ON A LINEAR FORM FOR CATALAN’S CONSTANT

C. KRATTENTHALER^{†} AND T. RIVOAL

Dedicated to George Andrews

Abstract. It is shown how Andrews’ multidimensional extension of Watson’s transformation between a very-well-poised _{8}ϕ_{7}-series and a balanced _{4}ϕ_{3}-series can be used to give a straightforward proof of a conjecture of Zudilin and the second author on the arithmetic behaviour of the coefficients of certain linear forms of 1 and Catalan’s constant. This proof is considerably simpler and more stream-lined than the first proof, due to the second author.

1. Introduction

Andrews’ multidimensional extension [1, Theorem 4] of Watson’s transformation between a very-well-poised _{8}ϕ_{7}-series and a balanced _{4}ϕ_{3}-series [6, (2.5.1); Appendix (III.18)] in its full beauty reads

∑_{k=0}^{n} \frac{(a; q)_{k} (q\sqrt{a}; q)_{k} (−q\sqrt{a}; q)_{k} (b_{1}; q)_{k} (c_{1}; q)_{k} ⋯ (b_{m+1}; q)_{k} (c_{m+1}; q)_{k} (q^{−n}; q)_{k}}{(\sqrt{a}; q)_{k} (−\sqrt{a}; q)_{k} (qa/b_{1}; q)_{k} (qa/c_{1}; q)_{k} ⋯ (qa/b_{m+1}; q)_{k} (qa/c_{m+1}; q)_{k} (q^{n+1}a; q)_{k}}

· (\frac{a^{m+1}q^{m+1+n}}{b_{1}c_{1} ⋯ b_{m+1}c_{m+1}} )^{k}

= \frac{(qa; q)_{n} (qa/b_{m+1}c_{m+1}; q)_{n}}{(qa/b_{m+1}; q)_{n} (qa/c_{m+1}; q)_{n}} ∑_{0≤i_{1}≤i_{2}≤···≤i_{m}≤n} \frac{a^{i_{1}+···+i_{m−1}}q^{i_{1}+···+i_{m}}}{(b_{2}c_{2})^{i_{1}} ⋯ (b_{m}c_{m})^{i_{m−1}}}

· \frac{(q^{−n}; q)_{i_{m}}}{(b_{m+1}c_{m+1}/aq^{n}; q)_{i_{m}}} ∏_{k=1}^{m} \frac{(qa/b_{k}c_{k}; q)_{i_{k}−i_{k−1}} (b_{k+1}; q)_{i_{k}} (c_{k+1}; q)_{i_{k}}}{(q; q)_{i_{k}−i_{k−1}} (qa/b_{k}; q)_{i_{k}} (qa/c_{k}; q)_{i_{k}}} , (1.1)

where, by definition, i_{0} := 0. Here, (α; q)_{k} = (1 − α)(1 − αq) ⋯ (1 − αq^{k−1}) if k ≥ 1 and (α; q)_{0} = 1. This formula has found important applications to the theory of partition identities (see [1]).

2 C. KRATTENTHALER^{†} AND T. RIVOAL

Remarkably, Andrews’ formula has started a surprising new life recently. Its utility for proving arithmetic properties of coefficients of certain linear forms for values of the Riemann zeta function at integers was discovered by the authors in [9], and was also exploited in [10] for proving the equality of certain multiple integrals and hypergeometric series. Closely related are the applications given by Zudilin in [16, 17]. The afore-mentioned articles make actually “only” use of the q = 1 special case of (1.1) (see (4.2) below for the explicit statement of that special case). The line of argument developed in [9] has been extended to the q-case by Jouhet and Mosaki in [8] to establish irrationality results for values of a q-analogue of the zeta function. Moreover, Guo, Jouhet and Zeng [7] have extended Zudilin’s work in [16] to the q-case, together with further applications of Andrews’ formula (1.1). In a completely different field, Beliakova, Bühler and Lê [3, 4, 11] have exploited (1.1) in the study of quantum invariants of manifolds. Finally, Andrews himself returned to his identity after over 30 years to prove deep partition theorems in [2].

2000 Mathematics Subject Classification. Primary 11J72; Secondary 11J82, 33C20.

Key words and phrases. Catalan’s constant, linear forms, hypergeometric series, Andrews’ identity.

†Research partially supported by the Austrian Science Foundation FWF, grant S9607-N13, in the framework of the National Research Network “Analytic Combinatorics and Probabilistic Number Theory”.

The purpose of the present paper is to add another item to this list of applications of Andrews’ formula. More precisely, we show how the ideas from [9] lead to an alternative proof of a conjecture from [13] on the arithmetic behaviour of the coefficients in certain linear forms of 1 and Catalan’s constant G = ∑_{k=1}^{∞} \frac{(−1)^{k−1}}{(2k−1)^{2}} . It is considerably simpler and more stream-lined than the first proof [12] by one of the authors, which used a somewhat indirect method based on Padé approximations. A partial, “asymptotic,” proof had been given earlier by Zudilin in [15].

We give a precise statement of the conjecture in the next section, where we also derive explicit expressions for the coefficients a_{n} and b_{n} in the linear forms of 1 and Catalan’s constant. The arithmetic claim for the coefficient a_{n} is then proved in Section 3 with the help of a limit case of Whipple’s transformation between a very-well-poised _{7}F_{6}-series and a balanced _{4}F_{3}-series (the latter being the q = 1 special case of the afore-mentioned transformation formula of Watson). The arithmetic claim for the coefficient b_{n} is proved in Section 4 with the help of the q = 1 special case of Andrews’ formula (1.1), given explicitly in (4.2).

2. A linear form for Catalan’s constant

Let us consider the series

G_{n} = n! ∑_{k=1}^{∞}(−1)^{k}(k + \frac{n − 1}{2}) \frac{(k − n)_{n}(k + n)_{n}}{(k − \frac{1}{2})_{n+1}^{3}} ,  (2.1)

where the Pochhammer symbol (α)_{k} is defined by (α)_{k} = α(α+1) ⋯ (α+k−1) if k ≥ 1 and (α)_{0} = 1. By applying a partial fraction decomposition with respect to k to the summand, and by performing the appropriate summations, it is not difficult to see (cf. [5, Sec. 1.4] for details on this kind of calculation) that

G_{n} = a_{n}G − b_{n},

where

a_{n} = 4(−1)^{n−1} ∑_{j=0}^{n} \frac{∂}{∂ε}( (\frac{n}{2} − j + ε) (\frac{n!}{(1 − ε)_{j} (1 + ε)_{n−j}} )^{3}

· \binom{n + j − ε − \frac{1}{2}}{n}\binom{2n − j + ε − \frac{1}{2}}{n})|_{ε=0}, (2.2)

and

b_{n} = (−1)^{n} ∑_{j=0}^{n} ∑_{e=1}^{3} \frac{1}{(3 − e)!} \frac{∂^{3−e}}{∂j^{3−e}} ( (\frac{n}{2} − j + ε) (\frac{n!}{(1 − ε)_{j} (1 + ε)_{n−j}} )^{3}

· \binom{n + j − ε − \frac{1}{2}}{n}\binom{2n − j + ε − \frac{1}{2}}{n})|_{ε=0} ∑_{k=1}^{j} \frac{(−1)^{k}}{(k − \frac{1}{2})^{e}} . (2.3)

Writing d_{n} for lcm(1, 2, …, n), it is easy to see by a standard approach (see [13, Sec. 5]) that 2^{4n}d_{2n}a_{n} and 2^{4n}d_{2n}^{3}b_{n} are integers. Based on computer calculations, the second author and Zudilin conjectured however (cf. [13, p. 720]) that in fact even 2^{4n}a_{n} and 2^{4n}d_{2n}^{2}b_{n} are integers. While this is still too weak for proving the irrationality of Catalan’s constant G, it is nevertheless an interesting and non-obvious observation which we shall prove in the two subsequent sections. This proof makes use of identities for (generalised) hypergeometric series, the latter being defined by

q+1F_{q}[\begin{matrix}α_{0}, α_{1}, …, α_{q} \\ β_{1}, …, β_{q} \end{matrix}; z] = ∑_{k=0}^{∞} \frac{(α_{0})_{k} (α_{1})_{k} ⋯ (α_{q})_{k}}{k! (β_{1})_{k} ⋯ (β_{q})_{k}} z^{k}.

As we already mentioned in the Introduction, an earlier (but more involved) proof is due to one of the authors [12].

3. The coefficient a_{n}

The purpose of this section is to prove the following theorem.

Theorem 1. For all positive integers n, the number 2^{4n}a_{n} is an integer.

For accomplishing the proof of this theorem (as well as the proof of Theorem 2 in the following section), we need the following two arithmetic auxiliary facts (cf. [14, Sec. 7] and [13, Lemma 6], respectively). Following [14] (where this is attributed to Nesterenko), we shall call the expressions R_{1}(α, β; t) and R_{2}(α, β; t) in the two lemmas below elementary bricks.

Lemma 1. Given integers α and β, let

R_{1}(α, β; t) = \begin{cases}\frac{(t + β)_{α−β}}{(α − β)!} & if α ≥ β, \\ \frac{(β − α − 1)!}{(t + α)_{β−α}} & if α < β. \end{cases}

Then, for all integers α, β, k, H with α ≥ β and H ≥ 0, the number

d_{α−β}^{H} · \frac{1}{H!} \frac{∂^{H}}{∂t^{H}} R_{1}(α, β; t)|_{t=−k}

is an integer. Furthermore, for all integers α, β, k, H with α ≤ k ≤ β − 1 and H ≥ 0, the number

d_{β−α−1}^{H} · \frac{1}{H!} \frac{∂^{H}}{∂t^{H}} R_{1}(α, β; t)(t + k)|_{t=−k}

is an integer.

Lemma 2. Given integers α and β with α ≥ β, let

R_{2}(α, β; t) = 2^{2(α−β)} \frac{(t + β − \frac{1}{2})_{α−β}}{(α − β)!} .

Then, for all integers k and H with H ≥ 0, the number

d_{2(α−β)}^{H} · \frac{1}{H!} \frac{∂^{H}}{∂t^{H}} R_{2}(α, β; t)|_{t=−k}

is an integer.

In order to apply these two lemmas, we need an alternative expression for the coefficient a_{n}, see the lemma below. The expression in (3.1) was already given in [12, Sec. 4.1]. Again, it was obtained there in a somewhat roundabout way. Here, the equality in the next-to-last displayed equation in [12, Sec. 4.2] is explained directly.

Lemma 3. For all non-negative integers n, we have

a_{n} = −4 ∑_{j=0}^{n} \binom{n}{j}\binom{n − \frac{1}{2}}{j}\binom{n + j − \frac{1}{2}}{j}.  (3.1)

Proof. We loosely follow analogous considerations in [9, Lemme 7].

Let H_{m} denote the m-th harmonic number, defined by H_{m} = ∑_{j=1}^{m} \frac{1}{j}. By abuse of notation, we “extend” harmonic numbers to half-integers m by defining H_{m} = ∑_{j=1}^{⌈m⌉} \frac{1}{m−j+1}. For example,

H_{5/2} = \frac{1}{5/2} + \frac{1}{3/2} + \frac{1}{1/2}.

ON A LINEAR FORM FOR CATALAN’S CONSTANT 5

We rewrite the expression for a_{n} given in (2.2) in the form

a_{n} = 4(−1)^{n−1} ∑_{j=0}^{n} (\frac{n}{2} − j) \binom{n}{j}^{3}\binom{n + j − \frac{1}{2}}{n}\binom{2n − j − \frac{1}{2}}{n}

· ( \frac{1}{\frac{n}{2} − j} + 3H_{j} − 3H_{n−j} + H_{2n−j− \frac{1}{2}} − H_{n+j− \frac{1}{2}} − H_{n−j− \frac{1}{2}} + H_{j− \frac{1}{2}} )

= 4(−1)^{n−1} lim_{ε→0} \frac{2}{ε} ∑_{j=0}^{∞} (\frac{n}{2} + \frac{ε}{2} − j) \binom{n}{j}

· \frac{(n − j − ε + 1)_{j}}{(1 − ε)_{j}} \frac{(n − j + ε + 1)_{j}}{(1 − 2ε)_{j}} \frac{(j + \frac{1}{2})_{n}}{(1 − ε)_{n}} \frac{(n − j + ε + \frac{1}{2})_{n}}{(1 + ε)_{n}} .

In hypergeometric notation, this reads

a_{n} = 4(−1)^{n−1} lim_{ε→0} \frac{(n + ε) (\frac{1}{2})_{n} (n + ε + \frac{1}{2})_{n}}{ε(1 − ε)_{n} (1 + ε)_{n}}

×_{6}F_{5}[\begin{matrix}−n − ε, 1 − \frac{n}{2} − \frac{ε}{2}, n + \frac{1}{2}, −n, −n + ε, \frac{1}{2} − n − ε \\ −\frac{n}{2} − \frac{ε}{2}, \frac{1}{2} − 2n − ε, 1 − ε, 1 − 2ε, \frac{1}{2} \end{matrix}; −1] .

To the _{6}F_{5}-series we apply the transformation formula (see [6, (3.10.4), q → 1])

6F_{5}[\begin{matrix}a, 1 + \frac{a}{2}, b, x, y, −N \\ \frac{a}{2}, 1 + a − b, 1 + a − x, 1 + a − y, 1 + a + N \end{matrix}; −1]

= \frac{(1 + a)_{N} (1 + a − x − y)_{N}}{(1 + a − x)_{N} 1 + a − y)_{N}} _{3}F_{2}[\begin{matrix}−N, x, y \\ −a − N + x + y, 1 + a − b \end{matrix}; 1] , (3.2)

where N is a non-negative integer. Thus, we obtain

a_{n} = 4(−1)^{n−1} lim_{ε→0} \frac{(−1)^{n} n!}{(1 − ε)_{n}}_{3}F_{2}[\begin{matrix}−n, n + \frac{1}{2}, \frac{1}{2} − n − ε \\ 1, 1 − 2ε \end{matrix}; 1]

= −4 ∑_{j=0}^{n} \binom{n}{j}\binom{n − \frac{1}{2}}{j}\binom{n + j − \frac{1}{2}}{j},

as we claimed. □

We are now in the position to prove Theorem 1. Proof of Theorem 1. By Lemma 2 with α = j, β = H = 0, and k = −n respectively k = −n − j, the numbers 2^{2n}\binom{n− \frac{1}{2}}{j} and 2^{2n}\binom{n+j− \frac{1}{2}}{j} are integers. Given the expression for a_{n} in Lemma 3, this implies the assertion of the theorem. □

4. The coefficient b_{n}

The purpose of this section is to prove the following theorem. Theorem 2. For all positive integers n, the number 2^{4n}d_{2n}^{2}b_{n} is an integer.

Proof. This proof follows loosely analogous considerations in [9, Prop. 7]. It depends on an arithmetic fact which is stated and proved separately in Lemma 4 below.

Let us start by reordering the summations in (2.3) to obtain

b_{n} = (−1)^{n} ∑_{e=1}^{3} ∑_{k=1}^{n} \frac{(−1)^{k}}{(k − \frac{1}{2})^{e}} \frac{1}{(3 − e)!} \frac{∂^{3−e}}{∂ε^{3−e}} ( ∑_{j=k}^{n} (\frac{n}{2} − j + ε) (\frac{n!}{(1 − ε)_{j} (1 + ε)_{n−j}} )^{3}

· \binom{n + j − ε − \frac{1}{2}}{n}\binom{2n − j + ε − \frac{1}{2}}{n})|_{ε=0}. (4.1)

Also for b_{n}, we need an alternative expression. It is provided for by the q = 1 special case of Andrews’ identity (1.1). More precisely, in (1.1) on replaces a by q^{a}, b_{i} by q^{b_{i}}, c_{i} by q^{c_{i}}, i = 1, 2, …, m + 1, and then lets q tend to 1. As a result, one obtains the transformation formula

2m+5F_{2m+4} [ \begin{matrix}a, \frac{a}{2} + 1, b_{1}, c_{1}, …, b_{m+1}, c_{m+1}, −n \\ \frac{a}{2}, 1 + a − b_{1}, 1 + a − c_{1}, …, 1 + a − b_{m+1}, 1 + a − c_{m+1}, 1 + a + n \end{matrix} ; 1]

= \frac{(1 + a)_{n} (1 + a − b_{m+1} − c_{m+1})_{n}}{(1 + a − b_{m+1})_{n} (1 + a − c_{m+1})_{n}} ∑_{0≤i_{1}≤i_{2}≤···≤i_{m}≤n} \frac{(−n)_{i_{m}}}{(b_{m+1} + c_{m+1} − a − n)_{i_{m}}}

· ( ∏_{k=1}^{m} \frac{(1 + a − b_{k} − c_{k})_{i_{k}−i_{k−1}} (b_{k+1})_{i_{k}} (c_{k+1})_{i_{k}}}{(i_{k} − i_{k−1})! (1 + a − b_{k})_{i_{k}} (1 + a − c_{k})_{i_{k}}} ), (4.2)

where again, by definition, i_{0} := 0. In this formula we put m = 3, a = −n + 2k − 2ε, b_{1} = −n+k −ε, b_{2} = −n+k −ε+ \frac{1}{2}, c_{2} = n+k −ε+ \frac{1}{2}, b_{3} = −n+k −ε, c_{3} = k −2ε−δ +1, b_{4} = −n + k − ε, c_{4} = 1, N = n − k, and then let δ tend to 0. This leads to the identity

∑_{j=k}^{n} (\frac{n}{2} − j + ε) (\frac{n!}{(1 − ε)_{j} (1 + ε)_{n−j}} )^{3} \binom{n + j − \frac{1}{2} − ε}{n}\binom{2n − j − \frac{1}{2} + ε}{n}

= −\frac{1}{2} (k − ε − \frac{1}{2}) ∑_{0≤i_{1}≤i_{2}≤i_{3}≤n−k}(−1)^{i_{2}} \frac{i_{3}!}{i_{1}! (i_{2} − i_{1})! (i_{3} − i_{2})!} \frac{(\frac{1}{2} − ε)_{n}}{(\frac{1}{2} − ε)_{k} (1 + ε)_{n−k}}

· \frac{(n − ε + \frac{1}{2})_{k+i_{1}}}{(1 − ε)_{k+i_{1}}} \frac{(n + i_{1} − i_{2} + ε + \frac{1}{2})_{n−k−i_{1}}}{(1 + ε)_{n−k−i_{1}}} \frac{n!}{(1 − ε)_{k+i_{2}} (1 + ε)_{n−k−i_{2}}}

· \frac{(\frac{1}{2} + n + i_{1} − i_{2})_{i_{2}−i_{1}}}{(\frac{1}{2} + n + ε + i_{1} − i_{2})_{i_{2}−i_{1}}} \frac{(n − \frac{1}{2} − i_{3} + ε)_{i_{3}+1}}{(n − \frac{1}{2} − i_{3} − ε)_{i_{3}+1}}

· \frac{(ε)_{i_{3}−i_{2}} (1 − 2ε)_{k+i_{2}} (\frac{1}{2} + ε)_{n−i_{3}−1}}{(1 − 2ε)_{k−1} (\frac{1}{2} + ε)_{n−k−i_{3}} (1 − ε)_{k+i_{3}}} . (4.3)

Using the notations R_{1}(α, β; t) and R_{2}(α, β; t) for elementary bricks that were introduced in Lemmas 1 and 2, and the notations

R_{3}(n, i_{1}, i_{2}, ε) = \frac{(\frac{1}{2} + n + i_{1} − i_{2})_{i_{2}−i_{1}}}{(\frac{1}{2} + n + ε + i_{1} − i_{2})_{i_{2}−i_{1}}} ,

R_{4}(n, i_{3}, ε) = \frac{(n − \frac{1}{2} − i_{3} + ε)_{i_{3}+1}}{(n − \frac{1}{2} − i_{3} − ε)_{i_{3}+1}} ,

R_{5}(n, k, i_{2}, i_{3}, ε) = 2^{2(k−1)} \frac{(ε)_{i_{3}−i_{2}} (1 − 2ε)_{k+i_{2}} (\frac{1}{2} + ε)_{n−i_{3}−1}}{(1 − 2ε)_{k−1} (\frac{1}{2} + ε)_{n−k−i_{3}} (1 − ε)_{k+i_{3}}}

for the special bricks R_{3}(n, i_{1}, i_{2}, ε), R_{4}(n, i_{3}, ε), and R_{5}(n, k, i_{2}, i_{3}, ε), use of (4.3) in (4.1) yields

2^{4n}d_{2n}^{2}b_{n} = −(−1)^{n}d_{2n}^{2} ∑_{e=1}^{3} ∑_{k=1}^{n} \frac{(−1)^{k}}{(k − \frac{1}{2})^{e}} \frac{1}{(3 − e)!} \frac{∂^{3−e}}{∂ε^{3−e}} ( (2k − 2ε − 1)

× ∑_{0≤i_{1}≤i_{2}≤i_{3}≤n−k}(−1)^{i_{2}} \frac{i_{3}!}{i_{1}! (i_{2} − i_{1})! (i_{3} − i_{2})!} · R_{2}(n, k; 1 − ε) · ε · R_{1}(0, n + 1 − k; ε)

· R_{2}(k + i_{1}, 0; n − ε + 1) · (−ε) · R_{1}(0, k + i_{1}; −ε)
· R_{2}(n − k − i_{1}, 0; n + i_{1} − i_{2} + 1) · ε · R_{1}(0, n − k − i_{1}; ε)

·(−1)^{k+i_{2}}ε·R_{1}(−k −i_{2}, n−k −i_{2} +1; ε)·R_{3}(n, i_{1}, i_{2}, ε)·R_{4}(n, i_{3}, ε)·R_{5}(n, k, i_{2}, i_{3}, ε))|_{ε=0}. (4.4)

We can rewrite this in the form

2^{4n}d_{2n}^{2}b_{n} = −(−1)^{n}d_{2n}^{2} ∑_{e=1}^{3} ∑_{k=1}^{n} \frac{(−1)^{k}}{(k − \frac{1}{2})^{e}} \frac{1}{(3 − e)!} \frac{∂^{3−e}}{∂ε^{3−e}} \{ (2k − 2ε − 1)

· ∑_{0≤i_{1}≤i_{2}≤i_{3}≤n−k} C(i_{1}, i_{2}, i_{3}) · R_{5}(n, k, i_{2}, i_{3}; ε) ∏_{h=1}^{M} t_{h}(n, k, i_{1}, i_{2}, i_{3}; ε)}|_{ε=0},

where each C(i_{1}, i_{2}, i_{3}) is an integer and each t_{h} is an expression R_{1}(α, β; ±ε+K) with α ≥ β, an expression R_{1}(α, β; ±ε) multiplied by ±ε with α < β, an expression R_{2}(α, β; ±ε+K) with α ≥ β, or one of R_{3}(n, i_{1}, i_{2}, ε) and R_{4}(n, i_{3}, ε).

By Leibniz’s formula, this last expression can be expanded into

2^{4n}d_{2n}^{2}b_{n} = −(−1)^{n} ∑_{e=1}^{3} ∑_{k=1}^{n} \frac{2(−1)^{k}d_{2n}^{e−1}}{(k − \frac{1}{2})^{e−1}}

· d_{2n}^{3−e}\{ ∑_{ℓ_{0}+···+ℓ_{M} =3−e} \frac{1}{ℓ_{0}! ℓ_{1}! ⋯ ℓ_{M}!} ∑_{0≤i_{1}≤i_{2}≤i_{3}≤n−k} C(i_{1}, i_{2}, i_{3})

· \frac{∂^{ℓ_{0}}}{∂ε^{ℓ_{0}}} R_{5}(n, k, i_{2}, i_{3}; ε) ∏_{h=1}^{M} \frac{∂^{ℓ_{h}}}{∂ε^{ℓ_{h}}} t_{h}(n, k, i_{1}, i_{2}, i_{3}; ε)\}|_{ε=0}

+ (−1)^{n} ∑_{e=1}^{3} ∑_{k=1}^{n} \frac{2(−1)^{k}d_{2n}^{e}}{(k − \frac{1}{2})^{e}}

· d_{2n}^{2−e}\{ ∑_{ℓ_{0}+···+ℓ_{M} =2−e} \frac{1}{ℓ_{0}! ℓ_{1}! ⋯ ℓ_{M}!} ∑_{0≤i_{1}≤···≤i_{3}≤n−k} C_{2}(i_{1}, i_{2}, i_{3})

· \frac{∂^{ℓ_{0}}}{∂ε^{ℓ_{0}}} R_{5}(n, k, i_{2}, i_{3}; ε) ∏_{h=1}^{M} \frac{∂^{ℓ_{h}}}{∂ε^{ℓ_{h}}} t_{h}(n, k, i_{1}, i_{2}, i_{3}; ε)\}|_{ε=0}. (4.5)

Now, for any h with 1 ≤ h ≤ M, we claim that

\frac{d_{2n}^{ℓ_{h}}}{ℓ_{h}!} \frac{∂^{ℓ_{h}}}{∂ε^{ℓ_{h}}} t_{h}(n, k, i_{1}, i_{2}, i_{3}; ε)|_{ε=0}

is an integer. Indeed, if t_{h}(n, k, i_{1}, i_{2}, i_{3}; ε) is one of the elementary bricks R_{1}(…) (possibly multiplied by ±ε) or R_{2}(…), then this follows directly from Lemmas 1 and 2. If t_{h}(n, k, i_{1}, i_{2}, i_{3}; ε) is one of the special bricks R_{3}(…) or R_{4}(…), this can be seen directly. Since 2(k − \frac{1}{2}) divides d_{2n}, Identity (4.5) would imply the assertion of the theorem once we could prove that

\frac{d_{2n}^{ℓ_{0}}}{ℓ_{0}!} \frac{∂^{ℓ_{0}}}{∂ε^{ℓ_{0}}} R_{5}(n, k, i_{2}, i_{3}; ε)|_{ε=0}  (4.6)

is an integer as well.

To accomplish this, we distinguish between two cases. If i_{2} = i_{3}, then R_{5}(n, k, i_{2}, i_{3}; ε) can be factored as follows:

R_{5}(n, k, i_{2}, i_{3}; ε) = R_{5}(n, k, i_{3}, i_{3}; ε)

= 2^{2(k−1)} \frac{(1 − 2ε)_{k+i_{3}} (\frac{1}{2} + ε)_{n−i_{3}−1}}{(1 − 2ε)_{k−1} (1 − ε)_{k+i_{3}} (\frac{1}{2} + ε)_{n−k−i_{3}}}

= R_{1}(k + i_{3}, 0; 1 − 2ε)

· (−ε) · R_{1}(0, k + i_{3} + 1; −ε) · R_{2}(k − 1, 0; n − k − i_{3} + ε)
· (−2ε) · R_{1}(0, k; −2ε).

Another application of Leibniz’s formula and of Lemmas 1 and 2 show that (4.6) is an integer for i_{2} = i_{3}.

If i_{2} < i_{3}, one observes that

R_{5}(n, k, i_{2}, i_{3}; ε) = ε · R_{6}(n, k, i_{2}, i_{3}; ε),

where R_{6}(…) is the special brick defined in Lemma 4. Consequently, for ℓ_{0} ≥ 1, we have

\frac{1}{ℓ_{0}!} \frac{∂^{ℓ_{0}}}{∂ε^{ℓ_{0}}} R_{5}(n, k, i_{2}, i_{3}; ε)|_{ε=0} = \frac{1}{(ℓ_{0} − 1)!} \frac{∂^{ℓ_{0}−1}}{∂ε^{ℓ_{0}−1}} R_{6}(n, k, i_{2}, i_{3}; ε)|_{ε=0}.

The above relation together with Lemma 4 with m_{1} = i_{3} and m_{2} = i_{2} then shows that (4.6) is also an integer for i_{2} < i_{3}.

This completes the proof of the theorem. □

Lemma 4. Let

R_{6}(n, k, m_{1}, m_{2}; ε) = 2^{2(k−1)} \frac{(1 + ε)_{m_{1}−m_{2}−1} (1 − 2ε)_{k+m_{2}} (\frac{1}{2} + ε)_{n−m_{1}−1}}{(1 − 2ε)_{k−1} (1 − ε)_{k+m_{1}} (\frac{1}{2} + ε)_{n−k−m_{1}}} .

Then, for all integers n, k, m_{1}, m_{2}, H with H ≥ 0 and 0 ≤ m_{2} < m_{1} ≤ n − k, the number

d_{2n}^{H+1} · \frac{1}{H!} \frac{∂^{H}}{∂ε^{H}} R_{6}(n, k, m_{1}, m_{2}; ε)|_{ε=0}  (4.7)

is an integer.

Proof. We loosely follow analogous arguments in the proof of [9, Lemme 11]. In fact, the arguments given in the last paragraph here show that that proof could have been simplified.

We shall show that, for all integers 1 ≤ f_{1} ≤ f_{2} ≤ ⋯ ≤ f_{H+1} ≤ 2n, the number

d_{2n}^{H+1} · \frac{1}{H!} 2^{2(k−1)} \frac{(m_{1} − m_{2} − 1)! (k + m_{2})! (\frac{1}{2})_{n−m_{1}−1}}{(k − 1)! (k + m_{1})! (\frac{1}{2})_{n−k−m_{1}}} \frac{1}{f_{1}f_{2} ⋯ f_{H}}  (4.8)

is an integer. In view of the definition of R_{6}(n, k, m_{1}, m_{2}; ε), this implies that (4.7) is an integer.

We prove the above claim by verifying that the p-adic valuation of (4.8) is non-negative for all prime numbers p. Writing [α] for the greatest integer less than or equal to α, this p-adic valuation is equal to

(H +1)·[log_{p}(2n)]+∑_{ℓ=1}^{∞} ( [\frac{k + m_{2}}{p^{ℓ}} ]+[\frac{m_{1} − m_{2} − 1}{p^{ℓ}}]+[\frac{2n − 2m_{1} − 2}{p^{ℓ}}]−[\frac{n − m_{1} − 1}{p^{ℓ}}]

− [\frac{k − 1}{p^{ℓ}}] − [\frac{k + m_{1}}{p^{ℓ}} ] − [\frac{2n − 2k − 2m_{1}}{p^{ℓ}} ] + [\frac{n − k − m_{1}}{p^{ℓ}} ] ) − ∑_{h=1}^{H} v_{p}(f_{h}) (4.9)

for any prime number p (also for p = 2!). If p > 2n, it is obvious that this expression is non-negative since all terms vanish. Hence, from now on we assume that p ≤ 2n.

In fact, the conditions on k, n, m_{1}, m_{2} imply that the terms of the infinite series in (4.9) vanish for ℓ > [log_{p}(2n)]. The expression (4.9) can therefore be rewritten in the form

[log_{p}(2n)] + ∑_{ℓ=1}^{[log_{p}(2n)]} ( [\frac{k + m_{2}}{p^{ℓ}} ] + [\frac{m_{1} − m_{2} − 1}{p^{ℓ}}] + [\frac{2n − 2m_{1} − 2}{p^{ℓ}}] − [\frac{n − m_{1} − 1}{p^{ℓ}}]

−[\frac{k − 1}{p^{ℓ}}]−[\frac{k + m_{1}}{p^{ℓ}} ]−[\frac{2n − 2k − 2m_{1}}{p^{ℓ}} ]+[\frac{n − k − m_{1}}{p^{ℓ}} ] )−∑_{h=1}^{H} (v_{p}(f_{h}) − [log_{p}(2n)]) . (4.10)

Since, by definition, 1 ≤ f_{h} ≤ 2n for all h, the terms in the sum over h are non-positive. Hence, it suffices to show that the summands in the sum over ℓ are all at least −1.

In order to accomplish this, we write N = {n/p^{ℓ}}, K = {k/p^{ℓ}}, M_{1} = {m_{1}/p^{ℓ}}, M_{2} = {m_{2}/p^{ℓ}} for the fractional parts of n/p^{ℓ}, k/p^{ℓ}, m_{1}/p^{ℓ} and m_{2}/p^{ℓ}, respectively. With these notations, the summand of the sum over ℓ becomes

[K + M_{2}] + [M_{1} − M_{2} − \frac{1}{p^{ℓ}} ] + ([2N − 2M_{1} − \frac{2}{p^{ℓ}} ] − [N − M_{1} − \frac{1}{p^{ℓ}} ])

− [K − \frac{1}{p^{ℓ}} ] − [K + M_{1}] − ( [2N − 2K − 2M_{1}] − [N − K − M_{1}] ). (4.11)

We first discuss the case K = 0. For this special choice of K, the expression in (4.11) reduces to

[M_{1} − M_{2} − \frac{1}{p^{ℓ}} ] + ([2N − 2M_{1} − \frac{2}{p^{ℓ}} ] − [N − M_{1} − \frac{1}{p^{ℓ}} ])

+ 1 − ( [2N − 2M_{1}] − [N − M_{1}] ). (4.12)

Since, by elementary properties of the (weakly) increasing function x ↦ [2x] − [x], we have

([2N − 2M_{1} − \frac{2}{p^{ℓ}} ] − [N − M_{1} − \frac{1}{p^{ℓ}} ]) − ( [2N − 2M_{1}] − [N − M_{1}] ) ≥ −1,

the expression in (4.12) is indeed ≥ −1.

From now on let K > 0, i.e., K ≥ \frac{1}{p^{ℓ}}. In this case, clearly, [K − \frac{1}{p^{ℓ}} ] = 0 and

([2N − 2M_{1} − \frac{2}{p^{ℓ}} ] − [N − M_{1} − \frac{1}{p^{ℓ}} ]) − ( [2N − 2K − 2M_{1}] − [N − K − M_{1}] ) ≥ 0.

Hence, if the expression in (4.11) wants to be ≤ −2, then we must have [K + M_{2}] = 0, [M_{1} − M_{2} − \frac{1}{p^{ℓ}} ] = −1 and [K + M_{1}] = 1, that is

K + M_{2} < 1,  (4.13)
M_{1} − M_{2} − \frac{1}{p^{ℓ}} < 0,  (4.14)
K + M_{1} ≥ 1.  (4.15)

But a combination of (4.13) and (4.15) yields M_{1} − M_{2} > 0, which contradicts (4.14) since the denominators of the rational numbers M_{1} and M_{2} are both p^{ℓ}. □

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C. Krattenthaler, Fakultät für Mathematik, Universität Wien, Nordbergstraße 15,

A-1090 Vienna, Austria. WWW: http://www.mat.univie.ac.at/~kratt.

T. Rivoal, Institut Fourier, CNRS UMR 5582, Université Grenoble 1, 100 rue des Maths, BP 74, 38402 Saint-Martin d’Hères cedex, France.

WWW: http://www-fourier.ujf-grenoble.fr/~rivoal.