On Particular Families of Hyperquadratic Continued Fractions in Power Series Fields

of Odd Characteristic

Alain Lasjaunias Institut de Mathématiques de Bordeaux CNRS-UMR 5251 Université de Bordeaux 33405 Talence France

              Alain.Lasjaunias@math.u-bordeaux.fr

Abstract We discuss the form of certain algebraic continued fractions in the field of power series over 𝔽_{p}, where p is an odd prime number. This leads to giving explicit continued fractions in these fields, satisfying an explicit algebraic equation of arbitrary degree d ≥ 2 and having an irrationality measure equal to d. Our results are based on a mysterious finite sequence of rational numbers.

1 Introduction

We are concerned with continued fractions in the fields of formal power series over a finite field. For a general account on this matter the reader may consult Schmidt’s article [13] and also Thakur’s book [14]. Let p be a prime number and 𝔽_{q} the finite field of characteristic p, having q elements. Given a formal indeterminate T, we consider the ring 𝔽_{q}[T], the field 𝔽_{q}(T) and 𝔽_{q}((T^{−1})), here simply denoted by 𝔽(q), the field of power series in 1/T over the finite field 𝔽_{q}. A non-zero element of 𝔽(q) can be written as

α = ∑_{i≤i_{0}} u_{i}T^{i} where i ∈ ℤ, u_{i} ∈ 𝔽_{q} and u_{i_{0}} ≠ 0.

An ultrametric absolute value is defined over this field by |0| = 0 and |α| = |T |^{i_{0}} where |T | is a fixed real number greater than 1. We also consider the subset 𝔽(q)^{+} = {α ∈ 𝔽(q) s.t. |α| > 1}. Note that 𝔽(q) is the completion of the field 𝔽_{q}(T) for this absolute value.

In power series fields over a general finite field 𝔽_{q}, where q is a power of p, contrarily to the case of real numbers, the continued fraction expansion for many algebraic elements can be explicitly given. This phenomenon is due to the existence of the Frobenius isomorphism in these fields. A particular subset of 𝔽(q), denoted by H(q), containing certain algebraic elements called hyperquadratic, has been considered by Bluher and the author [2]. Let t ≥ 0 be an integer and r = p^{t}, an irrational element of 𝔽(q) is called hyperquadratic of order t if it satisfies a non-trivial algebraic equation of the following form:

uX^{r+1} + vX^{r} + wX + z = 0 where (u, v, w, z) ∈ (𝔽_{q}[T])^{4}.

Note that a hyperquadratic element of order 0 is simply quadratic. Quadratic power series, as quadratic real numbers, have an ultimately periodic continued fraction expansion. However H(q) contains power series of arbitrary large algebraic degree over 𝔽_{q}(T) and the continued fraction expansion for various elements in this class has also been given explicitly. Note that if a hyperquadratic element has order t, it also has order kt for all integers k ≥ 1 [2, p. 258]. The consideration of this subset of algebraic elements was first put forward in the study of Diophantine approximation, beginning with Mahler’s article [11]. For more on this topic and more references, the reader may also consult the survey of the author [3].

Let us recall that, for an irrational element α ∈ 𝔽(q), the irrationality measure is defined by:

ν(α) = lim sup_{|Q|→∞}(− log |α − P/Q|/ log |Q|),

where P and Q belong to 𝔽_{q}(T). Then ν(α) is a real number greater or equal to 2. By adapting a theorem on rational approximation for real numbers, due to Liouville in the 19th century, Mahler [11] proved that, if α is an algebraic element of degree d > 1 over 𝔽_{q}(T), then we have ν(α) ∈ [2; d]. Furthermore if α is any irrational number in 𝔽(q), having the infinite continued fraction expansion α = [a_{1}, a_{2}, …, a_{n}, …], then the irrationality measure is directly connected to the sequence of degrees of the partial quotients [3, p. 214] and we have

ν(α) = 2 + lim sup_{n≥1}(deg(a_{n+1})/ ∑_{1≤i≤n} deg(a_{i})).

In this note, we present certain particular algebraic continued fractions in 𝔽(p) with odd p, which can be fully described. In Section 2, we recall several generalities on continued fractions. In Section 3, we describe a large family of algebraic continued fractions and, in Section 4, we exhibit two particular sub-families where the sequence of partial quotients is regularly distributed. The importance of these last continued fractions is highlighted in the last section, bringing families of hyperquadratic elements, in 𝔽(p) with odd p, having a prescribed algebraic degree, an explicit continued fraction expansion and an irrationality measure equal to the algebraic degree.

The reader will observe that this paper contains mainly conjectures based on computer calculations. Some results (see Theorem 4 below) were established in previous works, but the aim of this exposition is to point out several mathematical statements remaining largely mysterious and longing for clearness.

2 Continued fractions

Concerning continued fractions in the area of function fields, we use classical notation, as it can be found for instance in the survey of the author [8, pp. 3–8]. Throughout the paper we are dealing with finite sequences (or words), consequently we recall the following notation on sequences in 𝔽_{q}[T]. Let W = w_{1}, w_{2}, …, w_{n} be such a finite sequence, then we set |W | = n for the length of the word W. If we have two words W_{1} and W_{2}, then W_{1}, W_{2} denotes the word obtained by concatenation.

As usual, we let [W] = [w_{1}, …, w_{n}] ∈ 𝔽_{q}(T) denote the finite continued fraction w_{1} + 1/(w_{2} + 1/(…)). In this formula the w_{i}, called the partial quotients, are non-constant polynomials. Still, we will also use the same notation if the w_{i} are constant and the resulting quantity is in 𝔽_{q}. However in this last case, by writing [w_{1}, w_{2}, …, w_{n}] we assume that this quantity is well defined in 𝔽_{q}, i.e., w_{n} ≠ 0, [w_{n−1}, w_{n}] ≠ 0, …, [w_{2}, …, w_{n}] ≠ 0.

For n ≥ 0, a continuant X_{n} is a polynomial, in the n variables x_{1}, x_{2}, …, x_{n}, defined recursively by X_{0} = 1, X_{1} = x_{1} and X_{k} = x_{k}X_{k−1} + X_{k−2} for 2 ≤ k ≤ n. We use the notation ⟨W ⟩ for the continuant built from W = w_{1}, w_{2}, …, w_{n}. In the sequel the w′_{i}s are in 𝔽_{q}[T], then the degree in T of ⟨W ⟩ is clearly equal to the sum of the degrees in T of the w′_{i}s.

We let W′ (resp., W′′) denote the word obtained from W by removing the first (resp., last) letter of W. Hence, we recall that we have [W] = ⟨W ⟩/⟨W′⟩. We let W^{∗} = w_{n}, w_{n−1}, …, w_{1}, be the word W written in reverse order. We have ⟨W^{∗}⟩=⟨W ⟩ and also [W^{∗}] = ⟨W ⟩/⟨W′′⟩.

Moreover, if y ∈ 𝔽_{q}^{∗}, then we define y · W as the following sequence

y · W = yw_{1}, y^{−1}w_{2}, …, y^{(−1)^{n−1}}w_{n}.

Then, it is also known that ⟨y · W ⟩ = y⟨W ⟩ (resp., = ⟨W ⟩) if |W | is odd (resp., if |W | is even) and [y · W] = y[W].

If α ∈ 𝔽(q) is an infinite continued fraction, α = [a_{1}, a_{2}, …, a_{n}, …], we set x_{n} = ⟨a_{1}, a_{2}, …, a_{n}⟩ and y_{n} = ⟨a_{2}, …, a_{n}⟩. In this way, we have x_{n}/y_{n} = [a_{1}, a_{2}, …, a_{n}], with x_{1} = a_{1}, y_{1} = 1 and by convention x_{0} = 1, y_{0} = 0. We introduce α_{n+1} = [a_{n+1}, a_{n+2}, …] as the tail of the expansion of the complete quotient (α_{1} = α), and we have

α = (x_{n}α_{n+1} + x_{n−1})/(y_{n}α_{n+1} + y_{n−1}) for n ≥ 1.

We recall the following general result [4, p. 332].

Proposition 1. Let p be a prime number, q = p^{s} and r = p^{t} with s, t ≥ 1. Let ℓ ≥ 1 be an integer and (a_{1}, a_{2}, …, a_{ℓ}) ∈ (𝔽_{q}[T])^{l}, with deg(a_{i}) > 0 for 1 ≤ i ≤ ℓ. Let (P, Q) ∈ (𝔽_{q}[T])^{2} with deg(Q) < deg(P) < r. Then there exists a unique infinite continued fraction α in 𝔽(q)^{+} satisfying

α = [a_{1}, a_{2}, …, a_{ℓ}, α_{ℓ+1}] and α^{r} = P α_{ℓ+1} + Q.  (∗)

This element α is the unique root in 𝔽(q)^{+} of the following algebraic equation:

y_{ℓ}X^{r+1} − x_{ℓ}X^{r} + (P y_{ℓ−1} − Qy_{ℓ})X − P x_{ℓ−1} + Qx_{ℓ} = 0.  (∗∗)

3 P_{k}-continued fractions

In this note, to avoid unnecessary sophistication in a first stage, the base field will simply be the finite field 𝔽_{p} having p elements. Moreover, for the subject treated here we need have odd characteristic. Actually, in many areas concerning power series over a finite field, both cases, even and odd characteristic, must be considered separately. Hence, here p is a prime number with p > 2 and throughout this note, k is an integer with 1 ≤ k < p/2.

We shall now describe several numbers and polynomials which appeared in earlier papers. These were introduced in a paper by the author [4], to where the reader is invited to refer for the proof of the properties stated below.

In 𝔽_{p}[T], we define P_{k}(T) = (T^{2} − 1)^{k}. Next, we consider the following euclidean division in 𝔽_{p}[T]:

T^{p} = A_{1}P_{k} + R_{k}.

The polynomial A_{1} is the integer part of the rational T^{p}/P_{k} and R_{k} is the remainder. Hence we have A_{1} = [T^{p}/P_{k}], here the brackets denote the integer part (i.e., polynomial part) of the rational function. Note that we have deg(A_{1}) = p − 2k and also A_{1} = T if k = (p − 1)/2. We define

ω_{k} = (−1)^{k−1} ∏_{1≤i≤k}(1 − 1/2i) ∈ 𝔽_{p}^{∗}.

In previous works another polynomial closely related to R_{k} was used. Defining Q_{k} = (2kω_{k})^{−1}R_{k}, it was proved that the following holds in 𝔽_{p}[T]:

Q_{k}(T) = ∑_{0≤i≤k−1}(−1)^{k−1−i}\binom{k − 1}{i}(2i + 1)^{−1}T^{2i+1}.

These formulas are the key to obtain the following. In 𝔽_{p}(T), we have the remarkable continued fraction expansion [4, pp. 338–341]:

T^{p}/P_{k} = [A_{1}, w_{1}T, w_{2}T, …, w_{2k}T],  (1)

where the w_{i} ∈ 𝔽_{p}^{∗} are defined by : w_{1} = (2k − 1)(2kω_{k})^{−1} and

w_{i+1}w_{i} = (2k − 2i − 1)(2k − 2i + 1)(i(2k − i))^{−1} for 1 ≤ i ≤ 2k − 1.

Note that the w_{i} are rational numbers which, for all primes p with the condition 2k < p, exist by reduction in 𝔽_{p}^{∗}. We also have

w_{2k+1−i} = −w_{i} for 1 ≤ i ≤ 2k.  (2)

In the sequel, we consider the finite word W = w_{1}T, w_{2}T, …, w_{2k}T. Hence, with our notation, (1) implies T^{p}/P_{k} = A_{1} + 1/[W], and we can write

P_{k}/R_{k} = [w_{1}T, w_{2}T, …, w_{2k}T] = [W].  (3)

Let us do a thorough investigation of the extremal case: k = (p − 1)/2. Then, we clearly have A_{1} = [T^{p}/P_{k}] = T and consequently

R_{k} = T^{p} − T (T^{2} − 1)^{k} = T^{p} − T (T^{2k} − kT^{2k−2} + ⋯) = kT^{2k−1} + ⋯

Hence, we get [P_{k}/R_{k}] = (1/k)T = w_{1}T. Since 2k = −1, we have w_{1} = 1/k = −2. We obtain ω_{k} = (2k − 1)/(2kw_{1}) = −2/(−w_{1}) = −1. But we also have w_{i+1}w_{i} = −4. It follows that W = −2T, 2T, −2T, ⋯, 2T [4, Corollary 4.2 with a different notation].

Applying Proposition 1, we consider particular algebraic continued fractions in 𝔽(p) defined as follows.

Definition 2. An infinite continued fraction α ∈ 𝔽(p) is a P_{k}-expansion if there exist an integer ℓ ≥ 1, an ℓ-tuple, (λ_{1}, λ_{2}, …, λ_{ℓ}) ∈ (𝔽_{p}^{∗})^{ℓ} and (ϵ_{1}, ϵ_{2}) ∈ 𝔽_{p}^{∗} × 𝔽_{p} such that

α = [λ_{1}T, λ_{2}T, …, λ_{ℓ}T, α_{ℓ+1}] and α^{p} = ϵ_{1}P_{k}α_{ℓ+1} + ϵ_{2}R_{k}.

For given integers k, p and ℓ, chosen as above, a P_{k}-expansion is defined by an (ℓ+2)-tuple: Λ = (λ_{1}, λ_{2}, …, λ_{ℓ}, ϵ_{1}, ϵ_{2}). Moreover, this continued fraction satisfies the algebraic equation (∗∗), introduced in Proposition 1, with (P, Q) = (ϵ_{1}P_{k}, ϵ_{2}R_{k}), whose four coefficients in 𝔽_{p}[T] are only depending on k, p, ℓ and Λ.

These particular continued fraction expansions, introduced by the author [4], have been considered in several articles. They were particularly studied in a more general setting by the author [6]. In previous publications, we used in this definition the polynomial Q_{k} instead of R_{k}. This has no consequence, since both polynomials are proportional. Originally, a particular and simple continued fraction expansion in 𝔽(13), solution of an algebraic equation of degree four, was observed by Mills and Robbins [12]. In order to understand this particular and remarkable pattern, we developed this notion of P_{k}-expansion. So we could show that Mills and Robbins’ example was a particular case, defined as above with p = 13, k = 4 and ℓ = 6. This implied its full description [5] and a generalization concerning Robbins’ quartic [7, 1]. The possibility of describing the sequence of partial quotients for an arbitrary P_{k}-expansion is yet out of reach. However, using a technical lemma stated below, we will see in the next section that the sequence of partial quotients for certain P_{k}-expansions can be given explicitly. This lemma is the following (its origin can be found in a paper of the author [4, p. 336 and p. 343]).

Lemma 3. Let A ∈ 𝔽_{p}[T], δ ∈ 𝔽_{p}^{∗} and X ∈ 𝔽(p). Then we have

[A + δR_{k}P_{k}^{−1}, X] = [A, δ^{−1} · W, XP_{k}^{−2} + δ^{−1}R_{k}P_{k}^{−1}].

Proof. According to (3), we have P_{k}/R_{k} = [w_{1}T, w_{2}T, …, w_{2k}T] = [W]. By (2), we also have W^{∗} = −1 · W. We observe the following links between the polynomials P_{k}, R_{k} and the continuants built from W. We have

P_{k}/R_{k} = [W] = ⟨W ⟩/⟨W′⟩.

We note that ⟨W ⟩ = ⟨w_{1}T, w_{2}T, …, w_{2k}T ⟩ and P_{k} = (T^{2} − 1)^{k} have the same degree in T equal to 2k. Besides, the constant term of P_{k} is (−1)^{k} while, since W has even length, the constant term of ⟨W ⟩ is 1. Consequently we obtain

⟨W ⟩ = (−1)^{k}P_{k} and ⟨W′⟩ = (−1)^{k}R_{k}.  (4)

Hence, we can write

A + δR_{k}/P_{k} = A + 1/(δ^{−1}[W]) = A + 1/[δ^{−1} · W] = [A, δ^{−1} · W].

We set U = A, δ^{−1}·W = u_{1}, u_{2}, …, u_{n}. Here we have n = 2k+1, u_{1} = A and u_{i+1} = δ^{(−1)^{i}}w_{i}T for 1 ≤ i ≤ 2k. We set x_{n} = ⟨U⟩ and y_{n} = ⟨U′⟩, so that we have [U] = x_{n}/y_{n}. Let us consider an arbitrary element b ∈ 𝔽(p). Since x_{n}y_{n−1} − y_{n}x_{n−1} = (−1)^{n}, we can write

[U, b] − [U] = (x_{n}b + x_{n−1})/(y_{n}b + y_{n−1}) − x_{n}/y_{n} = (−1)^{n−1}(y_{n}(y_{n}b + y_{n−1}))^{−1}.

Then we have y_{n} = ⟨U′⟩ = ⟨δ^{−1} · W ⟩ = ⟨W ⟩, since |W |, the length of W, is even. But also y_{n−1} = ⟨(U′)′′⟩ = ⟨δ^{−1} · W′′⟩ = δ^{−1}⟨W′′⟩, since |W′′| is odd. Due to the properties of continuants and by (2), we have

⟨W′′⟩ = ⟨w_{1}T, w_{2}T, …, w_{2k−1}T ⟩ = ⟨−w_{2k}T, …, −w_{2}T ⟩ = −⟨W′⟩.

Hence, by (4), we get

y_{n} = ⟨W ⟩ = (1)^{k}P_{k} and y_{n−1} = −δ^{−1}⟨W′⟩ = (−1)^{k+1}δ^{−1}R_{k}.

Therefore, we obtain

[U, b] − [U] = (P_{k}^{2}b − δ^{−1}P_{k}R_{k})^{−1}.

Choose b = XP_{k}^{−2} + δ^{−1}R_{k}P_{k}^{−1} or equivalently X = P_{k}^{2}b − δ^{−1}P_{k}R_{k}. The last formula gives [U, b] = [U] + X^{−1} and therefrom, we obtain

[A, δ^{−1} · W, XP_{k}^{−2} + δ^{−1}R_{k}P_{k}^{−1}] = [U] + X^{−1} = [A + δR_{k}P_{k}^{−1}, X].

4 Perfect P_{k}-continued fractions

Our goal is to describe certain P_{k}-expansions explicitly. This will be done in two different cases, each corresponding to a particular choice of Λ. Because of some change of notation and to highlight the similarity between both cases, we repeat the results concerning the first case, which have already appeared in previous publications. To describe the partial quotients in these continued fractions, we need to introduce in 𝔽_{p}[T] two sequences of polynomials (A_{n})_{n≥0} and (B_{n})_{n≥0} as follows [1, p. 134]. Note that all these polynomials are monic. The first sequence is defined by

A_{0} = T and recursively A_{n+1} = [A_{n}^{p}/P_{k}] for n ≥ 0.

Note that, in agreement to the beginning of the previous section, we have A_{1} = [T^{p}/P_{k}]. While the second sequence is defined by

B_{0} = A_{0} = T and B_{1} = A_{1} = [T^{p}/P_{k}]

and recursively

B_{n+1} = B_{n}^{p}P_{k}^{(−1)^{n+1}} for n ≥ 1.

We are particularly interested in the degrees of these polynomials. We set u_{n} = deg(A_{n}) and v_{n} = deg(B_{n}). From the recursive definition of these polynomials, we get u_{0} = v_{0} = 1 and also

u_{n+1} = pu_{n} − 2k and v_{n+1} = pv_{n} + 2k(−1)^{n+1} for n ≥ 0.

Note that the sequence (u_{n})_{n≥0} is constant if 2k = p − 1, then we have A_{n} = A_{0} = T for n ≥ 0. Otherwise, both sequences (u_{n})_{n≥0} and (v_{n})_{n≥0} are strictly increasing.

The first case is described in the following theorem.

Theorem 4. Let p be an odd prime and k, ℓ integers chosen as above. Let α ∈ 𝔽(p) be a P_{k}-expansion, depending as above on an (ℓ + 2)-tuple Λ. We assume that ϵ_{2} ≠ 0 and Λ satisfies the following equality:

[λ_{ℓ}, λ_{ℓ−1}, …, λ_{2}, λ_{1} − ϵ_{2}] = ϵ_{1}/(ϵ_{2}ω_{k}).  C(A)

Then there exist a sequence (λ_{n})_{n≥1} in 𝔽_{p}^{∗} and a sequence (i(n))_{n≥1} in ℕ, such that

a_{n} = λ_{n}A_{i(n)} for n ≥ 1.

The proof of this theorem appears in two papers of the author [5, p. 1111] and [6, pp. 256–257]; the second paper corresponds to a much larger context. It also has a remark [6, p. 257] concerning a mistake in the writing of a formula in the first paper [5]. This proof is derived from Lemma 3 presented in Section 3. Both sequences (λ_{n})_{n≥1} and (i(n))_{n≥1} have been described in the papers cited above [5, 6], sometimes with different notation. We recall here below the description of these sequences, when the base field is prime, which is the only case we consider here.

For n ≥ 1, we set f(n) = (2k + 1)n + ℓ − 2k. At first, according to C(A), we can define in 𝔽_{p}^{∗} : z_{1} = λ_{1} − ϵ_{2} and z_{i} = λ_{i} + z_{i−1}^{−1} for 2 ≤ i ≤ ℓ. Then we consider the sequence (z_{n})_{n≥1} in 𝔽_{p}^{∗} defined, from the initial values z_{1}, z_{2}, …, z_{ℓ}, by the recursive formulas

z_{f(n)+i} = θ_{i}ϵ_{1}^{(−1)^{n+i}}z_{n}^{(−1)^{i}} for n ≥ 1 and 0 ≤ i ≤ 2k,

where

θ_{0} = −ω_{k} and θ_{i} = −iw_{i}/(2k − 2i + 1) for 1 ≤ i ≤ 2k.

Then the sequence (λ_{n})_{n≥1} in 𝔽_{p}^{∗} is defined recursively, from the initial values λ_{1}, λ_{2}, …, λ_{ℓ}, by the formulas

λ_{f(n)} = ϵ_{1}^{(−1)^{n}}λ_{n} and λ_{f(n)+i} = ϵ_{1}^{(−1)^{n+i}}w_{i}z_{n}^{(−1)^{i}},

for n ≥ 1 and for 1 ≤ i ≤ 2k.

It is worth mentioning that the complexity of this sequence (λ_{n})_{n≥1} in 𝔽_{p} has been studied. Indeed, it was proved to be (2k + 1)-automatic. For a full account on this matter, in the most general setting, the reader is advised to refer to a paper of Yao and the author [10, Section 5].

Concerning the sequence (i(n))_{n≥1}, we have the following description:

i(n) = 0 if n ∉ f(ℕ^{∗}) and i(f(n)) = i(n) + 1 for n ≥ 1.

We want to compute the irrationality measure for a continued fraction described in Theorem 4. We only need to know the sequence (i(n))_{n≥1} and the sequence (deg(A_{n}))_{n≥0}. The proof of the following corollary is based on another equivalent description [1, p. 143] of this sequence (i(n))_{n≥1}. In the sequel, we will use the following notation. For n ≥ 0, if we have the word w, w, …, w of length n, then we denote it shortly by w^{[n]} with w^{[0]} = ∅. In the same way W^{[n]} denotes the word W, W, …, W where W is a finite word repeated n times and W^{[0]} = ∅. Let (I_{n})_{n≥0} be the sequence of finite words of integers defined recursively by

I_{0} = 0 and I_{n} = n, I_{0}^{[2k]}, I_{1}^{[2k]}, …, I_{n−1}^{[2k]} for n ≥ 1.

Then the sequence I = (i(n))_{n≥1} in ℕ, introduced in Theorem 4, is given by the infinite word:

I = I_{0}^{[ℓ]}, I_{1}^{[ℓ]}, I_{2}^{[ℓ]}, …, I_{n}^{[ℓ]}, …

By mean of the last formula, on the irrationality measure, given in the introduction, we get the following result [1, p. 148].

Corollary 5. Let α be a P_{k}-expansion, depending on the (ℓ + 2)-tuple Λ satisfying C(A), then we have

ν(α) = 2 + (p − 2k − 1)/ℓ.

Before going on to the second kind of P_{k}-expansion, we need to point at an exceptional element in 𝔽(p) for all p ≥ 3 which is quadratic but also a perfect P_{k}-expansion of the type described in Theorem 4. This element appears in the extremal case k = (p − 1)/2. We have observed that, in this case, the sequence (A_{n})_{n≥0} is constant and A_{n} = T for n ≥ 0. Consequently, the elements described in Theorem 4.1 will have all partial quotients of the form a_{n} = λ_{n}T (note that, in agreement to Corollary 5, the irrationality measure of the continued fraction will be equal to 2). In these cases the sequence (λ_{n})_{n≥1} will not generally be periodic and this is why these non-quadratic examples were brought to light [12]. However at the bottom of these examples lies a universal element which is the formal golden mean. In 𝔽(p), for p ≥ 3, let us consider the following infinite continued fractions

ϕ(T) = [T, T, …, T, …] and ρ(T) = [−2T, 2T, …, −2T, 2T, …].

Note that we have ρ(T) = (u/2)ϕ(uT) with u^{2} = −4. Elementary computations [4, pp. 331–332] and [9, pp. 267–268] show that we have

ρ^{p} = (−1)^{ℓ}P_{k}ρ_{ℓ+1} − R_{k} for k = (p − 1)/2 and ℓ ≥ 1.

Since ω_{k} = −1, we observe that C(A) reduces to

[(−1)^{ℓ}2, (−1)^{ℓ−1}2, …, −2 + 1] = (−1)^{ℓ},

which is easily verified by induction. This element ρ will appear again in the last section (Proposition 9).

If we have reported Theorem 4 here, this is due to its somehow mysterious proof and also to the apparent closeness with the following conjecture. There is a second case, corresponding to a different choice of Λ, where a P_{k}-expansion can be described but only partially and conjecturally.

Conjecture 6. Let p be an odd prime and k, ℓ integers chosen as above. Let α ∈ 𝔽(p) be a P_{k}-expansion, depending as above on an (ℓ + 2)-tuple Λ. We assume that Λ satisfies the following equality:

[λ_{ℓ}, λ_{ℓ−1}, …, λ_{1} − ϵ_{2}] = 0.  C(B)

Then there exist a sequence (λ_{n})_{n≥1} in 𝔽_{p}^{∗} and a sequence (j(n))_{n≥1} in ℕ, such that

a_{n} = λ_{n}B_{j(n)} for n ≥ 1.

It may be worth noticing that condition C(B) (and also C(A)) can be stated in a different way. Indeed condition C(B) is simply equivalent to: ϵ_{2} = [λ_{1}, …, λ_{ℓ−1}, λ_{ℓ}]. This conjecture results from broad computer observations, letting the parameters p, k and ℓ vary. We also have a conjectural description of the sequence (j(n))_{n≥1} as follows.

Let (J_{n})_{n≥1} be the sequence of finite words of integers defined recursively by

J_{0} = 0 and J_{n} = n, J_{n−1}^{[2k−1]}, n − 1 for n ≥ 1.

Then the sequence J = (j(n))_{n≥1} in ℕ is given by the infinite word:

J = J_{0}^{[ℓ−1]}, 0, J_{1}^{[ℓ−1]}, 1, …, J_{n}^{[ℓ−1]}, n, …

The irrationality measure, for a continued fraction described in Conjecture 6, only depends on the sequences (j(n))_{n≥1} and (deg(B_{n}))_{n≥0}. Accordingly, using arguments similar to the ones given in a paper by Ayadi and the author [1, pp. 150–151]), we obtain the following.

Conjecture 7. Let α be a P_{k}-expansion, depending on the (ℓ + 2)-tuple Λ satisfying C(B), then we have

ν(α) = 2 + (p − 2k + 1)(p − 1)/(ℓ(p + 1) − 2k).

The description of the sequence (λ_{n})_{n≥1} in 𝔽_{p}^{∗} is in general yet out of reach. Note that in the simplest case ℓ = 1, it can be proved that J = ℕ and a_{n} = λ_{n}B_{n−1} where λ_{2n} = λ_{1}ϵ_{1}^{−1} and λ_{2n+1} = λ_{1} for n ≥ 1. Moreover, in this case, the irrationality measure is equal to p + 1 which is the maximal possible value. The sequence of partial quotients has also been fully described [8, pp. 20–21] in the simple case: k = 1 and ℓ = 2.

The reader may wonder why, in the two cases presented above, the P_{k}-expansions are special. In these cases we say that the expansion is perfect of type A or of type B. First, a computer observation, outside these cases, shows a certain irregularity in the sequence of partial quotients. Moreover the specificity of these perfect expansions of type A and B is pointed out in the next and last section. Finally, the possible connection between these two cases is an open question (regarding the similarity between conditions C(A) and C(B)).

Before concluding this section, we must make a remark on the origin of these particular continued fractions. As pointed out above, the P_{k}-expansions appeared from the study of Mills and Robbins quartic over 𝔽_{13} [12, pp. 403–404]. This quartic equation has been generalized [7] for all primes p ≥ 5. The solution has a continued fraction expansion with a different pattern according to the residue modulo 3 of the prime p. In the case p ≡ 1 mod 3 (particularly for p = 13), the pattern of this continued fraction is perfect of type A. In the case p ≡ 2 mod 3, the solution is a hyperquadratic element of order 2, but not of order 1 [2]. The continued fraction for this solution has a different pattern, described by Ayadi and Lasjaunias [1], and the partial quotients are all proportional to elements of the sequence (B_{n})_{n≥0}. This has given rise to the following generalization concerning this sequence (B_{n})_{n≥0}, in connection with hyperquadratic elements of higher order.

Returning to Proposition 1, let us take r = p^{n} with n ≥ 1. Let us make the following choice for the pair (P, Q) ∈ (𝔽_{p}[T])^{2}:

P = ϵ_{1}P_{k}^{(p^{n}+(−1)^{n−1})/(p+1)} and Q = ϵ_{2}R_{k}^{p^{n−1}} where (ϵ_{1}, ϵ_{2}) ∈ 𝔽_{p}^{∗} × 𝔽_{p}.

We consider the continued fraction α in 𝔽(p) so defined by (∗), which is a hyperquadratic element of order n. We observed the following. If the (ℓ + 2)-tuple Λ = (a_{1}, ⋯, a_{ℓ}, ϵ_{1}, ϵ_{2}) is well chosen, then all partial quotients for α appear to be proportional to certain polynomials B_{n} as in Conjecture 6 (corresponding to the case n = 1). Again this observation is just based on broad computer calculations. Amazingly, this is what happens for the solution of the generalized quartic when p ≡ 2 mod 3, in this case we have n = 2 and we may say that the expansion is perfect of type B and order 2. We have described [1, pp. 141–143] a choice of Λ to obtain the above conjecture in the case n = 2 and the description of the corresponding analogue of the sequence (j(n))_{n≥1}.

Here below, to support the observation which has just been made, we have collected a few examples of presumably perfect P_{k}-expansions of type B and of order 3. To allow easier computer calculations, we have only considered the values p = 3, n = 3, k = 1 and (ϵ_{1}, ϵ_{2}) = (1, 1).

Conjecture 8. Let ℓ ≥ 1 be an integer. Let α = [a_{1}, a_{2}, …, a_{n}, …] ∈ 𝔽(3) be the expansion defined by

α = [a_{1}, a_{2}, …, a_{ℓ}, α_{ℓ+1}] and α^{27} = (T^{2} − 1)^{7}α_{ℓ+1} + T^{9}

where we have ℓ = 3 and (a_{1}, a_{2}, a_{3}) = (2T, T, T) or ℓ = 4 and (a_{1}, a_{2}, a_{3}, a_{4}) = (T, 2T, 2T, T) or ℓ = 4 and (a_{1}, a_{2}, a_{3}, a_{4}) = (2T, T^{5} + 2T^{3}, T, T) or ℓ = 5 and (a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) = (2T, T, T, T, T^{5} + 2T^{3}). Then, in each of these four cases, there exist a sequence (λ_{n})_{n≥1} in 𝔽_{3}^{∗} and a sequence (j(n))_{n≥1} in ℕ, such that

a_{n} = λ_{n}B_{j(n)} for n ≥ 1.

Let us make a comment about this last statement. There are several ways [7, p. 34] to obtain the first partial quotients of an algebraic element α in 𝔽(q). A natural way is to obtain the beginning of the continued fraction expansion from a rational approximation to α. In the present case (this is valid for all hyperquadratic elements), we can write the algebraic equation (∗∗) satisfied by α in the following way: α = (aα^{r} + b)/(cα^{r} + d) = g(α). Then we build rational approximations to α, starting from R_{1} = x_{ℓ}/y_{ℓ}, by the recursive formula R_{n+1} = g(R_{n}) for n ≥ 1. In the cases indicated above, these rational numbers have a special form, showing a particular expansion with partial quotients as wanted. We guess a proof of our conjecture could be derived from the study of the numerators and denominators of R_{n}.

Finally, we notice the existence of another different generalization for P_{k}-expansions of type A, also leading to hyperquadratic elements of higher order [6]. In the next and last section, we return to P_{k}-expansions of order 1 (i.e., r = p).

5 Hyperquadratic elements having a low algebraic degree

In this section, we introduce a family of polynomials in the variable X with coefficients in 𝔽_{p}[T]. These polynomials H have the particular form of hyperquadratic type:

H(X) = uX^{p+1} + vX^{p} + wX + z with u, v, w, z ∈ 𝔽_{p}[T].

The four coefficients are depending on parameters coming from the finite word W introduced in Section 3. Moreover H(X) = 0 will have solutions in 𝔽(p) having a continued fraction expansion of the type discussed above. Under a particular choice of these parameters, H will be reducible and consequently these solutions will have a particular algebraic degree smaller than p + 1. The consideration of these polynomials comes from a quartic equation, generalizing Mills and Robbins example, introduced by the author [7]. To describe these coefficients, we use some more notation appearing there [7, pp. 30–31].

Here p and k are as above and W = w_{1}T, w_{2}T, …, w_{2k}T is the finite word introduced in Section 3.

Let n, t and m be integers with 1 ≤ n ≤ t ≤ m ≤ 2k. Then we introduce the continuant:

K_{n,m} = ⟨w_{n}T, w_{n+1}T, …, w_{m}T ⟩ ∈ 𝔽_{p}[T].

By convention we extend this notation with K_{n,n−1} = 1 and K_{n,n−2} = 0. From general properties of continuants [8, p. 7], we have the following formula

K_{n,m}K_{t,m−1} − K_{n,m−1}K_{t,m} = (−1)^{m−t}K_{n,t−2}.  (5)

Since, by (2), we have W^{∗} = −1 · W, ⟨A^{∗}⟩ = ⟨A⟩ and ⟨−1 · A⟩ = (−1)^{|A|}⟨A⟩ for any finite word A, we also get

K_{n,m} = (−1)^{m−n+1}K_{2k+1−m,2k+1−n}.  (6)

Let us consider the vector V = (p, k, j, ϵ) where p, k are as above, j is an integer with 1 ≤ j ≤ 2k − 1 and ϵ ∈ 𝔽_{p}^{∗}. To each such vector V, we associate a polynomial H_{V} in 𝔽_{p}[T][X] defined by

H_{V} (X) = K_{j+2,2k}X^{p+1} − K_{j+1,2k}X^{p} + ϵ(K_{1,j}X + K_{1,j−1}).

We have the following proposition.

Proposition 9. Let V = (p, k, j, ϵ) and H_{V} ∈ 𝔽_{p}[T][X] be defined as above. For each choice of V, there are two infinite continued fractions α and β in 𝔽(p)^{+}, both P_{k}-expansions, such that we have H_{V} (α) = 0 and H_{V} (1/β) = 0. The first one α is defined by

α = [w_{j+1}T, w_{j+2}T, …, w_{2k}T, α_{2k−j+1}]

and

α^{p} = ϵ(−1)^{k+j}(P_{k}α_{2k−j+1} − R_{k}).

The second one β is defined by

β = [w_{2k−j+1}T, w_{2k−j+2}T, …, w_{2k}T, β_{j+1}]

and

β^{p} = ϵ^{−1}(−1)^{k+j}(P_{k}β_{j+1} − R_{k}).

Moreover, corresponding to each triple (p, k, j), we define two different values ϵ_{A} and ϵ_{B} in 𝔽_{p}^{∗}, as follows:

ϵ_{A} = (−1)^{k+j+1}[w_{j+1}, …, w_{2k}, ω_{k}] and ϵ_{B} = (−1)^{k+j+1}[w_{j+1}, …, w_{2k}].

If ϵ = ϵ_{A} then both P_{k}-expansions α and β are perfect of type A (i.e., condition C(A) in Theorem 4 is satisfied). If ϵ = ϵ_{B} then both P_{k}-expansions α and β are perfect of type B (i.e., condition C(B) in Conjecture 6 is satisfied). Finally, let k = (p − 1)/2 and V = (p, k, j, ϵ_{A}). Then H_{V} has a factor P of degree 2. We have α = β, P (α) = 0 and P (X) = X^{2} + 2(−1)^{j}X + 1.

Proof. According to (6), we can write

K_{1,j} = (−1)^{j}K_{2k+1−j,2k} and K_{1,j−1} = (−1)^{j−1}K_{2k+2−j,2k},
K_{j+1,2k} = (−1)^{j}K_{1,2k−j} and K_{j+2,2k} = (−1)^{j−1}K_{1,2k−j−1}.

From these formulas, we can write (−1)^{j−1}H_{V} (X) =

K_{1,2k−j−1}X^{p+1} + K_{1,2k−j}X^{p} + ϵ(−K_{2k+1−j,2k}X + K_{2k+2−j,2k}).

Therefore, we get ϵ^{−1}(−1)^{j−1}X^{p+1}H_{V} (1/X) =

K_{2k+2−j,2k}X^{p+1} − K_{2k+1−j,2k}X^{p} + ϵ^{−1}(K_{1,2k−j}X + K_{1,2k−j−1}).

We set V^{∗} = (p, k, 2k − j, ϵ^{−1}). Hence, we have obtained

X^{p+1}H_{V} (1/X) = ϵ(−1)^{j−1}H_{V^{∗}}(X).  (7)

By Proposition 1, we know that α ∈ 𝔽(p) defined by

α = [a_{1}, a_{2}, ⋯, a_{ℓ}, α_{ℓ+1}] and α^{p} = P α_{ℓ+1} + Q

satisfies I(α) = 0 where I ∈ 𝔽_{p}[T][X] is defined by

I(X) = y_{ℓ}X^{p+1} − x_{ℓ}X^{p} + (P y_{ℓ−1} − Qy_{ℓ})X − P x_{ℓ−1} + Qx_{ℓ}.

In order to prove that H_{V} (α) = 0, for the element α ∈ 𝔽(p) defined in this proposition, we will show that I = H_{V} if

(a_{1}, …, a_{ℓ}, P, Q) = (w_{j+1}T, …, w_{2k}T, ϵ_{1}P_{k}, −ϵ_{1}R_{k}) with ϵ_{1} = (−1)^{k+j}ϵ.

First, recall that, according to (4) in Section 3, we have

P_{k} = (−1)^{k}K_{1,2k} and R_{k} = (−1)^{k}K_{2,2k}.

Consequently, by (6) and ϵ_{1} = (−1)^{k+j}ϵ, we obtain

P = ϵ(−1)^{j}K_{1,2k} and Q = ϵ(−1)^{j}K_{1,2k−1}.  (8)

Since (a_{1}, …, a_{ℓ}) = (w_{j+1}T, …, w_{2k}T), we also have

x_{ℓ} = K_{j+1,2k}, y_{ℓ} = K_{j+2,2k}, x_{ℓ−1} = K_{j+1,2k−1}, y_{ℓ−1} = K_{j+2,2k−1}.  (9)

Hence, by (8), (9) and (5), we get

P y_{ℓ−1} − Qy_{ℓ} = ϵ(−1)^{j}(K_{1,2k}K_{j+2,2k−1} − K_{1,2k−1}K_{j+2,2k}) = ϵK_{1,j}

and also

Qx_{ℓ} − P x_{ℓ−1} = ϵ(−1)^{j}(K_{1,2k−1}K_{j+1,2k} − K_{1,2k}K_{j+1,2k−1}) = ϵK_{1,j−1}.

Finally, we obtain the desired outcome:

I(X) = K_{j+1,2k}X^{p+1} − K_{j+2,2k}X^{p} + ϵK_{1,j} + ϵK_{1,j−1} = H_{V} (X).

Changing j into 2k − j and ϵ into ϵ^{−1}, in the definition for α, we get the definition for β. Since this means changing V into V^{∗}, we have just proved that H_{V^{∗}}(β) = 0 and according to (7) this gives H_{V} (1/β) = 0. Let us now consider the (2k − j + 2)-tuple Λ_{α} defining the P_{k}-expansion α. We have Λ_{α} = (w_{j+1}, …, w_{2k}, −ϵ_{2}, ϵ_{2}) where ϵ_{2} = (−1)^{k+j+1}ϵ. Hence C(A) is satisfied if and only if we have [w_{2k}, w_{2k−1} …, w_{j+1} − ϵ_{2}] = −1/ω_{k}. This is equivalent to ϵ_{2} = [w_{j+1}, …, w_{2k} + 1/ω_{k}]. Consequently C(A) is satisfied by Λ_{α} if and only if we have

ϵ = ϵ_{A} = (−1)^{j+k+1}[w_{j+1}, …, w_{2k}, ω_{k}].

Considering the same (2k − j + 2)-tuple Λ_{α}, we see in the same way that Λ_{α} satisfies C(B) if and only if we have

ϵ = ϵ_{B} = (−1)^{j+k+1}[w_{j+1}, …, w_{2k}].

Now, we need to consider β instead of α. Here the (j + 2)-tuple Λ_{β} defining β is Λ_{β} = (w_{2k−j+1}, …, w_{2k}, −ϵ_{2}, ϵ_{2}) where ϵ_{2} = (−1)^{k+j+1}ϵ^{−1}. In the same way, we obtain that Λ_{β} satisfies C(A) if

ϵ = \overline{ϵ_{A}} = (−1)^{j+k+1}[w_{2k−j+1}, …, w_{2k}, ω_{k}]^{−1}

and also that Λ_{β} satisfies C(B) if

ϵ = \overline{ϵ_{B}} = (−1)^{j+k+1}[w_{2k−j+1}, …, w_{2k}]^{−1}.

We only need to prove that ϵ_{A} = \overline{ϵ_{A}} and ϵ_{B} = \overline{ϵ_{B}}. Let us compare ϵ_{B} and \overline{ϵ_{B}}. According to (2), we can write [w_{2k−j+1}, …, w_{2k}] = [−w_{j}, ⋯, −w_{1}]. Hence ϵ_{B} = \overline{ϵ_{B}} is equivalent to

[w_{j+1}, …, w_{2k}] + [w_{j}, …, w_{1}]^{−1} = 0.

With our notation on continuants, this can be written as

(K_{j+1,2k}/K_{j+2,2k})(1) + (K_{1,j−1}/K_{1,j})(1) = 0.

This comes from K_{1,2k}(1) = (−1)^{k}(1 − 1)^{k} = 0 and the following general formula on continuants [8, p. 7]: K_{1,2k} = K_{1,j}K_{j+1,2k} + K_{1,j−1}K_{j+2,2k}. Turning to ϵ_{A} = \overline{ϵ_{A}}, we see, using the same arguments as above that this is equivalent to

[w_{j+1}, …, w_{2k}, ω_{k}] + [w_{j}, …, w_{1}, −ω_{k}]^{−1} = 0.

Using the same general formula on continuants, this is again equivalent to

⟨−ω_{k}, w_{1}, w_{2}, …, w_{2k}, ω_{k}⟩ = 0.

The truth of that is derived from the following two equalities:

⟨w_{2}, …, w_{2k}⟩ = (−1)^{k} and ⟨w_{2}, …, w_{2k−1}⟩ = 2(−1)^{k+1}ω_{k}.

The first one is obtained from R_{k}(1) = (−1)^{k}K_{2,2k}(1) = 1, while the proof for the second one is more mysterious and it is left to the reader.

We finish the proof of this proposition by considering the extremal case k = (p − 1)/2. In this particular case, we have seen in Section 3, that we have W = −2T, 2T, −2T, …, 2T and ω_{k} = −1. Hence, for all j, we get

ϵ_{A} = (−1)^{k+j+1}[2(−1)^{j+1}, …, 2, −1] = (−1)^{k}.

Consequently, setting ℓ = 2k − j, α is defined by

α = [2(−1)^{ℓ−1}T, …, 2T, α_{ℓ+1}] and α^{p} = (−1)^{j}(P_{k}α_{ℓ+1} − R_{k}).

Note that the same definition holds for β with ℓ = j. Let us consider in 𝔽(p) the infinite continued fraction γ = (−1)^{j}α. We have γ^{p} = (−1)^{j}α^{p} and γ_{ℓ+1} = (−1)^{j}α_{ℓ+1} for all ℓ ≥ 1. Since (−1)^{j} = (−1)^{ℓ}, we see that γ is defined by

γ = [−2T, …, 2(−1)^{ℓ}T, γ_{ℓ+1}] and γ^{p} = (−1)^{ℓ}P_{k}γ_{ℓ+1} − R_{k}.

Comparing to the definition of the quadratic continued fraction ρ, derived from the formal golden mean, described in Section 4 and because of the uniqueness in this definition, we obtain γ = ρ, α = (−1)^{j}ρ and in the same way β = α. A basic computation shows that the minimal polynomial of (−1)^{j}ρ is P (X) = X^{2} + 2(−1)^{j}T X + 1.

Concerning this polynomial H_{V}, we are interested in its quality of being or not being reducible over 𝔽_{p}(T). In other words, we are interested in the exact algebraic degree of α and β over 𝔽_{p}(T). We have checked the reducibility of H_{V} by computer calculations. In the vector V, we first fix p and k. Then, according to formula (7), we observe that the reducibility of H_{V} need only be studied for 1 ≤ j ≤ k.

To illustrate our purpose, let us consider the simplest, and somehow trivial, case : p = 3. Then, we only have to consider k = j = 1. Here we have ω_{1} = 1/2 = −1 and W = T, −T. There are two polynomials H_{1} and H_{2} corresponding to V_{1} = (3, 1, 1, 1) and V_{2} = (3, 1, 1, 2):

H_{1}(X) = H_{V_{1}}(X) = X^{4} + T X^{3} + T X + 1

and

H_{2}(X) = H_{V_{2}}(X) = X^{4} + T X^{3} − T X − 1 = (X^{2} − 1)(X^{2} + T X + 1).

Note that reducibility of the polynomials H_{V} over 𝔽_{p}(T) has been tested by computer, using Maple programming software. The polynomial H_{1} is irreducible over 𝔽_{3}(T). According to Proposition 5.1, in both cases, α is defined by α = [−T, α_{2}] and α^{3} = ϵ((T^{2} − 1)α_{2} − T) and we have α = β (hence α and 1/α are both solutions in 𝔽(3) of H_{V}). Moreover, we get ϵ_{A} = (−1)^{3}[2, −1] = −1 and ϵ_{B} = (−1)^{3}[2] = 1. If ϵ = 1, the solution α ∈ 𝔽(3) of H_{1}(X) = 0 is algebraic of degree 4, we have α = −[B_{0}, B_{1}, ⋯, B_{n}, ⋯] and ν(α) = 4 (see the comment after Conjecture 7). If ϵ = −1, the solution α of H_{2}(X) = 0 satisfies α^{2} + T α + 1 = 0 and therefrom we get α = [−T, T, ⋯, −T, T, ⋯]. Indeed, we have α = uϕ(uT) where ϕ(T) = [T, T, T, …, T, …] is the formal golden mean and u^{2} = −1.

Given the prime p, an easy computation shows that there are (p^{2} − 1)/8 pairs (j, k) with 1 ≤ j ≤ k ≤ (p − 1)/2. Hence, for a given odd prime number p, there are (p^{2} − 1)(p − 1)/8 polynomials H_{V} to be considered. We have investigated the reducibility of these polynomials in all the possible cases, for the small primes p. The three conjectures presented below are based on computer calculations, with limitation on the size of the prime p. From this observation, we can make the first and main conjecture.

Conjecture 10. Let p be an odd prime number. Let V = (p, k, j, ϵ) and H_{V} be defined as above. If ϵ ≠ ϵ_{A} and ϵ ≠ ϵ_{B}, then the polynomial H_{V} is irreducible over 𝔽_{p}(T).

The irreducibility of H_{V}, outside the two values for ϵ given in the proposition, may be surprising. Another surprise is that the reducibility may sometimes happen, for a particular value of ϵ, but only if the triple (p, k, j) is well chosen. A last surprise is the following: if the reducibility appears for a certain triple (p, k, j) and a particular choice of ϵ, then this is most of the times if ϵ = ϵ_{A}, and sometimes if ϵ = ϵ_{A} or ϵ = ϵ_{B}, but apparently never if ϵ = ϵ_{B} alone. To illustrate this, we give below three tables where we have collected the data obtained by computer for p = 7, p = 11 and p = 13. Each cell (j, k) with j ≤ k, corresponds to p − 1 tested polynomials. In each cell, we have written Ir. if the polynomials H_{V} are irreducible for all ϵ. If H_{V} is reducible (for ϵ = ϵ_{A} or ϵ = ϵ_{B}), we indicate the degrees of its irreducible factors. For instance, A : 5^{2}.2 means 2 factors of degree 5 and 1 of degree 2, if ϵ = ϵ_{A} and B : 8.4 means 1 factor of degree 8 and 1 of degree 4, if ϵ = ϵ_{B}. Note that, in all cases, the sum of these degrees is equal to p + 1.

Reducibility of H_{V}: p = 7

1  2  3
3 A : 3^{2}.2 A : 3^{2}.2 A : 2^{3}.1^{2}
2 A : 4^{2} A : 3^{2}.1^{2}
B : 4^{2}
1 A : 6.1^{2}

Reducibility of H_{V}: p = 11

1  2  3  4  5
5 A : 5^{2}.2 A : 5^{2}.2 A : 5^{2}.2 A : 5^{2}.2 A : 2^{5}.1^{2}
4 Ir. Ir. Ir. A : 5^{2}.1^{2}

B : 6.3^{2}

3 Ir. Ir. A : 10.1^{2}
B : 8.4
2 Ir. A : 5^{2}.1^{2}

B : 6^{2}

1 A : 10.1^{2}

Reducibility of H_{V}: p = 13

1  2  3  4  5  6
6 A : 6^{2}.2 A : 6.3^{2}.2 A : 4^{2}.2^{3} A : 6.3^{2}.2 A : 6^{2}.2 A : 2^{6}.1^{2}
B : 7^{2}
5  Ir.  Ir.  Ir.  Ir.  A : 12.1^{2}
4 A : 6.4^{2} A : 6.4^{2}  Ir.  A : 3^{4}.1^{2}
B : 7^{2}
3 A : 8.6  Ir.  A : 4^{3}.1^{2}
2  Ir.  A : 6^{2}.1^{2}
B : 7^{2}
1 A : 12.1^{2}

Our second conjecture is the following. Note that the irrationality measure for α, given in both conjectures below, can be directly derived applying Corollary 5, with ℓ = 2k − j.

Conjecture 11. Let k ≥ 1 and m ≥ 2 be integers. Let p be a prime number such that p = km + 1. Let V = (p, k, k, ϵ_{A}) and H_{V} be defined as above. Then the polynomial H_{V} has an irreducible factor P in 𝔽_{p}[T][X] such that P (α) = 0, deg_{X}(P) = m and ν(α) = m. More precisely, we have

P (X) = ∑_{0≤i≤m}(−1)^{ki}\binom{m}{i}T^{(1−(−1)^{i})/2}X^{m−i}.

Finally, we give a last conjecture allowing us to return to the quartic equation which has been the starting point of this investigation.

Conjecture 12. Let p be a prime number such that p ≡ 1 mod 3 and k = (p − 1)/3. Let V = (p, k, k/2, ϵ_{A}) and H_{V} be defined as above. Then the polynomial H_{V} has an irreducible factor P in 𝔽_{p}[T][X] such that P (α) = 0, deg_{X}(P) = 4 and ν(α) = 8/3. More precisely, with j = k/2, we have

P (X) = X^{4} − w_{j+1}T X^{3} − (w_{j+1}/w_{j+2})X^{2} − (w_{j+1}/w_{j+2})^{2}/12.

If V = (13, 4, 2, −1) then we have P = X^{4} +6T X^{3} +2X^{2} +4. By elementary transformations on P and α [5, pp. 1114–1115], we return to the quartic equation X^{4}+X^{2}−T X +1 = 0, having a solution in 𝔽(13), with predictable continued fraction expansion, introduced by Mills and Robbins [12, p. 403].

Lastly, we want to add that the interested reader can willingly obtain the code for the programs supporting these conjectures, by writing to the author.

6 Acknowledgments

I would like to warmly thank my friend Nicolas Brisebarre for his steady and generous support during the preparation of this note. Let me also express my gratitude for the skillful advice on computer programming given by Bill Allombert. I thank the anonymous referee for a very carefull reading of this note. Finally, let me have a kind thought for Professor Michel Mendès France (1936–2018), a great lover of continued fractions, who led me to study this subject.

References

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2010 Mathematics Subject Classification: Primary 11J70; Secondary 11T55. Keywords: continued fraction, field of power series, finite field.

Received March 8 2018; revised version received September 12 2018; November 4 2018. Published in Journal of Integer Sequences, November 24 2018.

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