On the Partitions of a Number into Arithmetic Progressions

Augustine O. Munagi and Temba Shonhiwa

The John Knopfmacher Centre for Applicable Analysis and Number Theory

School of Mathematics University of the Witwatersrand Private Bag 3, Wits 2050 South Africa

                  Augustine.Munagi@wits.ac.za
                Temba.Shonhiwa@maths.wits.ac.za

Abstract

The paper investigates the enumeration of the set AP(n) of partitions of a positive integer n in which the nondecreasing sequence of parts form an arithmetic progression. We establish formulas for such partitions, and characterize a class of integers n with the property that the length of every member of AP(n) divides n. We prove that the number of such integers is small.

1 Introduction

A partition of a positive integer n is a nondecreasing sequence of positive integers whose sum is n. The summands are called parts of the partition. We will denote the partition n_{1}, n_{2}, …, n_{k} as the k-tuple (n_{1}, n_{2}, …, n_{k}).

We consider the problem of enumerating the set AP(n) of partitions of n in which the nondecreasing sequence of parts form an arithmetic progression (AP). Our investigation was in part motivated by sequence \underline{A049988} in Sloane’s table [6], that is, the number of arithmetic progressions of positive, integers, nondecreasing with sum n. Cook and Sharpe [3] obtained necessary and sufficient conditions for a positive integer to possess a partition into arithmetic progressions with a prescribed common difference. Nyblom and Evans [5] undertook the enumeration problem and found the following representation for the number p_{d}(n) of partitions of n with common difference d.

p_{d}(n) = \{ \begin{matrix}τ_{1}(n) − 2 − f(n), & if n = d\frac{m(m+1)}{2} for some m > 1; \\ τ_{1}(n) − 1 − f(n), & otherwise, \end{matrix}

where f(n) = |A_{n}| with A_{n} = {c : c|n, c odd, c^{2} < d(2n − c), 2n < dc(c − 1)}, and τ_{1}(n) is the number of odd positive divisors of n. The authors also state a closed formula for p_{2}(n).

In what follows we count partitions directly, and mostly according to the length k of an AP, an approach which reduces our domain to the natural subclass of k-partitions of n. This makes it possible to obtain simpler results with certain new consequences.

Let AP(n, k) = {π | π ∈ AP(n) and π has k parts }, and let ap(n) = |AP(n)|, ap(n, k) = |AP(n, k)| denote the cardinalities of the respective sets.

Note that AP(n, 1) ≠ ∅, n > 0, and AP(n, 2) ≠ ∅, n > 1, since (n) ∈ AP(n, 1) and (1, n − 1) ∈ AP(n, 2) respectively.

The set DP(n) of partitions using a single integer (divisor of n) forms a distinguished subset of AP(n). The cardinality dp(n) = |DP(n)| is given by the number f(n, 2) of ordered factorizations of n into two factors plus 2 (counting the partitions (1, 1, …, 1) and (n)): each factorization n = rs, r, s > 0, gives the s-tuple (r, r, …, r) ∈ DP(n), and n = sr gives the r-tuple (s, s, …, s) ∈ DP(n). Since the first factor runs over the divisors of n, we have

dp(n) = f(n, 2) + 2 = τ(n),  (1)

where τ(n) is the number of positive integral divisors of n.

If p is an odd prime then each π ∈ AP(p) can have at most two distinct parts since the sum of a k-term AP with a positive common difference is composite if k ≥ 3. Since dp(n) = 2 for prime p, observe that

|AP(p) \ DP(p)| = |\{(i, p − i) | 1 ≤ i ≤ \frac{p − 1}{2}\}| = \frac{p − 1}{2}.

Consequently ap(p) = 2 + (p − 1)/2 = (p + 3)/2, from which we obtain the following result on the enumeration of AP(n) for prime n.

Proposition 1. ap(2) = 2, and if p is an odd prime, then ap(p) = \frac{p + 3}{2}.

Proposition 1 has the following extension. Let AP_{t}(n) denote the subset of AP(n) containing partitions with at most t distinct parts and ap_{t}(n) = |AP_{t}(n)|. Then ap_{2}(n) = τ(n)+\frac{n}{2}−1 if n is even, and ap_{2}(n) = τ(n) + \frac{(n−1)}{2} if n is odd. Hence we have the following result.

Proposition 2. If n is a positive integer, then ap_{2}(n) = τ(n) + ⌊\frac{n − 1}{2}⌋ , where ⌊x⌋ is the

greatest integer ≤ x.

The next-step result follows from the summation of the parts of a partition in AP(n).

a + (a + d) + ⋯ + (a + (k − 1)d) = ka + \binom{k}{2}d = n, d ≥ 0,  (2)

for some integers a, d, k, a ≥ 1, d ≥ 0, 1 ≤ k ≤ n.

Then k = 3 implies 3a + 3d = n. Hence ap(n, 3) > 0 if and only if n is a multiple of 3. If 3|n, the solution set of a + d = n/3 is clearly {(a, d) = (r, (n − 3r)/3) | 1 ≤ r ≤ n/3}. Thus ap_{3}(n) = ap_{2}(n) + n/3.

Proposition 3. If n is a positive integer such that 3|n, then ap_{3}(n) = τ(n) + ⌊\frac{(5n − 3)}{6}⌋ .

The formula for ap_{t}(n), t ≥ 4, is uneconomical, (2, 5, 8, 11) ∈ AP(26) but 4 ∤ 26. Let Div(n) denote the set of divisors of n. Since k ∈ Div(n) implies ap(n, k) > 0, we define the set APDiv(n) = {k | ap(n, k) > 0}. Then Div(n) ⊆ APDiv(n) in general.

In Section 2, we derive the general formula for ap(n, k). This will make it possible, in Section 3, to obtain more inclusive formulas in the spirit of those stated above, bearing in mind Div(n) ⊆ APDiv(n). In particular, we characterize the class of numbers n for which Div(n) = APDiv(n), and show that such numbers are few.

2 The formula for ap(n, k)

We present the main theorem of this section.

Theorem 4. 1. Let n be a positive integer and k > 0 an even number such that ap(n, k) >

0. Then

ap(n, k) = ⌊\frac{n + k(k − 2)}{k(k − 1)}⌋ , if k|n and  (3)

ap(n, k) = ⌊\frac{2n + k(k − 3)}{2k(k − 1)}⌋ , if k ∤ n.  (4)

2. Let n be a positve integer and k an odd number such that ap(n, k) > 0, k > 1. Then

ap(n, k) = ⌊\frac{2n + k(k − 3)}{k(k − 1)}⌋ .  (5)

Proof. The proof makes use of the following result [2]. The linear Diophantine equation ax + by = c has a solution if and only if g|c, where g = gcd(a, b). If (x_{0}, y_{0}) is any particular solution of this equation, then all other solutions are given by,

x = x_{0} + (\frac{b}{g} ) t, y = y_{0} − (\frac{a}{g}) t

where t is an arbitrary integer.

In view of equation (2), the enumerative function ap(n, k) is given by

ap(n, k) = ∑_{{ka+ℓd=n \atop a≥1, d≥0}} 1,

where ℓ = k \frac{(k − 1)}{2}. However, ka + ℓd = n has a solution if and only if gcd(k, ℓ)|n. For the case k odd, it follows that gcd(k, ℓ) = k leading to a solution if and only if k|n, that is, n is a multiple of k. In which case one particular solution is d_{0} = 0 and a_{0} = \frac{n}{k}, an integer. And hence, the other solutions are given by

a = \frac{n}{k} + \frac{ℓ}{gcd(k, ℓ)}t ≥ 1 and d = −\frac{k}{gcd(k, ℓ)}t ≥ 0,

where t is an integer. Thus,

ap(n, k) = ∑_{{a= \frac{n}{k} + \frac{ℓ}{gcd(k,ℓ)} t≥1 \atop d=− \frac{k}{gcd(k,ℓ)} t≥0}} 1

= ∑_{(1− \frac{n}{k}) \frac{gcd(k,ℓ)}{ℓ} ≤t≤0} 1 = ⌊\frac{2(n − k)}{k(k − 1)}⌋ + 1,

a result equivalent to formula (5).

Next we consider the case when k is even and note that

gcd(k, ℓ) = gcd (2\frac{k}{2}, \frac{k}{2}(k − 1)) = \frac{k}{2},

that is, ka + ℓd = n has a solution if and only if \frac{k}{2} | n, which implies n = s\frac{k}{2}, for some positive integer s. Clearly, for s even, the previous argument goes through save for a minor adjustment, leading to the result,

ap(n, k) = ⌊\frac{(n − k)}{k(k − 1)}⌋ + 1,

where n is a multiple of k, a result equivalent to formula (3).

On the other hand, if s is odd, then one valid solution is

d_{0} = 1 which implies a_{0} = \frac{(s\frac{k}{2} − ℓ)}{k} = \frac{\frac{k}{2}(s − (k − 1))}{k}, which is an integer when s is odd .

The condition a ≥ 1 implies the condition n ≥ \frac{k}{2}(k + 1). Therefore,

ap(n, k) = ∑_{{a= \frac{n−ℓ}{k} + \frac{ℓ}{gcd(k,ℓ)} t≥1 \atop d=1− \frac{k}{gcd(k,ℓ)} t≥0}} 1

= ∑_{\frac{gcd(k,ℓ)}{ℓ}(1− \frac{(n−ℓ)}{k})≤t≤ \frac{gcd(k,ℓ)}{k}} 1.

Since gcd(k, ℓ) = k/2, this gives t ≤ \frac{1}{2} and hence,

ap(n, k) = ⌊\frac{(2n − k(k + 1))}{2k(k − 1)}⌋ + 1,

where n = s\frac{k}{2} ≥ \frac{k}{2}(k + 1), a result equivalent to formula (4). This exhausts all possible outcomes.

3 The formula for ap(n)

Using the results of Section 2, we can write the sum of ap(n, k) over all divisors k of n, denoted divap(n). Thus, Theorem 5.

divap(n) = 1 + ∑_{k|n, k>1 is even} ⌊\frac{n + k(k − 2)}{k(k − 1)}⌋ + ∑_{k|n, k>1 is odd} ⌊\frac{2n + k(k − 3)}{k(k − 1)}⌋ .

Alternatively,

divap(n) = τ(n) + n − σ(n) + ∑_{{k|n \atop k>1, k even}} ⌊\frac{n − 1}{k − 1}⌋ + ∑_{{k|n \atop k>1, k odd}} ⌊\frac{n + k(n − 2)}{k(k − 1)}⌋ .

Proof.

divap(n) = ∑_{k|n} AP(n, k) = ∑_{k|n, k odd} AP(n, k) + ∑_{k|n, k even} AP(n, k)
= 1 + ∑_{{k|n \atop k>1, k odd}} (⌊\frac{2(n − k)}{k(k − 1)}⌋ + 1) + ∑_{{k|n \atop k>1, k even}} (⌊\frac{n − k}{k(k − 1)}⌋ + 1)
= 1 + ∑_{{k|n \atop k>1, k odd}} 1 + ∑_{{k|n \atop k>1, k even}} 1 + ∑_{{k|n \atop k>1, k odd}} ⌊\frac{2(n − k)}{k(k − 1)}⌋ + ∑_{{k|n \atop k>1, k even}} ⌊\frac{n − k}{k(k − 1)}⌋

= τ(n) − ( ∑_{{k|n \atop k<n, k odd}} \frac{n}{k} + ∑_{{k|n \atop k<n, k even}} \frac{n}{k})

+ ∑_{{k|n \atop k>1, k odd}} ⌊\frac{2(n − 1)}{k − 1} − \frac{n}{k}⌋ + ∑_{{k|n \atop k>1, k even}} ⌊\frac{n − 1}{k − 1}⌋

= τ(n) − (σ(n) − n) + ∑_{{k|n \atop k>1, k even}} ⌊\frac{n − 1}{k − 1}⌋ + ∑_{{k|n \atop k>1, k odd}} ⌊\frac{n + k(n − 2)}{k(k − 1)}⌋ .

Note that since n = \frac{k}{2}(2a + (k − 1)d) by (2), it follows that if k is odd and ap(n, k) > 0, then k|n, and if k ∤ n and ap(n, k) > 0, then k is even and n ≡ 0 (mod \frac{k}{2}).

Hence the set-difference APDiv(n) \ Div(n) is given by

Ek(n) = APDiv(n) \ Div(n) = \{k = 2v | k ∤ n, v|n, n ≥ \binom{k + 1}{2}\} .

That is,

Ek(n) = {2v ∈ 2^{α+1}Div(m) | n ≥ v(2v + 1)},  (6)

where m is the unique odd number satisfying n = 2^{α}m, α ≥ 0, and rS = {rs | s ∈ S}. The importance of Ek(n) lies in the following statement.

ap(n) = divap(n) if and only if Ek(n) = ∅.  (7)

In the case Ek(n) ≠ ∅ we have

ap(n) = divap(n) + extap(n), where, extap(n) = ∑_{k∈Ek(n)} ⌊\frac{2n + k(k − 3)}{2k(k − 1)}⌋ .  (8)

Since 1 ∈ Div(m), we obtain

Ek(n) = ∅ if and only if m < 2^{α+1} + 1.  (9)

In particular Ek(2^{α}) = ∅, α ≥ 0. Hence the next result follows from Theorem 5 and (7).

Proposition 6. If α is a nonnegative integer, then

ap(2^{α}) = 1 + ∑_{j=1}^{α} ⌊\frac{2^{α−j} + 2^{j} − 2}{2^{j} − 1}⌋ = 1 + α + ∑_{j=1}^{α} ⌊\frac{2^{α−j} − 1}{2^{j} − 1}⌋ .

Proposition 6 is a special case of the next result.

Theorem 7. The following assertions are equivalent for any even integer n.

(i) ap(n) = divap(n).

(ii) n can be expressed in the form n = 2^{j}(r + 2^{j−1}), r = 0, 1, …, 3 · 2^{j−1}, where j is a positive integer.

Proof. Let n = 2^{j}(r + 2^{j−1}) = 2^{α}m, α ≥ j, where m is odd. m = 2^{j−α}(r +2^{j−1}) ≤ 2^{j−α}(3·2^{j−1} +2^{j−1}) = 2^{2j−α+1} < 2^{α+1} +1. So Ek(n) = ∅ by (9). Hence (ii) ⇒ (i).

Conversely, notice that 3 · 2^{j−1} < r = 3 · 2^{j−1} + 1 gives n = 2^{j}(2^{j+1} + 1) ≡ 2^{j}m, which implies Ek(n) ≠ ∅ or ap(n) > apdiv(n), a contradiction of (i).

Remarks

(i) The special even numbers of Theorem 7 form an AP from 2^{2j−1} to 2^{2j+1} for each j (with common difference 2^{j}), say R(j). For example, R(1) = (2, 4, 6, 8) and R(2) = (8, 12, 16, 20, 24, 28, 32).

(ii) Theorem 7 implies Proposition 6 since 2^{2j−1}, 2^{2j} ∈ R(j), when r = 0, 2^{j−1}.

(iii) For fixed j, there are integers n = 2^{j}(r +2^{j−1}) with r ∉ {0, 1, …, 3·2^{j−1}} which satisfy Theorem 7(i). However, this is not a violation of the theorem since n ∈ R(j) for some (legal) j and r. For example if j = 1, then n = 12 corresponds to r = 5 > 3. So 12 ∉ R(1) even though ap(12) = apdiv(12). But note that 12 ∈ R(2). This phenomenon is explained by removing the restriction on r and observing that i < j ⇒ R_{∞}(i) ⊃ R_{∞}(j), where R_{∞}(j) = {2^{j}(r + 2^{j−1}) | r ≥ 0}.

But if n is odd, (9) reduces to n < 3, owing to the fact that ap(n, 2) > 0 for each n > 1, including prime n. So, for odd numbers, we can skip the 1 ∈ Div(m) and use the least prime factor of n, to obtain the adjusted version of (9).

Let n > 1 be an odd positive integer, and let p denote the least prime divisor of n.

Ek(n) = \{2\} if and only if \frac{n}{p} < 2p + 1.  (10)

The next theorem characterizes odd numbers n satisfying (10), i.e., ap(n) = divap(n) + ap(n, 2).

Theorem 8. The following assertions are equivalent for any odd integer n > 1.

(i) ap(n) = divap(n) + \frac{(n−1)}{2}.

(ii) n is prime, or n = p_{1}p_{2}, where p_{1}, p_{2} are primes such that p_{2} < 2p_{1} + 1.

Proof. Clearly (i) is true if n is prime, since there are \frac{(n−1)}{2} partitions of n into two parts (see Proposition 1). The proof follows from (10) and the observation that n = p_{1}p_{2} ⋯ p_{r}, r > 2, implies \frac{n}{p_{1}} ≥ 2p_{1} + 1, which implies that Ek(n) properly contains \{2\}.

Corollary 9. If p is an odd prime, then

ap(p^{2}) = \frac{p^{2} + 9}{2}.

Corollary 9 is also a corollary of the next theorem.

Theorem 10. If p is an odd prime and α is a positive integer, then

ap(p^{α}) = 1 + ∑_{i=1}^{α} ⌊\frac{2p^{α−i} + p^{i} − 3}{p^{i} − 1}⌋ + \begin{matrix}⌊ & \frac{α−1}{2} & ⌋ \\ ∑ \\ i=0 \end{matrix} ⌊\frac{p^{α−i} + 2p^{i} − 3}{2(2p^{i} − 1)}⌋ .

Proof. Div(p^{α}) = {1, p, …, p^{α}}. So for each i, 0 ≤ i ≤ α, we have p^{α−i} ≥ 2p^{i} + 1 if and only if α > 2i if and only if 0 ≤ i ≤ ⌊(α − 1)/2⌋. Thus Ek(p^{α}) = {2p^{i} | 0 ≤ i ≤ ⌊(α − 1)/2⌋ }. Substituting for p^{i} in (7) and (8), and simplifying the following summations gives the theorem.

ap(p^{α}) = 1 + ∑_{i=1}^{α} divap(p^{α}, p^{i}) + \begin{matrix}⌊ & \frac{α−1}{2} & ⌋ \\ ∑ \\ i=0 \end{matrix} extap(p^{α}, 2p^{i}).

Given an odd prime p let α and c be nonnegative integers, 0 ≤ c ≤ α. We claim that

⌊\frac{2p^{α−c} + p^{c} − 3}{p^{c} − 1}⌋ = 1 + \frac{2p^{r}(p^{(q−1)c} − 1)}{p^{c} − 1}, α = qc + r, 0 ≤ r < c.  (11)

Denote the left side of (11) by u(α, c), and write h(α, c) = \frac{(2p^{α−c}−2p^{c})}{(p^{c}−1)}, so that ⌊h(α, c) + 3⌋ =

u(α, c). Then clearly, u(2c, c) = 1 and α < 2c implies α − c < c which implies ⌊\frac{h(α,c)}{2}⌋ = −1

which implies u(α, c) = 1. But if 2c < α < 3c, we have ⌊\frac{h(α,c)}{2}⌋ = p^{α−2c}+⌊\frac{h(α,2c)}{2}⌋ which gives

⌊\frac{h(α,c)}{2}⌋ = p^{α−2c} − 1. Iterating the procedure we obtain an expression of the form ⌊\frac{h(α,c)}{2}⌋ =

p^{α−2c} + p^{α−3c} + ⋯ + p^{α−qc} + ⌊\frac{h(α,qc)}{2}⌋, where q = ⌊\frac{α}{c}⌋, giving ⌊\frac{h(α,qc)}{2}⌋ = −1. If we compute

\frac{α}{c} and reverse the order of summation, we obtain ⌊\frac{h(α,c)}{2}⌋ = p^{r} + p^{c+r} + ⋯ + p^{(q−2)c+r} − 1, where α = qc + r. Summing the finite geometric series gives the result.

Hence we obtain from Theorem 10,

divap(p^{α}) = 1 + α + ∑_{{c=1 \atop α=qc+r, 0≤r<c}}^{α} \frac{2p^{r}(p^{(q−1)c} − 1)}{p^{c} − 1}.  (12)

This results in a sequence of polynomials in p over the positive integers for apdiv(p^{α}), α = 0, 1, 2, …, and consequently yielding simplified forms of ap(p^{α}). The degree of apdiv(p^{α}) is clearly max(α − 2c | c > 0) = α − 2, α > 1.

The polynomials apdiv(p^{α}) are given below for α = 2, 3, …, 9.

divap(p^{2}) = 5
divap(p^{3}) = 2p + 6
divap(p^{4}) = 2p^{2} + 2p + 9
divap(p^{5}) = 2p^{3} + 2p^{2} + 4p + 8
divap(p^{6}) = 2p^{4} + 2p^{3} + 4p^{2} + 2p + 13
divap(p^{7}) = 2p^{5} + 2p^{4} + 4p^{3} + 2p^{2} + 6p + 10
divap(p^{8}) = 2p^{6} + 2p^{5} + 4p^{4} + 2p^{3} + 6p^{2} + 2p + 15
divap(p^{9}) = 2p^{7} + 2p^{6} + 4p^{5} + 2p^{4} + 6p^{3} + 2p^{2} + 6p + 14.

On the other hand, given an odd prime p, the set of numbers n = pp_{2}, p_{2} ≥ p (p_{2} a prime) which satisfy Theorem 8 is nicely bounded: p^{2} ≤ n < 2p^{2}. So let S(p) denote the set of all numbers n between p^{2} and 2p^{2} inclusive which satisfy Theorem 8. We deduce that

|S(p)| = |{p_{2} prime | p ≤ p_{2} < 2p}| + |{p_{2} prime | p^{2} ≤ p_{2} < 2p^{2}}|.  (13)

In terms of the sequence a(n) = number of primes between n and 2n inclusive [4, A035250], a concise expression is |S(p)| = a(p) + a(p^{2}).

We examine the size of the set of numbers which satisfy the “closure” relation ap(n) = divap(n). By Theorem 7 all such numbers (> 1) are even. For a fixed positive integer j define AE(j) = {2c | 2^{2j−2} ≤ c ≤ 2^{2j}}. Recalling the sets R(j) defined in the remarks following Theorem 7, we have

|R(j)| = 3 · 2^{j−1} + 1, and |AE(j) − R(j)| = 3 · 2^{j−1}(2^{j−1} − 1).

Note that AE(j) \ R(j) is the set of even numbers n within the range of elements of R(j) which satisfy ap(n) > divap(n). It follows that, for sufficiently large j,

\frac{|R(j)|}{|AE(j)|} ≤ \frac{1}{2^{j−1}} ⟶ 0, and \frac{|AE(j) \ R(j)|}{|AE(j)|} ≤ \frac{2^{j−1} − 1}{2^{j−1}} ⟶ 1.

We conclude that practically all even numbers satisfy the strict inequality ap(n) > divap(n). Thus more readily so for all positive integers.

4 Conclusion

We close with some remarks on the set Ek(n). It follows from (6) that if k ∈ Ek(n), then

n = M(\frac{k}{2}) for some odd integer M. Writing n = 2^{α}m as previously, and k = 2^{α+1}m_{i}

(m_{i} odd), we have n = M \frac{k}{2} = M(2^{α}m_{i}) = Mm_{i} · 2^{α}, where Mm_{i} = m, m_{i} ≥ 1. Thus each element of Ek(n) corresponds to a decomposition of m into two factors. This gives

|Ek(n)| ≤ \frac{τ(m)}{2}.

That is,

|Ek(n)| ≤ ⌊\frac{τ(m)}{2}⌋ , where m = \frac{n}{2^{α}} is odd, α ≥ 0.

The case of prime powers (see Theorem 10) shows that this upper bound is sharp. The determination of the exact value of |Ek(n)| remains an open problem.

Acknowledgment

We wish to thank Jeffrey Shallit for bringing the Nyblom-Evans paper to our attention.

References

[1] G. E. Andrews,The Theory of Partitions, Encyclopedia of Mathematics and its Applications, Vol. 2, Addison-Wesley, 1976.

[2] T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York Inc., 1976.

[3] R. Cook and D. Sharp, Sums of arithmetic progressions, Fib. Quart. 33 (1995), 218–221.

[4] R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Second Edition, Addison-Wesley, 1994.

[5] M. A. Nyblom and C. Evans, On the enumeration of partitions with summands in arithmetic progression, Australas. J. Combin. 28 (2003), 149–159.

[6] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, published electronically

    at http://www.research.att.com/∼njas/sequences/.

2000 Mathematics Subject Classification: Primary 11P81; Secondary 05A15, 05A17.

Keywords: partition, arithmetic progression, divisor function, Diophantine equation.

(Concerned with sequences \underline{A004119}, \underline{A035250}, and \underline{A049988}.)

Received October 8 2007; revised version received October 10 2007; December 5 2008. Published in Journal of Integer Sequences, December 13 2008.

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