemma A.5. If (f_{n})_{n=0}^{∞} ∈ A^{∗} then for n ≥ 0,

f_{n}(x) = ∑^{(n)} f_{0} (\frac{p_{n} + b^{a_{n}}p_{n−1}(x − 1)}{q_{n} + b^{a_{n}}q_{n−1}(x − 1)}) \frac{b^{∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(x − 1))^{2}} .  (A.4)

Proof. For n = 0, we just have a single interval, so

∑^{(0)}f_{0} (\frac{p_{0} + b^{a_{0}}p_{−1}(x − 1)}{q_{0} + b^{a_{0}}q_{−1}(x − 1)}) \frac{b^{a_{0}}}{(q_{0} + b^{a_{0}}q_{−1}(x − 1))^{2}}

= f_{0} (\frac{1 + (1)(1)(x − 1)}{1 + (1)(0)(x − 1)}) \frac{1}{(1 + (1)(0)(x − 1))^{2}} = f_{0}(x).

Now suppose (A.4) holds for n. Then

f_{n+1}(x)

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} f_{n} (1 + \frac{b^{−k}}{x + ℓ − 1})

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} ∑^{(n)} f_{0} (\frac{p_{n} + b^{a_{n}}p_{n−1}(1 + \frac{b^{−k}}{x+ℓ−1})}{q_{n} + b^{a_{n}}q_{n−1}(1 + \frac{b^{−k}}{x+ℓ−1})}) \frac{b^{∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(1 + \frac{b^{−k}}{x+ℓ−1}))^{2}}

= ∑^{(n)} ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} f_{0} (\frac{p_{n}b^{k}(x + ℓ − 1) + b^{a_{n}}p_{n−1}}{q_{n}b^{k}(x + ℓ − 1) + b^{a_{n}}q_{n−1}} ) \frac{b^{∑_{j=0}^{n} a_{j}}b^{k}}{(q_{n}b^{k}(x + ℓ − 1) + b^{a_{n}}q_{n−1})^{2}}

= ∑^{(n)} ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} f_{0} (\frac{ℓb^{k}p_{n} + b^{a_{n}}p_{n−1} + b^{k}p_{n}(x − 1)}{ℓb^{k}q_{n} + b^{a_{n}}q_{n−1} + b^{k}q_{n}(x − 1)}) \frac{b^{∑_{j=0}^{n} a_{j}}b^{k}}{(ℓb^{k}q_{n} + b^{a_{n}}q_{n−1} + b^{k}q_{n}(x − 1))^{2}}

= ∑^{(n+1)} f_{0} (\frac{c_{n+1}b^{a_{n+1}}p_{n} + b^{a_{n}}p_{n−1} + b^{a_{n+1}}p_{n}(x − 1)}{c_{n+1}b^{a_{n+1}}q_{n} + b^{a_{n}}q_{n−1} + b^{a_{n+1}}q_{n}(x − 1)}) \frac{b^{∑_{j=0}^{n} a_{j}}b^{a_{n+1}}}{(c_{n+1}b^{a_{n+1}}q_{n} + b^{a_{n}}q_{n−1} + b^{a_{n+1}}q_{n}(x − 1))^{2}}

= ∑^{(n+1)} f_{0} (\frac{p_{n+1} + b^{a_{n+1}}p_{n}(x − 1)}{q_{n+1} + b^{a_{n+1}}q_{n}(x − 1)}) \frac{b^{∑_{j=0}^{n+1} a_{j}}}{(q_{n+1} + b^{a_{n+1}}q_{n}(x − 1))^{2}} ,

so the result follows by induction. Lemma A.6. If (f_{n})_{n=0}^{∞} ∈ A^{∗∗}, then for n ≥ 0,

|f′_{n}(x)| ≤ \frac{3μ}{2^{n/2}} + 4M.

Proof. Differentiate (A.4) termwise, letting u = \frac{p_{n}+b^{an}p_{n−1}(x−1)}{q_{n}+b^{an}q_{n−1}(x−1)}, to get

f′_{n}(x) = ∑^{(n)} f′_{0}(u)\frac{(−1)^{n−1}b^{2 ∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(x − 1))^{4}} − 2 ∑^{(n)} f_{0}(u)\frac{b^{a_{n}}q_{n−1}b^{∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(x − 1))^{3}} .  (A.5)

The validity of termwise differentiation follows from the uniform convergence of both sums on the right hand side for 1 ≤ x ≤ 2. Notice that

|\frac{(−1)^{n−1}b^{2 ∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(x − 1))^{4}} | ≤ \frac{q_{n}b^{∑_{j=0}^{n} a_{j}}}{q_{n}^{4}} ≤ \frac{2b^{∑_{j=0}^{n} a_{j}}}{2^{(n−1)/2}q_{n}(q_{n} + b^{a_{n}}q_{n−1})}  (A.6)

by Lemma 24, Lemma 23, and the fact that q_{n} + b^{a_{n−1}}q_{n−1} ≤ 2q_{n}. Additionally,

\frac{b^{a_{n}}q_{n−1}b^{∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(x − 1))^{3}} ≤ \frac{b^{a_{n}}q_{n−1}b^{∑_{j=0}^{n} a_{j}}}{q_{n}^{3}} ≤ \frac{2b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})}  (A.7)

since b^{a_{n}}q_{n−1} ≤ q_{n} and q_{n} + b^{a_{n}}q_{n−1} ≤ 2q_{n}. Since (f_{n})_{n=0}^{∞} ∈ A^{∗∗}, we have by Definition A.3 that |f_{0}(x)| < M and |f′_{0}(x)| < μ for all x ∈ (1, 2). Thus we have by (A.5), (A.6), (A.7), and Lemma A.4,

|f′_{n}(x)| ≤ ∑^{(n)} |f′_{0}(u)| |\frac{(−1)^{n−1}b^{2 ∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(x − 1))^{4}} | + 2 ∑^{(n)} |f_{0}(u)| |\frac{b^{a_{n}}q_{n−1}b^{∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(x − 1))^{3}} |

≤ \frac{2μ}{2^{(n−1)/2}} ∑^{(n)} \frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})} + 4M ∑^{(n)} \frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})}

= \frac{2μ}{2^{(n−1)/2}} + 4M = \frac{2\sqrt{2}μ}{2^{n/2}} + 4M < \frac{3μ}{2^{n/2}} + 4M.

Lemma A.7. If (f_{n})_{n=0}^{∞} ∈ A^{∗} and for some constants T > t > 0,

\frac{t}{x(x + b − 1)} < f_{n}(x) < \frac{T}{x(x + b − 1)} ∀x ∈ (1, 2),

then

\frac{t}{x(x + b − 1)} < f_{n+1}(x) < \frac{T}{x(x + b − 1)} ∀x ∈ (1, 2).

Proof. By (A.3) and Lemma A.1 we have

f_{n+1}(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} f_{n} (1 + \frac{b^{−k}}{x + ℓ − 1})

> ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} \frac{t}{(1 + (x + ℓ − 1)^{−1}b^{−k})(b + (x + ℓ − 1)^{−1}b^{−k})}

= \frac{t}{x(x + b − 1)},

and a similar derivation shows

f_{n+1}(x) < \frac{T}{x(x + b − 1)},

from which the result follows.

Lemma A.8. If (f_{n})_{n=0}^{∞} ∈ A^{∗} then for all n ≥ 0,

∫_{1}^{2} f_{n}(z) dz = ∫_{1}^{2} f_{0}(z) dz.

Proof. Notice that

(1, 2] = ⋃_{k=0}^{∞} ⋃_{ℓ=1}^{b−1}(1 + (ℓ + 1)^{−1}b^{−k}, 1 + ℓ^{−1}b^{−k}],

where the intervals are pairwise disjoint. We then have, by (A.3) that

∫_{1}^{2} f_{n}(z) dz = ∫_{1}^{2} ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(z + ℓ − 1)^{2}} f_{n−1} (1 + \frac{b^{−k}}{z + ℓ − 1}) dz

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} ∫_{1+ℓ^{−1}b^{−k}}^{1+(ℓ+1)^{−1}b^{−k}} −f_{n−1}(u) du

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} f_{n−1}(u) du

= ∫_{1}^{2} f_{n−1}(u) du,

from which the result follows by induction. Lemma A.9. Suppose (f_{n})_{n=0}^{∞} ∈ A^{∗∗} and there are constants g, G > 0 such that for all x ∈ (1, 2),

\frac{g}{x(x + b − 1)} < f_{0}(x) < \frac{G}{x(x + b − 1)}.  (A.8)

Then there exist n ∈ ℕ and g_{1}, G_{1} > 0 such that

\frac{g_{1}}{x(x + b − 1)} < f_{n}(x) < \frac{G_{1}}{x(x + b − 1)},  (A.9)
g < g_{1} < G_{1} < G,  (A.10)

and

G_{1} − g_{1} < (G − g)δ + 2^{−n/2}(μ + G),  (A.11)

where δ = 1 − \frac{1}{4(b−1)} log \frac{2b}{b+1}. Proof. First define

φ_{n}(x) = f_{n}(x) − \frac{g}{x(x + b − 1)},  ψ_{n}(x) = \frac{G}{x(x + b − 1)} − f_{n}(x),  (A.12)

which are both positive functions by (A.8) and Lemma A.7. Notice that for the functions h(x) = \frac{g}{x(x+b−1)} and H(x) = \frac{G}{x(x+b−1)}, we have by Lemma A.1 that

h(x) = \frac{g}{x(x + b − 1)} = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} \frac{g}{(1 + b^{−k}(x + ℓ − 1)^{−1})(b + b^{−k}(x + ℓ − 1)^{−1})}

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} h (1 + \frac{b^{−k}}{x + ℓ − 1}) ,

and similarly

H(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} H (1 + \frac{b^{−k}}{x + ℓ − 1}) .

Thus for n ≥ 1, we have

φ_{n+1}(x) = f_{n+1}(x) − h(x)

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} f_{n} (1 + \frac{b^{−k}}{x + ℓ − 1}) − ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} h (1 + \frac{b^{−k}}{x + ℓ − 1})

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} φ_{n} (1 + \frac{b^{−k}}{x + ℓ − 1}) ,

and similarly

ψ_{n+1}(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{b^{−k}}{(x + ℓ − 1)^{2}} ψ_{n} (1 + \frac{b^{−k}}{x + ℓ − 1}) .

Thus (φ_{n})_{n=0}^{∞}, (ψ_{n})_{n=0}^{∞} ∈ A^{∗}, so by Lemma A.5, setting u = \frac{p_{n}+b^{an}p_{n−1}(x−1)}{q_{n}+b^{an}q_{n−1}(x−1)}, we get

φ_{n}(x) = ∑^{(n)} φ_{0}(u)\frac{b^{∑_{j=0}^{n} a_{j}}}{(q_{n} + b^{a_{n}}q_{n−1}(x − 1))^{2}} ≥ \frac{1}{4} ∑^{(n)} φ_{0}(u)\frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}^{2}} ,  (A.13)

and similarly

ψ_{n}(x) =≥ \frac{1}{4} ∑^{(n)} ψ_{0}(u)\frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}^{2}} ,  (A.14)

since

q_{n} + b^{a_{n}}q_{n−1}(x − 1) ≤ 2q_{n} ∀x ∈ (1, 2).

On the other hand, the mean value theorem gives

\frac{1}{4} ∫_{1}^{2} φ_{0}(z) dz = \frac{1}{4} ∑^{(n)} φ_{0}(u_{1})\frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})},  (A.15)

and

\frac{1}{4} ∫_{1}^{2} ψ_{0}(z) dz = \frac{1}{4} ∑^{(n)} ψ_{0}(u_{2})\frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})},  (A.16)

where for each interval (\frac{p_{n}}{q_{n}} , \frac{p_{n}+b^{an}p_{n−1}}{q_{n}+b^{an}q_{n−1}} ) of rank n, u_{1} and u_{2} are points in the interval and the length of the interval is ^{\frac{b^{∑{n \atop j=0} a_{j}}}{q_{n}(q_{n}+b^{an}q_{n−1})}}. From (A.13) and (A.15) we then get

φ_{n}(x) − \frac{1}{4} ∫_{1}^{2} φ_{0}(z) dz ≥ \frac{1}{4} ∑^{(n)} (φ_{0}(u) − φ_{0}(u_{1})) \frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})},  (A.17)

and from (A.14) and (A.16), we get

ψ_{n}(x) − \frac{1}{4} ∫_{1}^{2} ψ_{0}(z) dz ≥ \frac{1}{4} ∑^{(n)} (ψ_{0}(u) − ψ_{0}(u_{2})) \frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})}.  (A.18)

Now for 1 ≤ x ≤ 2, we have |φ′_{0}(x)| ≤ |f′_{0}(x)| + g ≤ μ + g and |ψ′_{0}(x)| ≤ |f′_{0}(x)| + G ≤ μ + G, so it follows by Lemma 24 that

|φ_{0}(u_{1}) − φ_{0}(u)| ≤ (μ + g)|u_{1} − u| ≤ (μ + g)\frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})} ≤ \frac{μ + g}{q_{n}} ≤ 2\frac{μ + g}{2^{n/2}} , (A.19)

and similarly

|ψ_{0}(u_{2}) − ψ_{0}(u)| ≤ 2\frac{μ + G}{2^{n/2}}.  (A.20)

Then by Lemma A.4, (A.17) and (A.19) give

φ_{n}(x) > \frac{1}{4} ∫_{1}^{2} φ_{0}(z) dz − \frac{1}{4} ∑^{(n)} (φ_{0}(u_{1}) − φ_{0}(u)) \frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})}

≥ ℓ − \frac{1}{4} ∑^{(n)} |φ_{0}(u_{1}) − φ_{0}(u)|\frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})}

≥ ℓ − \frac{1}{2} \frac{μ + g}{2^{n/2}} ∑^{(n)} \frac{b^{∑_{j=0}^{n} a_{j}}}{q_{n}(q_{n} + b^{a_{n}}q_{n−1})} = ℓ − \frac{1}{2} \frac{μ + g}{2^{n/2}} = ℓ − \frac{μ + g}{2^{n/2+1}} ,

where ℓ = \frac{1}{4} ∫_{1}^{2} φ_{0}(z) dz. Similarly, (A.18) and (A.20) give

ψ_{n}(x) ≥ L − \frac{G + μ}{2^{n/2+1}},

where L = \frac{1}{4} ∫_{1}^{2} ψ_{0}(z) dz. Now by (A.12), we have

f_{n}(x) = \frac{g}{x(x + b − 1)} + φ_{n}(x) > \frac{g}{x(x + b − 1)} + ℓ − \frac{μ + g}{2^{n/2+1}}
> \frac{g + ℓ − 2^{−n/2−1}(μ + g)}{x(x + b − 1)} = \frac{g_{1}}{x(x + b − 1)},  (A.21)

where g_{1} = g + ℓ − 2^{−n/2}(μ + g), and

f_{n}(x) = \frac{G}{x(x + b − 1)} − ψ_{n}(x) < \frac{G}{x(x + b − 1)} − L + \frac{μ + G}{2^{n/2+1}}
< \frac{G − ℓ + 2^{−n/2−1}(μ + G)}{x(x + b − 1)} = \frac{G_{1}}{x(x + b − 1)},  (A.22)

where G_{1} = G − L + 2^{−n/2−1}(μ + G). Now since ℓ, L > 0, we can choose n sufficiently large so that 2^{−n/2−1}(μ + g) < ℓ and 2^{−n/2−1}(μ + G) < L, so that we get

g < g_{1} < G_{1} < G.  (A.23)

Thus by (A.21), (A.22), and (A.23), we have found g_{1}, G_{1}, and n that satisfy (A.9) and (A.10). Notice that we also have

G_{1} − g_{1} = G − g − (L + ℓ) + 2^{−n/2−1}(2μ + g + G) < G − g − (L + ℓ) + 2^{−n/2}(μ + G). (A.24)

Now since

ℓ + L = \frac{1}{4} ∫_{1}^{2} \frac{G − g}{x(x + b − 1)} dx = (G − g)\frac{1}{4(b − 1)} log \frac{2b}{b + 1},

(A.24) becomes

G_{1} − g_{1} < (1 − \frac{1}{4(b − 1)} log \frac{2b}{b + 1}) (G − g) + 2^{−n/2}(μ + G) = δ(G − g) + 2^{−n/2}(μ + G),

so we see that g_{1}, G_{1}, and n also satisfy (A.11), completing the proof.

Remark A.10. Notice that the value of n chosen depends only on the values of μ and G, and that if we make 0 < μ_{1} < μ and 0 < G_{1} < G, the value of n chosen for μ and G will also work for μ_{1} and G_{1}. In other words, we can make μ and G smaller without having to increase n. This will be useful in the proof of the Theorem A.11.

Theorem A.11. Suppose (f_{n})_{n=0}^{∞} ∈ A^{∗∗}. Then there exist constants λ, A > 0 such that for all n ≥ 0 and x ∈ (1, 2),

|f_{n}(x) − \frac{a}{x(x + b − 1)}| < Ae^{−λ\sqrt{n}},

where

a = \frac{b − 1}{log \frac{2b}{b+1}} ∫_{1}^{2} f_{0}(z) dz.

Proof. By assumption, f_{0} is differentiable and continuous on [1, 2], so there is some constant m > 0 such that m < f_{0}(x) < M for all x ∈ [1, 2]. Then since \frac{1}{2(b+1)} < \frac{1}{x(x+b−1)} < \frac{1}{b} for all x ∈ (1, 2), we have

\frac{bm}{x(x + b − 1)} < f_{0}(x) < \frac{2(b + 1)M}{x(x + b − 1)} ∀x ∈ (1, 2).

Thus let g = bm and G = 2(b + 1)M and apply Lemma A.9 to f_{0}, g, and G, to get g_{1}, G_{1}, and n such that

\frac{g_{1}}{x(x + b − 1)} < f_{n}(x) < \frac{G_{1}}{x(x + b − 1)} ∀x ∈ (1, 2),

g < g_{1} < G_{1} < G,

and

G_{1} − g_{1} < δ(G − g) + 2^{−n/2}(μ + G).

By Lemma A.6, |f′_{n}(x)| < μ_{1} = \frac{3μ}{2^{n/2}} + 4M, and we can arrange to have μ_{1} < μ by making μ and n sufficiently large. (By Remark A.10, the results above are still valid for the new values of μ and n.) We can then apply Lemma A.9 again with f_{n}, g_{1}, and G_{1} instead of f_{0}, g, and G. This gives us new constants g_{2} and G_{2} such that (again due to Remark A.10),

\frac{g_{2}}{x(x + b − 1)} < f_{2n}(x) < \frac{G_{2}}{x(x + b − 1)} ∀x ∈ (1, 2),

g < g_{1} < g_{2} < G_{2} < G_{1} < G,

and

G_{2} − g_{2} < δ(G_{1} − g_{1}) + 2^{−2n/2}(μ_{1} + G_{1}).

Repeating this in a similar fashion gives, in general, constants g_{r}, G_{r} such that

\frac{g_{r}}{x(x + b − 1)} < f_{nr}(x) < \frac{G_{r}}{x(x + b − 1)} ∀x ∈ (1, 2),

g < g_{1} < ⋯ < g_{r−1} < g_{r} < G_{r} < G_{r−1} < ⋯ < G_{1} < G,

and

G_{r} − g_{r} < δ(G_{r−1} − g_{r−1}) + 2^{−rn/2}(μ_{r−1} + G_{r−1}),

where μ_{r−1} is a constant such that |f′_{n(r−1)}(x)| < μ_{r−1} for all x ∈ (1, 2). By Lemma A.6, we can take μ_{r} = \frac{3μ}{2^{nr/2}} + 4M, and then can choose r_{0} ∈ ℕ such that μ_{r−1} < 5M for all r ≥ r_{0}. Then since G_{r} < G = 2(b + 1)M, we have

G_{r} − g_{r} < δ(G_{r−1} − g_{r−1}) + (2b + 7)M2^{−nr/2} = δ(G_{r−1} − g_{r−1}) + M_{1}2^{−nr/2},  (A.25)

for all r ≥ r_{0} where M_{1} = (2b + 7)M. We now claim that for all k ≥ 0,

G_{r_{0}+k} − g_{r_{0}+k} < δ^{k}(G − g) + δ^{k}M_{1}2^{−nr_{0}/2} ∑_{j=0}^{k}(2^{−nj/2}δ^{−j}).  (A.26)

For k = 0, from (A.25), we have

G_{r_{0}} − g_{r_{0}} < δ(G_{r_{0}−1} − g_{r_{0}−1}) + M_{1}2^{−nr_{0}/2} < (G − g) + M_{1}2^{−nr_{0}/2}

= δ^{0}(G − g) + M_{1}δ^{0}2^{−nr_{0}/2} ∑_{j=0}^{0} 2^{−nj/2}δ^{−j}.

Now suppose (A.26) holds for k. Notice that

M_{1}2^{−n(r_{0}+k+1)/2} = M_{1}δ^{k+1}2^{−nr_{0}/2}2^{−n(k+1)/2}δ^{−(k+1)},

so by (A.25),

G_{r_{0}+k+1} − g_{r_{0}+k+1} < δ(G_{r_{0}+k} − g_{r_{0}+k}) + M_{1}2^{−n(r_{0}+k+1)/2}

< δ (δ^{k}(G − g) + M_{1}δ^{k}2^{−nr_{0}/2} ∑_{j=0}^{k}(2^{−nj/2}δ^{−j})) + M_{1}2^{−n(r_{0}+k+1)/2}

= δ^{k+1}(G − g) + M_{1}δ^{k+1}2^{−nr_{0}/2} (∑_{j=0}^{k}(2^{−nj/2}δ^{−j}) + 2^{−n(k+1)/2}δ^{−(k+1)})

= δ^{k+1}(G − g) + M_{1}δ^{k+1}2^{−nr_{0}/2} ∑_{j=0}^{k+1} 2^{−nj/2}δ^{−j},

so (A.26) follows by induction.

Now notice that for k > 0,

∑_{j=0}^{k} 2^{−nj/2}δ^{−j} < ∑_{j=0}^{∞}(2^{n/2}δ)^{−j} ≤ ∑_{j=0}^{∞}(2^{1/2}δ)^{−j} = γ < ∞,

since 2^{1/2}δ = \sqrt{2}(1 − \frac{1}{4(b−1)} log \frac{2b}{b+1}) > \sqrt{2}(1 − \frac{1}{4} log 2) > 1. (A.26) then becomes

G_{r_{0}+k} − g_{r_{0}+k} < δ^{k}(G − g + M_{1}2^{−nr_{0}/2}γ) = δ^{k}c,

where c > 0 is a constant. Then for r ≥ r_{0}, we have

G_{r} − g_{r} < δ^{r−r_{0}}c = δ^{r}(δ^{−r_{0}}c) = δ^{r}d,

where again, d > 0 is a constant. Finally, since δ < 1, we can choose B, λ > 0 such that G_{r} − g_{r} < Be^{−λr}. Thus there is clearly some common limit

a = lim_{r→∞} g_{r} = lim_{r→∞} G_{r},

and we have (setting r = n) that

|f_{n^{2}}(x) − \frac{a}{x(x + b − 1)}| < Be^{−λn}  ∀x ∈ (1, 2).  (A.27)

Thus we have

lim_{n→∞} ∫_{1}^{2} f_{n^{2}}(z) dz = ∫_{1}^{2} \frac{a}{x(x + b − 1)} = \frac{a}{b − 1} log \frac{2b}{b + 1},

so by Lemma A.8, ∫_{1}^{2} f_{0}(z) dz = \frac{a}{b−1} log \frac{2b}{b+1} and thus

a = \frac{b − 1}{log \frac{2b}{b+1}} ∫_{1}^{2} f_{0}(z) dz.

Now for arbitrary N ≥ r_{0}^{2}, we can choose n ≥ r_{0} such that n^{2} ≤ N < (n + 1)^{2}. We then have, by (A.27),

\frac{a − 2(b + 1)Be^{−λn}}{x(x + b − 1)} < \frac{a}{x(x + b − 1)}−Be^{−λn} < f_{n^{2}}(x) < \frac{a}{x(x + b − 1)}+Be^{−λn} < \frac{a + 2(b + 1)Be^{−λn}}{x(x + b − 1)} ,

for all x ∈ (1, 2). Then by Lemma A.7,

\frac{a − 2(b + 1)Be^{−λn}}{x(x + b − 1)} < f_{N}(x) < \frac{a + 2(b + 1)e^{−λn}}{x(x + b − 1)} ,

so

|f_{N}(x) − \frac{a}{x(x + b − 1)}| < \frac{2(b + 1)Be^{−λn}}{x(x + b − 1)} < 2(b+1)Be^{−λn} = 2(b+1)Be^{λ}e^{−λ(n+1)} < A′e^{−λ\sqrt{N}},

where A′ = 2(b + 1)Be^{λ} is a constant. Now for 0 ≤ N < r_{0}^{2}, note that each f_{N} is continuous (since f_{0} is differentiable and thus continuous and f_{N+1} is an absolutely convergent sum of continuous transformations of f_{N}). Thus we can choose A_{0}, A_{1}, …, A_{r_{0}^{2}−1} such that for 0 ≤ N ≤ r_{0}^{2} − 1

|f_{N}(x) − \frac{a}{x(x + b − 1)}| < A_{n}e^{−λ\sqrt{N}} ∀x ∈ (1, 2), ∀N ∈ \{0, 1, …, r_{0}^{2} − 1\}

for all x ∈ (1, 2). Finally, take A = max{A_{0}, A_{1}, …, A_{r_{0}−1}, A′}, so we have

|f_{N}(x) − \frac{a}{x(x + b − 1)}| < Ae^{−λ\sqrt{N}} ∀x ∈ (1, 2) ∀N ∈ ℤ_{≥0},

proving the theorem.

Corollary A.12. There exist constants λ, A > 0 such that for all n ≥ 0 and x ∈ (1, 2),

|m′_{n}(x) − \frac{a}{x(x + b − 1)}| < Ae^{−λ\sqrt{n}},

where

a = \frac{b − 1}{log \frac{2b}{b+1}} .

Proof. By Theorem A.2, (m′_{n})_{n=0}^{∞} ∈ A^{∗∗}. Then Theorem A.11 gives constants A, λ > 0 such that

|m′_{n}(x) − \frac{a}{x(x + b − 1)}| < Ae^{−λ\sqrt{n}},

where

a = \frac{b − 1}{log \frac{2b}{b+1}} ∫_{1}^{2} m′_{0}(z) dz = \frac{b − 1}{log \frac{2b}{b+1}} ∫_{1}^{2} 1 dz = \frac{b − 1}{log \frac{2b}{b+1}} ,

proving the corollary.

Our main goals, Theorems 41 and 44, follow easily from Corollary A.12.

Proof of 41. First note that since m_{n}(1) = 0 for all n, so by the Fundamental Theorem of Calculus,

m_{n}(x) = m_{n}(1) + ∫_{1}^{x} m′_{n}(z) dz = ∫_{1}^{x} m′_{n}(z) dz.

Thus

∫_{1}^{x} m′_{n}(z) dz − \frac{b − 1}{log \frac{2b}{b+1}} ∫_{1}^{x} \frac{1}{z(z + b − 1)} dz = m_{n}(x) − \frac{b − 1}{log \frac{2b}{b+1}} \frac{log \frac{bx}{x+b−1}}{b − 1} = m_{n}(x) − \begin{matrix}log & \frac{bx}{x+b−1} \\ log & \frac{2b}{b+1} \end{matrix} .

(A.28)

Then by Theorem A.11, we have

|m_{n}(x) − \begin{matrix}log & \frac{bx}{x+b−1} \\ log & \frac{2b}{b+1} \end{matrix} | = |∫_{1}^{x} m′_{n}(z) − \frac{b − 1}{log \frac{2b}{b+1}} \frac{1}{z(z + b − 1)} dz|

≤ ∫_{1}^{x} |m′_{n}(z) − \frac{b − 1}{log \frac{2b}{b+1}} \frac{1}{z(z + b − 1)}| dz < ∫_{1}^{x} Ae^{−λ\sqrt{n}}

= (x − 1)Ae^{−λ\sqrt{n}} < Ae^{−λ\sqrt{n}}.

Proof of Theorem 44. By Theorem 43,

λD_{n}(k, ℓ) = m_{n−1}(1 + ℓ^{−1}b^{−k}) − m_{n−1}(1 + (ℓ + 1)^{−1}b^{−k}) = ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} m′_{n−1}(z) dz.

Then by Corollary A.12, it follows that there is are constants A, λ > 0 such that

| ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}}m′_{n−1}(z) dz − \frac{b − 1}{log \frac{2b}{b+1}} ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} \frac{1}{z(z + b − 1)} dz|

≤ ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} |m′_{n−1}(z) − \frac{b − 1}{log \frac{2b}{b+1}} \frac{1}{z(z + b − 1)}| dz

< ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} Ae^{−λ\sqrt{n−1}} dz = (ℓ^{−1}b^{−k} − (ℓ + 1)^{−1}b^{−k})Ae^{−λ\sqrt{n−1}} = \frac{Ae^{−λ\sqrt{n−1}}}{ℓ(ℓ + 1)b^{k}} .

Finally, since

\frac{b − 1}{log \frac{2b}{b+1}} ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} \frac{1}{z(z + b − 1)} dz = \frac{b − 1}{log \frac{2b}{b+1}} \frac{log \frac{(1+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−(k+1)})}{(1+ℓ^{−1}b^{−(k+1)})(1+(ℓ+1)^{−1}b^{−k})}}{b − 1}

= \begin{matrix}log & \frac{(ℓb^{k}+1)((ℓ+1)b^{k+1}+1)}{(ℓb^{k+1}+1)((ℓ+1)b^{k}+1)} \\ log & \frac{2b}{b+1} \end{matrix} ,

we have

|λD_{n}(k, ℓ) − \begin{matrix}log & \frac{(ℓb^{k}+1)((ℓ+1)b^{k+1}+1)}{(ℓb^{k+1}+1)((ℓ+1)b^{k}+1)} \\ log & \frac{2b}{b+1} \end{matrix} | = |∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} m′_{n−1}(z) − \frac{b − 1}{log \frac{2b}{b+1}} \frac{1}{z(z + b − 1)} dz|

< \frac{Ae^{−λ\sqrt{n−1}}}{ℓ(ℓ + 1)b^{k}} .

B Proof of the type III logarithmic Khinchin constant

This appendix is devoted to proving Theorems 47 and 48. Note that the proofs in this appendix rely on certain results from Appendix A. Definition B.1. Let n ∈ ℕ, j_{1}, j_{2}, …, j_{n} ∈ ℕ be distinct, k_{1}, k_{2}, …, k_{n} ∈ ℤ_{≥0}, and ℓ_{1}, ℓ_{2}, …, ℓ_{n} ∈ {1, 2, …, b − 1}. Define

E (\begin{matrix}j_{1}, & j_{2}, & …, & j_{n} \\ k_{1}, & k_{2}, & …, & k_{n} \\ ℓ_{1}, & ℓ_{2}, & …, & ℓ_{n} \end{matrix}) = \{α ∈ (1, 2) : \begin{matrix}a_{j_{1}} = k_{1}, & a_{j_{2}} = k_{2}, & …, & a_{j_{n}} = k_{n} \\ c_{j_{1}} = ℓ_{1}, & c_{j_{2}} = ℓ_{2}, & …, & c_{j_{n}} = ℓ_{n} \end{matrix} \} .

Remark B.2. We will always assume that j_{1} < j_{2} < ⋯ < j_{n}, in which case E (\begin{matrix}j_{1}, & …, & j_{n} \\ k_{1}, & …, & k_{n} \\ ℓ_{1}, & …, & ℓ_{n} \end{matrix})

is a countable union of intervals of rank j_{n}. Theorem B.3. There exist constants A, λ > 0 such that for arbitrary m ∈ ℕ, j_{1} < … < j_{m} < j ∈ ℕ, k_{1}, …, k_{m}, k ∈ ℤ_{≥0}, and ℓ_{1}, …, ℓ_{m}, ℓ ∈ {1, …, b − 1}, we have

|\frac{λE (\begin{matrix}j_{1}, & …, & j_{m}, & j \\ k_{1}, & …, & k_{m}, & k \\ ℓ_{1}, & …, & ℓ_{m}, & ℓ \end{matrix})}{λE (\begin{matrix}j_{1}, & …, & j_{m} \\ k_{1}, & …, & k_{m} \\ ℓ_{1}, & …, & ℓ_{m} \end{matrix})} − \begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} | < \frac{Ae^{−λ\sqrt{j−j_{m}−1}}}{ℓ(ℓ + 1)b^{k}} .

Proof. First fix some interval J = J_{n} (\begin{matrix}k_{1}, & …, & k_{m} \\ ℓ_{1}, & …, & ℓ_{m} \end{matrix}) of rank m. Let

M_{n}(x) = λ {α ∈ J : z_{m+n} < x} .

In order to have α ∈ M_{n}(x) with a_{m+n} = k and c_{m+n} = ℓ, we must have 1+(x+ℓ−1)^{−1}b^{−k} < z_{m+n−1} ≤ 1 + ℓ^{−1}b^{−k} (similar to in (13)). It follows that

M_{n}(x) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} M_{n−1}(1 + ℓ^{−1}b^{−k}) − M_{n−1}(1 + (x + ℓ − 1)^{−1}b^{−k}),

so that (M′_{n})_{n=0}^{∞} ∈ A^{∗}. Now by Lemma 36, an arbitrary α ∈ J can be written as

α = \frac{p_{m}r_{m+1} + b^{a_{m}}p_{m−1}}{q_{m}r_{m+1} + b^{a_{m}}q_{m−1}} ,

or since r_{m+1} = \frac{1}{z_{m}−1},

α = \frac{p_{m} + b^{a_{m}}p_{m−1}(z_{m} − 1)}{q_{m} + b^{a_{m}}q_{m−1}(z_{m} − 1)}.

To have 1 < z_{m} < x, we must have

α ∈ (\frac{p_{m}}{q_{m}} , \frac{p_{m} + b^{a_{m}}p_{m−1}(x − 1)}{q_{m} + b^{a_{m}}q_{m−1}(x − 1)}) .

Thus

M_{0}(x) = |\frac{p_{m}}{q_{m}} − \frac{p_{m} + b^{a_{m}}p_{m−1}(x − 1)}{q_{m} + b^{a_{m}}q_{m−1}(x − 1)}| = \frac{b^{∑_{j=0}^{m} a_{m}}(x − 1)}{q_{m}(q_{m} + b^{a_{m}}q_{m−1}(x − 1))}.  (B.1)

Now define

χ_{n}(x) = \frac{M_{n}(x)}{λJ},

and note that (χ′_{n})_{n=0}^{∞} ∈ A^{∗}, since (M′_{n})_{n=0}^{∞} ∈ A^{∗} and

λJ = |\frac{p_{m}}{q_{m}} − \frac{p_{m} + b^{a_{m}}p_{m−1}}{q_{m} + b^{a_{m}}q_{m−1}} | = \frac{b^{∑_{j=0}^{m} a_{j}}}{q_{m}(q_{m} + b^{a_{m}}q_{m−1})}  (B.2)

is a constant. Now by (B.1) and (B.2), we have

χ_{0}(x) = \frac{(q_{m} + b^{a_{m}}q_{m−1})(x − 1)}{q_{m} + b^{a_{m}}q_{m−1}(x − 1)},
χ′_{0}(x) = \frac{q_{m}(q_{m} + b^{a_{m}}q_{m−1})}{(q_{m} + b^{a_{m}}q_{m−1}(x − 1))^{2}} ,
χ′′_{0}(x) = −\frac{2q_{m}b^{a_{m}}q_{m−1}(q_{m} + b^{a_{m}}q_{m−1})}{(q_{m} + b^{a_{m}}q_{m−1}(x − 1))^{3}}.

Thus for 1 ≤ x ≤ 2, we have χ′_{0}(x) < \frac{2q_{m}^{2}}{q_{m}^{2}} = 2, χ′_{0}(x) > \frac{q_{m}^{2}}{(2q_{m})^{2}} = \frac{1}{4}, and |χ′′_{0}(x)| < \frac{4q_{m}^{3}}{q_{m}^{3}} = 4, so (χ′_{n})_{n=0}^{∞} ∈ A^{∗∗}. It then follows from Theorem A.11 that there are constants A, λ > 0 such that

|χ′_{n}(x) − \frac{a}{x(x + b − 1)}| < Ae^{−λ\sqrt{n}},

for all n ≥ 0 and x ∈ (1, 2), or equivalently there exist functions θ_{n} : (1, 2) → (−1, 1) such that

χ′_{n}(x) = \frac{a}{x(x + b − 1)} + θ_{n}(x)Ae^{−λ\sqrt{n}}

for all n ≥ 0 and x ∈ (1, 2). We then have, for k ∈ ℤ_{≥0} and ℓ ∈ {1, …, b − 1}, that

χ_{n}(1+ℓ^{−1}b^{−k}) − χ_{n}(1 + (ℓ + 1)^{−1}b^{−k})

= ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} χ′_{n}(x) dx

= ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} \frac{a}{x(x + b − 1)} + θ_{n}(x)Ae^{−λ\sqrt{n}} dx

= \frac{a}{b − 1} log \frac{(1 + ℓ^{−1}b^{−k})(b + (ℓ + 1)^{−1}b^{−k})}{(b + ℓ^{−1}b^{−k})(1 + (ℓ + 1)^{−1}b^{−k})} + Ae^{−λ\sqrt{n}} ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} θ_{n}(x) dx.

Now

|∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} θ_{n}(x) dx| ≤ ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} |θ_{n}(x)| dx < ∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} 1 dx = \frac{1}{ℓ(ℓ + 1)b^{k}} ,

so there exist functions γ_{n} : (1, 2) → (−1, 1) such that

∫_{1+(ℓ+1)^{−1}b^{−k}}^{1+ℓ^{−1}b^{−k}} θ_{n}(x) dx = \frac{γ_{n}(x)}{ℓ(ℓ + 1)b^{k}} .

Then since λE (\begin{matrix}1, & …, & m, & m + n \\ k_{1}, & …, & k_{m}, & k_{m+n} \\ ℓ_{1}, & …, & ℓ_{m}, & ℓ_{m+n} \end{matrix}) = M_{n−1}(1 + ℓ^{−1}b^{−k}) − M_{n−1}(1 + (ℓ + 1)^{−1}b^{−k}),

λE (\begin{matrix}1, & …, & m, & m + n \\ k_{1}, & …, & k_{m}, & k_{m+n} \\ ℓ_{1}, & …, & ℓ_{m}, & ℓ_{m+n} \end{matrix})

= (\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} + \frac{γ_{n}(x)Ae^{−λ\sqrt{n−1}}}{ℓ(ℓ + 1)b^{k}} ) λE (\begin{matrix}1, …, m \\ k_{1}, …, k_{m} \\ ℓ_{1}, …, ℓ_{m} \end{matrix}) .

Now we can sum this relationship for k_{j} from 0 to ∞ and ℓ_{j} from 1 to b − 1 for certain indices j ≤ m. The indices we sum over will cancel from both sides, and we are left with an arbitrary sequence of subscripts 1 ≤ j_{1} < j_{2} < ⋯ < j_{t} = m. Then if we let j = m + n, we get

|\frac{λE (\begin{matrix}j_{1}, & …, & j_{m}, & j \\ k_{1}, & …, & k_{m}, & k \\ ℓ_{1}, & …, & ℓ_{m}, & ℓ \end{matrix})}{λE (\begin{matrix}j_{1}, & …, & j_{m} \\ k_{1}, & …, & k_{m} \\ ℓ_{1}, & …, & ℓ_{m} \end{matrix})} − \begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} | < \frac{Ae^{−λ\sqrt{j−j_{m}−1}}}{ℓ(ℓ + 1)b^{k}} ,

completing the proof.

Theorem B.4. Suppose f : ℤ_{≥0} × {1, …, b − 1} → ℝ is a positive function for which there exist constants C, δ > 0 such that

f(s, t) < C(tb^{s})^{\frac{1}{2} −δ}

for all s ∈ ℤ_{≥0} and t ∈ {1, …, b − 1}. Then for almost every α ∈ (1, 2),

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} f(a_{n}, c_{n}) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} f(k, ℓ)\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} .

Proof. First define

u_{k} = ∫_{1}^{2} f(a_{k}, c_{k}) dα,  b_{k} = ∫_{1}^{2}(f(a_{k}, c_{k}) − u_{k})^{2} dα,
g_{ik} = ∫_{1}^{2}(f(a_{i}, c_{i}) − u_{i})(f(a_{k}, c_{k}) − u_{k}) dα,  S_{n}(α) = ∑_{k=1}^{n}(f(a_{k}, c_{k}) − u_{k}).

Notice that the integral u_{k} is finite for all k, since

u_{k} = ∫_{1}^{2} f(a_{k}, c_{k}) dα = ∑_{s=0}^{∞} ∑_{t=1}^{b−1} f(s, t)λD_{n}(s, t)

< ∑_{s=0}^{∞} ∑_{t=1}^{b−1} C(tb^{b})^{\frac{1}{2} −δ}(2t^{−2}b^{−s}) = 2C ∑_{s=0}^{∞} ∑_{t=1}^{b−1} t^{−1}(tb^{s})^{−1−δ} < ∞.

Furthermore,

∫_{1}^{2} f_{n}(a_{k}, c_{k})^{2} dα = ∑_{s=0}^{∞} ∑_{t=1}^{b−1} f(s, t)^{2}λD_{n}(s, t) < ∑_{s=0}^{∞} ∑_{t=1}^{b−1} C^{2}(tb^{s})^{1−2δ}(2t^{−2}b^{−s})

= 2C^{2} ∑_{s=0}^{∞} ∑_{t=1}^{b−1} t^{−1}(tb^{s})^{−2δ} = C_{1} < ∞,

so

b_{k} = ∫_{1}^{2}(f(a_{k}, c_{k})−u_{k})^{2} dα = ∫_{1}^{2} f(a_{k}, c_{k})^{2} dα−2u_{k} ∫_{1}^{2} f(a_{k}, c_{k}) dα+u_{k}^{2} < C_{1}−u_{k} ≤ C_{1} < ∞,

(B.3)

and by the Cauchy-Schwarz Inequality,

u_{k} = ∫_{1}^{2} f(a_{k}, c_{k}) dα < \sqrt{∫_{1}^{2} f(a_{k}, c_{k})^{2} dα} < \sqrt{C_{1}}.  (B.4)

Furthermore, for k > i, we have

g_{ik} = ∫_{1}^{2} f(a_{i}, c_{i})f(a_{k}, c_{k}) dα − u_{i}u_{k} = ∑_{s_{1}=0}^{∞} ∑_{t_{1}=1}^{b−1} ∑_{s_{2}=0}^{∞} ∑_{t_{2}=1}^{b−1} f(s_{1}, t_{1})f(s_{2}, t_{2})λE (\begin{matrix}i & k \\ s_{1} & s_{2} \\ t_{1} & t_{2} \end{matrix}) − u_{i}u_{k}.

(B.5)

Now by Theorem B.3 and Corollary 38,

|λE (\begin{matrix}i & k \\ s_{1} & s_{2} \\ t_{1} & t_{2} \end{matrix}) − \begin{matrix}log & \frac{(1+t_{2}^{−1}b^{−s_{2}})(b+(t_{2}+1)^{−1}b^{−s_{2}})}{(b+t_{2}^{−1}b^{−s_{2}})(1+(t_{2}+1)^{−1}b^{−s_{2}})} \\ log & \frac{2b}{b+1} \end{matrix} λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix})| < \frac{Ae^{−λ\sqrt{k−i−1}}}{t_{2}(t_{2} + 1)b^{s_{2}}} λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix})

< 4Ae^{−λ\sqrt{k−i−1}}λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix}) , (B.6)

and by Theorem 44 and Corollary 38,

|λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix}) − \begin{matrix}log & \frac{(1+t_{2}^{−1}b^{−s_{2}})(b+(t_{2}+1)^{−1}b^{−s_{2}})}{(b+t_{2}^{−1}b^{−s_{2}})(1+(t_{2}+1)^{−1}b^{−s_{2}})} \\ log & \frac{2b}{b+1} \end{matrix} | < \frac{Ae^{−λ\sqrt{k−1}}}{t_{2}(t_{2} + 1)b^{s_{2}}} .  (B.7)

Now by (B.6) and (B.7), letting v = ^{\begin{matrix}log & \frac{(1+t{−1 \atop 2}b^{−s_{2}} )(b+(t_{2}+1)^{−1}b^{−s_{2}} )}{(b+t{−1 \atop 2}b^{−s_{2}} )(1+(t_{2}+1)^{−1}b^{−s_{2}} )} \\ log & \frac{2b}{b+1} \end{matrix}} , we get

|λE (\begin{matrix}i & k \\ s_{1} & s_{2} \\ t_{1} & t_{2} \end{matrix}) − λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix})|

≤ |λE (\begin{matrix}i & k \\ s_{1} & s_{2} \\ t_{1} & t_{2} \end{matrix}) − vλE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix})| + |vλE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) − λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix})|

< (4Ae^{−λ\sqrt{k−i−1}} + 4Ae^{−λ\sqrt{k−1}})λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix})

≤ 8Ae^{−λ\sqrt{k−i−1}}λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix}) .  (B.8)

Then by (B.5) and (B.8), we get

|g_{ik} − ∑_{s_{1}=0}^{∞} ∑_{t_{1}=1}^{b−1} ∑_{s_{2}=0}^{∞} ∑_{t_{2}=1}^{b−1} f(s_{1}, t_{1})f(s_{2}, t_{2})λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix}) + u_{i}u_{k}|

< 8Ae^{−λ\sqrt{k−i−1}} ∑_{s_{1}=0}^{∞} ∑_{t_{1}=1}^{b−1} ∑_{s_{2}=0}^{∞} ∑_{t_{2}=1}^{b−1} f(s_{1}, t_{1})f(s_{2}, t_{2})λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix})

= 8Ae^{−λ\sqrt{k−i−1}}u_{i}u_{k}.  (B.9)

But since

∑_{s_{1}=0}^{∞} ∑_{t_{1}=1}^{b−1} ∑_{s_{2}=0}^{∞} ∑_{t_{2}=1}^{b−1} f(s_{1}, t_{1})f(s_{2}, t_{2})λE (\begin{matrix}i \\ s_{1} \\ t_{1} \end{matrix}) λE (\begin{matrix}k \\ s_{2} \\ t_{2} \end{matrix}) = u_{i}u_{k},

(B.9) is just

|g_{ik}| < 8Ae^{−λ\sqrt{k−i−1}}u_{i}u_{k} < 8AC_{1}e^{−λ\sqrt{k−i−1}}.  (B.10)

From (B.3) and (B.10), we have for n > m > 0,

∫_{1}^{2}(S_{n}(α) − S_{m}(α))^{2} dα

= ∫_{1}^{2} ( ∑_{k=m+1}^{n} f(a_{k}, c_{k}) − u_{k})^{2} dα

= ∑_{k=m+1}^{n}(f(a_{k}, c_{k}) − u_{k})^{2} dα + 2 ∑_{i=m+1}^{n−1} ∑_{k=i+1}^{n} ∫_{1}^{2}(f(a_{i}, c_{i}) − u_{i})(f(a_{k}, c_{k}) − u_{k}) dα

= ∑_{k=m+1}^{n} b_{k} + 2 ∑_{i=m+1}^{n−1} ∑_{k=i+1}^{n} g_{ik} < C_{1}(n − m) + 16AC_{1} ∑_{i=m+1}^{n−1} ∑_{k=i+1}^{n} e^{−λ\sqrt{k−i−1}}

< C_{1}(n − m) + 16AC_{1} ∑_{i=m+1}^{n} ∑_{j=0}^{∞} e^{−λ\sqrt{j}} = C_{1}(n − m) + 16AC_{1}(n − m) ∑_{j=0}^{∞} e^{−λ\sqrt{j}}

= C_{2}(n − m),  (B.11)

where C_{2} = C_{1} + 16AC_{1} ∑_{j=0}^{∞} e^{−λ\sqrt{j}} is a constant. Now let ε > 0 and define

e_{n} = {α ∈ (1, 2) : |S_{n}(α)| ≥ εn}.

Clearly

∫_{1}^{2} S_{n}(α)^{2} dα ≥ ∫_{e_{n}} S_{n}(α)^{2} dα ≥ ε^{2}n^{2}λe_{n},

so that if we let m = 0 in (B.11) we get

λe_{n^{2}} ≤ \frac{∫_{1}^{2} S_{n^{2}}(α)^{2} dα}{ε^{2}n^{4}} < \frac{C_{2}}{ε^{2}n^{3}} .

Thus the series ∑_{n=1}^{∞} λe_{n^{2}} converges, so almost every α ∈ (1, 2) belongs to e_{n^{2}} for only finitely many n ∈ ℕ. Therefore for almost every α ∈ (1, 2) and for sufficiently large n,

\frac{S_{n^{2}}(α)}{n^{2}} < ε.

Now since ε > 0 was arbitrary, we can conclude that

lim_{n→∞} \frac{S_{n^{2}}(α)}{n^{2}} = 0  (B.12)

for almost every α ∈ (1, 2).

Now let N ∈ ℕ be arbitrary and choose n such that n^{2} ≤ N < (n + 1)^{2}, so that

∫_{1}^{2}(S_{N}(α) − S_{n^{2}}(α))^{2} dα < C_{2}(N − n^{2}) < C_{2}((n + 1)^{2} − n^{2}) = C_{2}(2n + 1) ≤ 3C_{2}n.

Let ε > 0 and define

e_{n,N} = {α ∈ (1, 2) : |S_{N}(α) − S_{n^{2}}(α)| ≥ εn^{2}}

and

E_{n} = ⋃_{N=n^{2}}^{(n+1)^{2}−1} e_{n,N}.

We then have for n^{2} ≤ N < (n + 1)^{2} that

∫_{1}^{2}(S_{N}(α) − S_{n^{2}}(α))^{2} dα ≥ ∫_{e_{n,N}} (S_{N}(α) − S_{n^{2}}(α))^{2} > ε^{2}n^{4}λe_{n,N},

and

λe_{n,N} < \frac{∫_{1}^{2}(S_{N}(α)^{2} − S_{n^{2}}(α))^{2}}{ε^{2}n^{4}} < \frac{3C_{2}}{ε^{2}n^{3}} ,

so

λE_{n} ≤ ∑_{N=n^{2}}^{(n+1)^{2}−1} λe_{n,N} < ((n + 1)^{2} − n^{2})\frac{3C_{2}}{ε^{2}n^{3}} ≤ \frac{9C_{2}}{ε^{2}n^{2}} .

Thus the series ∑_{n=1}^{∞} λE_{n} converges, so almost every α ∈ (1, 2) belongs to E_{n} for only finitely many n ∈ ℕ. In other words, for almost every α, sufficiently large N, and n = ⌊\sqrt{N}⌋, we have

\frac{|S_{N}(α) − S_{n^{2}}(α)|}{n^{2}} < ε.

Since ε > 0 was arbitrary, we can conclude

lim_{N→∞} \frac{S_{N}(α)}{n^{2}} − \frac{S_{n^{2}}}{n^{2}} = 0

for almost every α ∈ (1, 2), where n = ⌊\sqrt{N}⌋. By (B.12),

lim_{N→∞} \frac{S_{N}(α)}{n^{2}} = 0,

where n = ⌊\sqrt{N}⌋. Now since 0 < \frac{S_{N} (α)}{N} < \frac{S_{N} (α)}{n^{2}}, it follows that \frac{S_{N} (α)}{N} → 0 as n → ∞. Equivalently, by the definition of S_{N},

\frac{1}{N} ∑_{k=1}^{N} f(a_{k}, c_{k}) − \frac{1}{N} ∑_{k=1}^{N} u_{k} → 0  (B.13)

as N → ∞. Now by Theorem 44,

|u_{n}− ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} f(k, ℓ)\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} |

= ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} f(k, ℓ) |λD_{n} (\begin{matrix}k \\ ℓ \end{matrix}) − \begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} |

< Ae^{−λ\sqrt{n−1}} ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{f(k, ℓ)}{ℓ(ℓ + 1)b^{k}} < Ae^{−λ\sqrt{n−1}} ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{C(ℓb^{k})^{\frac{1}{2} −δ}}{ℓ(ℓ + 1)b^{k}}

= ACe^{−λ\sqrt{n−1}} ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} \frac{1}{(ℓ + 1)(ℓb^{k})^{\frac{1}{2} +δ}} < A_{1}e^{−λ\sqrt{n}}

for some constant A_{1}. Thus for almost every α ∈ (1, 2),

lim_{n→∞} u_{n} = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} f(k, ℓ)\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} ,

so indeed,

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} u_{n} = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} f(k, ℓ)\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} ,

at which point (B.13) gives

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} f(a_{n}, c_{n}) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} f(k, ℓ)\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} ,

for almost every α ∈ (1, 2).

We can now prove the desired theorems. Proof of Theorem 47. Fix k ∈ ℤ_{≥0} and ℓ ∈ {1, 2, …, b − 1}. Let

f(s, t) = \begin{cases}1, & if s = k and t = ℓ; \\ 0; & otherwise. \end{cases}

Clearly f(s, t) < 2 < 3(tb^{s})^{1/4} so f satisfies the conditions of Theorem B.4. Now

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} f(a_{n}, c_{n}) = lim_{N→∞} \frac{|\{n ∈ ℕ : a_{n} = k, c_{n} = ℓ\}|}{N},

so Theorem B.4 immediately gives, for almost every α ∈ (1, 2) that

P_{α}(k, ℓ) = lim_{N→∞} \frac{|\{n ∈ ℕ : a_{n} = k, c_{n} = ℓ\}|}{N} = \begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} ,

proving the theorem. Proof of Theorem 48. Define f(s, t) = log_{b}(tb^{s}) = s + log_{b} t. Notice that we can choose C > 0 such that log_{b}(x) < Cx^{1/3} for all x ≥ 1. Then if we take δ = \frac{1}{6}, we get

f(s, t) = log_{b}(tb^{s}) < C(tb^{s})^{1/3} = C(tb^{s})^{\frac{1}{2} −δ},

so f satisfies the conditions of Theorem B.4. We then get that for almost every α ∈ (1, 2).

lim_{N→∞} \frac{1}{N} log_{b}(c_{n}b^{a_{n}}) = ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} log_{b}(ℓb^{k})\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix} .  (B.14)

Now let u(k, ℓ) = log(1 + ℓ^{−1}b^{−k}) and v(k) = u(k, ℓ) − u(k, ℓ + 1). Notice that u(k, b) = u(k + 1, 1). Then

∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} log_{b}(ℓb^{k})\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix}

= \frac{1}{log \frac{2b}{b+1}} ∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1}(k + log_{b} ℓ)[u(k, ℓ) + u(k + 1, ℓ + 1) − u(k + 1, ℓ) − u(k, ℓ + 1)]

= \frac{1}{log \frac{2b}{b+1}} ∑_{ℓ=1}^{b−1} ∑_{k=0}^{∞}(k + log_{b} ℓ)[v(k) − v(k + 1)] = \frac{1}{log \frac{2b}{b+1}} (A + \frac{B}{log b}) ,

where

A = ∑_{ℓ=1}^{b−1} ∑_{k=0}^{∞} k[v(k) − v(k + 1)] = ∑_{ℓ=1}^{b−1} ∑_{k=1}^{∞} v(k) = ∑_{k=1}^{∞} ∑_{ℓ=1}^{b−1} u(k, ℓ) − u(k, ℓ + 1)

= ∑_{k=1}^{∞} u(k, 1) − u(k, b) = ∑_{k=1}^{∞} u(k, 1) − u(k + 1, 1) = u(1, 1) − lim_{k→∞} u(k, 1) = log (1 + \frac{1}{b}) ,

and

B = ∑_{ℓ=1}^{b−1} log ℓ ∑_{k=0}^{∞} v(k) − v(k + 1) = ∑_{ℓ=1}^{b−1} log ℓ(v(0) − lim_{k→∞} v(k))

= ∑_{ℓ=1}^{b−1} log ℓ (log \frac{1 + ℓ^{−1}}{1 + (ℓ + 1)^{−1}} − lim_{k→∞} log \frac{1 + ℓ^{−1}b^{−k}}{1 + (ℓ + 1)^{−1}b^{−k}} )

= ∑_{ℓ=1}^{b−1} log ℓ (log (1 + \frac{1}{ℓ}) − log (1 + \frac{1}{ℓ + 1}))

= ∑_{ℓ=1}^{b−1} log ℓ log (1 + \frac{1}{ℓ}) − ∑_{ℓ=2}^{b} log(ℓ − 1) log (1 + \frac{1}{ℓ})

= log 1 log 2 − ∑_{ℓ=2}^{b−1} (log(ℓ − 1) − log ℓ) log (1 + \frac{1}{ℓ}) − log(b − 1) log (1 + \frac{1}{b})

= − ∑_{ℓ=2}^{b−1} log (1 − \frac{1}{ℓ}) log (1 + \frac{1}{ℓ}) − log(b − 1) log (1 + \frac{1}{b}) .

Thus we have

∑_{k=0}^{∞} ∑_{ℓ=1}^{b−1} log_{b}(ℓb^{k})\begin{matrix}log & \frac{(1+ℓ^{−1}b^{−k})(b+(ℓ+1)^{−1}b^{−k})}{(b+ℓ^{−1}b^{−k})(1+(ℓ+1)^{−1}b^{−k})} \\ log & \frac{2b}{b+1} \end{matrix}

= \frac{1}{log b log \frac{2b}{b+1}} (log b log (1 + \frac{1}{b}) − log(b − 1) log (1 + \frac{1}{b}) − ∑_{ℓ=2}^{b−1} log (1 − \frac{1}{ℓ}) log (1 + \frac{1}{ℓ}))

= −\frac{1}{log b log \frac{2b}{b+1}} ∑_{ℓ=2}^{b} log (1 − \frac{1}{ℓ}) log (1 + \frac{1}{ℓ}) = \frac{1}{log b log \frac{b+1}{2b}} ∑_{ℓ=2}^{b} log (1 − \frac{1}{ℓ}) log (1 + \frac{1}{ℓ}) = A.

Thus (B.14) becomes

lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} log_{b}(c_{n}b^{a_{n}}) = A,

from which it follows that for almost all α ∈ (1, 2),

lim_{N→∞} (∏_{n=1}^{N} c_{n}b^{a_{n}})^{\frac{1}{N}} = b^{lim_{N→∞} \frac{1}{N} ∑_{n=1}^{N} log_{b}(c_{n}b^{an})} = b^{A},

as required.

References

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[2] J. M. Borwein, N. J. Calkin, S. B. Lindstrom, and A. Mattingly, Continued logarithms and associated continued fractions, Exp. Math. (2016), 1–18.

   http://dx.doi.org/10.1080/10586458.2016.1195307

[3] J. M. Borwein, A. van der Poorten, J. Shallit, and W. Zudilin, Neverending Fractions, Cambridge University Press, 2014.

[4] B. Gosper, Continued fraction arithmetic, Perl paraphernalia,

   http://perl.plover.com/classes/cftalk/INFO/gosper.txt.

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2010 Mathematics Subject Classification: Primary 11J70; Secondary 11K50. Keywords: continued fraction generalization, continued logarithm, Khinchin constant.

Received October 3 2016; revised versions received May 8 2017; May 9 2017. Published in Journal of Integer Sequences, May 21 2017.

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