Another Proof of Zagier’s Matrix Conjecture

Yawen Ma and Lee-Peng Teo Department of Mathematics Xiamen University Malaysia Jalan Sunsuria, Bandar Sunsuria 43900, Sepang, Selangor Malaysia

                     m1102439194@gmail.com
                       lpteo@xmu.edu.my

Abstract We prove a conjecture of Zagier about the inverse of a (K − 1) × (K − 1) matrix A = A_{K} using elementary methods. This formula allows one to express the product of single zeta values ζ(2r)ζ(2K + 1 − 2r) for 1 ≤ r ≤ K − 1 in terms of the double zeta values ζ(2r, 2K + 1 − 2r) for 1 ≤ r ≤ K − 1 and ζ(2K + 1).

Dedicated to Professor Don Zagier on the Occasion of His 70th Birthday.

1 Introduction

This paper addresses a conjecture of Zagier [5]. For positive integers k_{1}, …, k_{n}, with k_{n} ≥ 2, define the multiple zeta value ζ(k_{1}, k_{2}, …, k_{n}) by

ζ (k_{1}, k_{2}, …, k_{n}) = ∑_{1≤m_{1}<···<m_{n}} \frac{1}{m_{1}^{k_{1}} ⋯ m_{n}^{k_{n}}} .  (1)

Here k = k_{1} + k_{2} + ⋯ + k_{n} is called the weight of this multiple zeta value.

When n = 1, we have the classical Riemann zeta value

ζ(k) = ∑_{m=1}^{∞} \frac{1}{m^{k}} .

When n = 2, the double sum

ζ(k_{1}, k_{2}) = ∑_{m=2}^{∞} \frac{1}{m^{k_{2}}} ∑_{j=1}^{m−1} \frac{1}{j^{k_{1}}}  (2)

was considered by Euler.

Let H(0) = 1 and define

H(n) = ζ(\underbrace{2, 2, …, 2}_{n}) for n ≥ 1.

It is well known that for n ≥ 0,

H(n) = \frac{π^{2n}}{(2n + 1)!}.

When a and b are nonnegative integers, define

H(a, b) = ζ(\underbrace{2, …, 2}_{a}, 3, \underbrace{2, …, 2}_{b}).

Zagier [5] derived the formula

H(a, b) = 2 ∑_{r=1}^{K}(−1)^{r} (\binom{2r}{2a + 2} − (1 − \frac{1}{2^{2r}} ) \binom{2r}{2b + 1}) H(K − r)ζ(2r + 1), (3)

where K = a + b + 1. Here for a real number n and a nonnegative integer k, we define the

generalized binomial coefficient \binom{n}{k} by

\binom{n}{k} = \begin{cases}\frac{n(n − 1) ⋯ (n − k + 1)}{k!}, & if k ≥ 1; \\ 1, & if k = 0. \end{cases}

In particular, if n is an integer and n < k, then \binom{n}{k} = 0.

The formula (3) expresses H(a, b) as a rational linear combination of the H(K −r)ζ(2r + 1). It can be used to prove that the following two sets

ℬ_{1} = {H(a, K − 1 − a) | 0 ≤ a ≤ K − 1} ,
ℬ_{2} = {H(K − r)ζ(2r + 1) | 1 ≤ r ≤ K} ,

span the same vector space over ℚ.

On the other hand, Euler found that all double zeta values of odd weight can be expressed as rational linear combinations of products of Riemann zeta values. The following explicit formula was derived in [5]. For K ≥ 2 and 1 ≤ r ≤ K − 1, we have

ζ(2r, 2K + 1 − 2r) = −\frac{1}{2}ζ(2K + 1) + ∑_{s=1}^{K−1} A_{r,s}ζ(2s)ζ(2K + 1 − 2s),  (4)

where

A_{r,s} = \binom{2K − 2s}{2r − 1} + \binom{2K − 2s}{2K − 2r}.  (5)

Zagier [5] used an elementary argument to show that the (K − 1) × (K − 1) matrix

A = A_{K} = (A_{r,s})

has nonzero determinant, and thus it is invertible. Using the fact that both ζ(2s) and H(2s) are rational multiples of π^{2s}, this shows that the set

ℬ_{3} = {ζ(2r, 2K + 1 − 2r) | 1 ≤ r ≤ K − 1} ∪ {ζ(2K + 1)}

spans the same ℚ-vector space as the set ℬ_{2}.

Zagier [5] formulated three conjectures about the matrix A_{K}. The first one is about its determinant, the second is about its LU-decomposition, and the third is about its inverse. The main objective of this paper is to prove the third conjecture, which states a pair of conjectural formulas for A^{−1}.

Let P and Q be the (K − 1) × (K − 1) matrices with entries

\begin{aligned}P_{s,r} & = \frac{2}{2s − 1} ∑_{n=0}^{2K−2s} \binom{2r − 1}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}, \\ Q_{s,r} & = −\frac{2}{2s − 1} ∑_{n=0}^{2K−2s} \binom{2K − 2r}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}. \end{aligned}  (6)

Here B_{n} are the Bernoulli numbers.

Zagier conjectured that both P and Q are the inverse of A. This implies that P and Q must be the same matrix. This conjecture was proved by Ma [3] using generating functions. In this work, we use a different approach, which is interesting in its own right.

In Section 2, we prove that P = Q. In Section 3, we prove that they indeed give the inverse of A.

2 The equality of the two conjectural formulas

Recall that the Bernoulli numbers B_{n} are defined by the generating function [2, p. 12]:

\frac{t}{e^{t} − 1} = ∑_{n=0}^{∞} \frac{B_{n}}{n!} t^{n}.

One can compute B_{n} recursively by B_{0} = 1, and

B_{n} = −n! ∑_{k=0}^{n−1} \frac{B_{k}}{k!(n − k + 1)!}, for n ≥ 1.

It is an easy consequence of the definitions that B_{1} = −1/2, and for any odd integer n > 1 we have B_{n} = 0.

Let K be an integer greater than or equal to 2, and let P = (P_{s,r}) and Q = (Q_{s,r}) be the (K − 1) × (K − 1) matrices with entries P_{s,r} and Q_{s,r} given by (6). Our goal is to prove that P_{s,r} = Q_{s,r} for all 1 ≤ r, s ≤ K − 1, using a generating function technique that is quite different from that used by Ma [3]. In Zagier’s paper [5], the summations of n in (6) are taken to be until the term n = 2K − 2s + 1. Since s ≤ K − 1, we find that 2K − 2s + 1 is an odd integer greater than 2, and so B_{2K−2s+1} = 0. Thus the summations can be taken to be until n = 2K − 2s only.

We begin with the following lemma.

Lemma 1. Let s be a positive integer. Define the function f_{s}(t) by

f_{s}(t) = \frac{t^{2s−1}}{e^{t} − 1}.

If m is a positive integer, we have the following relation that relates the derivatives of f_{s} up to order m:

\begin{aligned}e^{t}f_{s}^{(m)}(t) & = ∑_{p=0}^{m}(−1)^{m−p}\binom{m}{p}f_{s}^{(p)}(t) \\ & + ∑_{p=0}^{min(m,2s−1)}(−1)^{m−p}\binom{m}{p}\frac{(2s − 1)!}{(2s − 1 − p)!}t^{2s−1−p}. \end{aligned}  (7)

Proof. By the definition of f_{s}(t), we have

f_{s}(t) = e^{−t}f_{s}(t) + t^{2s−1}e^{−t}.

Differentiate both sides m times and apply the Leibnitz rule. This gives

f_{s}^{(m)}(t) = ∑_{p=0}^{m} \binom{m}{p}f_{s}^{(p)}(t)\frac{d^{m−p}}{dt^{m−p}} e^{−t} + ∑_{p=0}^{m} \binom{m}{p}\frac{d^{p}}{dt^{p}} t^{2s−1} \frac{d^{m−p}}{dt^{m−p}} e^{−t}.

Since

\frac{d^{p}}{dt^{p}} t^{2s−1} = 0 if p > 2s − 1,

we find that

f_{s}^{(m)}(t) = ∑_{p=0}^{m}(−1)^{m−p}\binom{m}{p}e^{−t}f_{s}^{(p)}(t) + ∑_{p=0}^{min(m,2s−1)}(−1)^{m−p}\binom{m}{p}e^{−t} \frac{(2s − 1)!}{(2s − 1 − p)!}t^{2s−1−p}.

Multiplying both sides by e^{t} gives (7).

Now we can prove the main theorem in this section. Theorem 2. If K, r, s are integers with K ≥ 2 is an integer and r, s < K, then

\begin{matrix}\frac{(2s − 2)!}{(2K − 2r)!} ∑_{n=0}^{2K−2s} \binom{2K − 2r}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n} \\ = −\frac{(2s − 2)!}{(2K − 2r)!} ∑_{n=0}^{2K−2s} \binom{2r − 1}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}. \end{matrix}  (8)

In particular, this implies that the matrices P and Q defined by (6) are equal. Proof. The left hand side of (8) can be rewritten as

∑_{n=max(0,2r−2s+1)}^{2K−2s} \frac{1}{(2K − 2s − n + 1)!(2s − 2r + n − 1)!} \frac{(n + 2s − 2)!}{n!}B_{n},  (9)

and the right hand side of (8) can be rewritten as

− ∑_{n=max(0,2K−2s−2r+2)}^{2K−2s} \frac{(2r − 1)!}{(2K − 2s − n + 1)!(2r + 2s − 2K + n − 2)!} \frac{(n + 2s − 2)!}{(2K − 2r)!} \frac{B_{n}}{n!} . (10)

The equality (8) is proved by taking m = 2r − 1 in the equation (7) and comparing the coefficients of t^{2K−2r} on both sides. Namely, we want to compare the coefficients of t^{2K−2r} on both sides of the equation

\begin{aligned}e^{t}f_{s}^{(2r−1)}(t) & = − ∑_{p=0}^{2r−1}(−1)^{p}\binom{2r − 1}{p}f_{s}^{(p)}(t) \\ & − ∑_{p=0}^{min(2r−1,2s−1)}(−1)^{p}\binom{2r − 1}{p}\frac{(2s − 1)!}{(2s − 1 − p)!}t^{2s−1−p}. \end{aligned}  (11)

Notice that

f_{s}(t) = t^{2s−2} · \frac{t}{e^{t} − 1} = ∑_{n=0}^{∞} \frac{B_{n}}{n!} t^{n+2s−2}.

Therefore, we get

f_{s}^{(p)}(t) = ∑_{n=max(0,p−2s+2)}^{∞} B_{n} \frac{(n + 2s − 2)!}{(n + 2s − 2 − p)!n!}t^{n+2s−2−p}.

First we consider the coefficient of t^{2K−2r} in the left hand side of (11), namely, the coefficient of t^{2K−2r} in

e^{t}f_{s}^{(2r−1)}(t) = ∑_{k=0}^{∞} \frac{t^{k}}{k!} ∑_{n=max(0,2r−2s+1)}^{∞} B_{n} \frac{(n + 2s − 2)!}{(n + 2s − 2r − 1)!n!}t^{n+2s−2r−1}.  (12)

It is given by the expression (9).

The term on the right hand side of (11) can be written as T_{1} + T_{2}, where

T_{1} = − ∑_{p=0}^{2r−1}(−1)^{p}\binom{2r − 1}{p}f_{s}^{(p)}(t)

= − ∑_{p=0}^{2r−1}(−1)^{p} \frac{(2r − 1)!}{p!(2r − 1 − p)!} ∑_{n=max(0,p−2s+2)}^{∞} B_{n} \frac{(n + 2s − 2)!}{(n + 2s − 2 − p)!n!}t^{n+2s−2−p}

= − ∑_{n=0}^{∞} ∑_{p=0}^{min(2r−1,n+2s−2)}(−1)^{p} \frac{(2r − 1)!}{p!(2r − 1 − p)!}B_{n} \frac{(n + 2s − 2)!}{(n + 2s − 2 − p)!n!}t^{n+2s−2−p},

T_{2} = − ∑_{p=0}^{min(2r−1,2s−1)}(−1)^{p} \frac{(2r − 1)!}{p!(2r − 1 − p)!} \frac{(2s − 1)!}{(2s − 1 − p)!}t^{2s−1−p}.

The term T_{2} contains a term in t^{2K−2r} if and only if 2s − 1 ≥ 2K − 2r, or equivalently, r + s ≥ K + 1. In this case the coefficient of t^{2K−2r} in T_{2} is

\frac{(2r − 1)!}{(2r + 2s − 2K − 1)!(2K − 2s)!} \frac{(2s − 1)!}{(2K − 2r)!}.  (13)

For the term T_{1}, the coefficient of t^{2K−2r} is

\begin{matrix}− ∑_{n=max(0,2K−2s−2r+2)}^{2K−2s}(−1)^{n} \frac{(2r − 1)!}{(2K − 2s − n + 1)!(2r + 2s − 2K + n − 2)!} \frac{(n + 2s − 2)!}{(2K − 2r)!} \frac{B_{n}}{n!} . \\ (14) \end{matrix}

When r + s ≤ K, we find that 2K − 2s − 2r + 2 ≥ 2. Hence, the sum over n in (14) does not contain the n = 1 term. Since B_{n} = 0 when n > 2 is an odd integer, we find that (14) is equal to (10). Since there is no contribution from T_{2} to the term t^{2K−2r} when r + s ≤ K, this proves that when r + s ≤ K, the coefficient of t^{2K−2r} in the right hand side of (11) is (10).

When r + s ≥ K + 1, the n = 1 term is present in (14). Using the fact that B_{1} = −\frac{1}{2}, we find that this term is given by

−\frac{1}{2} \frac{(2r − 1)!}{(2r + 2s − 2K − 1)!(2K − 2s)!} \frac{(2s − 1)!}{(2K − 2r)!},

which is −1/2 of the term (13). Summing the coefficients of t^{2K−2r} from T_{1} and T_{2}, we find that the sum is equal to (10). Therefore, when r + s ≥ K + 1, the coefficient of t^{2K−2r} in the right hand side of (11) is (10).

Thus, we have shown that the coefficient of t^{2K−2r} in the left hand side of (11) is (9), and the coefficient of t^{2K−2r} in the right hand side of (11) is (10). This completes the proof of the theorem.

As we mentioned before, this theorem was proved by Ma [3] using a different method, with the help of Carlitz’s Bernoulli number identity [1, 4]. Our proof uses the generating function of the Bernoulli numbers directly. Remark 3. Carlitz’s identity says that for any nonnegative integers m and n, we have

(−1)^{m} ∑_{k=0}^{m} \binom{m}{k}B_{n+k} = (−1)^{n} ∑_{k=0}^{n} \binom{n}{k}B_{m+k}.  (15)

Prodinger [4] gave a short proof using an exponential generating function of two variables. Here we show that this identity can be derived directly from (7) by setting s = 1. Namely, we consider the generating function of the Bernoulli numbers

f(t) = \frac{t}{e^{t} − 1}.

Since (15) is symmetric in m and n, it is sufficient to consider the case 0 ≤ m < n. In this case, n ≥ 1. Equation (7) says that

e^{t}f^{(m)}(t) = (−1)^{m}t + (−1)^{m−1}m + ∑_{k=0}^{m}(−1)^{m−k}\binom{m}{k}f^{(k)}(t).

This gives

∑_{l=0}^{∞} \frac{t^{l}}{l!} ∑_{k=m}^{∞} \frac{B_{k}}{(k − m)!}t^{k−m} = (−1)^{m}t + (−1)^{m−1}m + ∑_{k=0}^{m}(−1)^{m−k}\binom{m}{k} ∑_{l=k}^{∞} \frac{B_{l}}{(l − k)!}t^{l−k}.

Compare the coefficients of t^{n} on both sides, we have

∑_{k=m}^{m+n} \frac{1}{(k − m)!(m + n − k)!}B_{k} = (−1)^{m}δ_{n,1} + ∑_{k=0}^{m}(−1)^{m−k}\binom{m}{k}\frac{B_{n+k}}{n!} .

If n = 1, the last sum contains the term (−1)^{m}B_{1}, which can be combined with (−1)^{m}δ_{n,1} to yield (−1)^{m−1}B_{1}. Since B_{k} = 0 when k is odd and greater than 2, we find that

∑_{k=m}^{m+n} \frac{n!}{(k − m)!(m + n − k)!}B_{k} = ∑_{k=0}^{m}(−1)^{m−n}\binom{m}{k}B_{n+k}.

Multiplying (−1)^{n} on both sides and shifting the summation variables on the left hand side, we obtain

(−1)^{n} ∑_{k=0}^{n} \binom{n}{k}B_{m+k} = (−1)^{m} ∑_{k=0}^{m} \binom{m}{k}B_{n+k},

which is Carlitz’s identity.

3 The proof of the conjecture

In this section, we prove that the inverse of the matrix A = A_{K} is the matrix P.

Theorem 4 (Zagier’s conjecture). Let K > 1 be an integer, and let A = A_{K} be the (K − 1) × (K − 1) matrix defined by

A_{r,s} = \binom{2K − 2s}{2r − 1} + \binom{2K − 2s}{2K − 2r}.  (16)

The inverse of A is the matrix P defined by one of the following two formulas that are equal.

\begin{aligned}P_{s,r} & = \frac{2}{2s − 1} ∑_{n=0}^{2K−2s} \binom{2r − 1}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}, \\ & = −\frac{2}{2s − 1} ∑_{n=0}^{2K−2s} \binom{2K − 2r}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}. \end{aligned}  (17)

Proof. Define the (K − 1) × (K − 1) matrices B and C by

B_{r,s} = \binom{2K − 2s}{2r − 1}, C_{r,s} = \binom{2K − 2s}{2K − 2r},

so that A = B + C. Notice that B_{r,s} = 0 if r + s > K, and C_{r,s} = 0 if r < s.

Our strategy of the proof is to show that the matrix P A = P B + P C is indeed the identity matrix, by showing that if s and s′ are positive integers less than K, then

(P B)_{s,s′} + (P C)_{s,s′} = \begin{cases}1, & if s = s′; \\ 0, & if s ≠ s′. \end{cases}  (18)

We use the following two elementary identities of binomial coefficients. If n ≥ 1, then

\begin{aligned}∑_{k=0}^{⌊n/2⌋} \binom{n}{2k} & = 2^{n−1}, \\ ∑_{k=0}^{⌊(n−1)/2⌋} \binom{n}{2k + 1} & = 2^{n−1}. \end{aligned}  (19)

First we compute (P B)_{s,s′} using the first formula in (17) for P_{s,r}. The cases where s ≤ s′ and s > s′ are considered separately.

If s ≤ s′, then for r ≤ K −s′, we have r ≤ K −s. Hence, 2K −2s−2r+2 ≥ 2. Therefore,

(P B)_{s,s′} = \frac{2}{2s − 1} ∑_{r=1}^{K−s′} \binom{2K − 2s′}{2r − 1} ∑_{n=2K−2s−2r+2}^{2K−2s} \binom{2r − 1}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}

= \frac{2}{2s − 1} ∑_{n=2s′−2s+2}^{2K−2s} ∑_{r=K−s− \frac{n}{2} +1}^{K−s′} \binom{2K − 2s′}{2r − 1}\binom{2r − 1}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}.

Notice that the summation over n only contains terms with n even, since 2s′ − 2s + 2 ≥ 2. Using the binomial identity

\binom{a}{b}\binom{b}{c} = \binom{a}{a − c}\binom{a − c}{b − c},

we have

\binom{2K − 2s′}{2r − 1}\binom{2r − 1}{2K − 2s − n + 1} = \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{2s − 2s′ + n − 1}{2r + 2s − 2K + n − 2}.

Therefore,

(P B)_{s,s′} = \frac{2}{2s − 1} ∑_{n=2s′−2s+2}^{2K−2s} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{n + 2s − 2}{n}B_{n}

× ∑_{r=K−s− \frac{n}{2} +1}^{K−s′} \binom{2s − 2s′ + n − 1}{2r + 2s − 2K + n − 2}.

For n ≥ 2s′ − 2s + 2, we have 2s − 2s′ + n − 1 ≥ 1. The first formula in (19) implies that

\begin{aligned}∑_{r=K−s− \frac{n}{2} +1}^{K−s′} \binom{2s − 2s′ + n − 1}{2r + 2s − 2K + n − 2} & = ∑_{r=0}^{s−s′+n/2−1} \binom{2s − 2s′ + n − 1}{2r} \\ & = 2^{2s−2s′+n−2}. \end{aligned}  (20)

This shows that when s ≤ s′, we have

(P B)_{s,s′} = \frac{2}{2s − 1} ∑_{n=2s′−2s+2}^{2K−2s} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{n + 2s − 2}{n}2^{2s−2s′+n−2}B_{n}.  (21)

When s > s′, we get

(P B)_{s,s′} = \frac{2}{2s − 1} ∑_{n=0}^{2K−2s} ∑_{K−s− \frac{n}{2} +1≤r≤K−s′} \binom{2K − 2s′}{2r − 1}\binom{2r − 1}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}.

Splitting out the n = 1 term, we have

(P B)_{s,s′} = \frac{2}{2s − 1} ∑_{\begin{matrix}0≤n≤2K−2s \\ n is even \end{matrix}} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{n + 2s − 2}{n}B_{n}

× ∑_{r=K−s− \frac{n}{2} +1}^{K−s′} \binom{2s − 2s′ + n − 1}{2r + 2s − 2K + n − 2}

+ \frac{2}{2s − 1}\binom{2K − 2s′}{2s − 2s′}\binom{2s − 1}{1}B_{1} ∑_{r=K−s+1}^{K−s′} \binom{2s − 2s′}{2r + 2s − 2K − 1}.

Notice that the second formula in (19) gives

∑_{r=K−s+1}^{K−s′} \binom{2s − 2s′}{2r + 2s − 2K − 1} = ∑_{r=0}^{s−s′−1} \binom{2s − 2s′}{2r + 1} = 2^{2s−2s′−1}.

Together with (20), we find that when s > s′, we have

(P B)_{s,s′} = \frac{2}{2s − 1} ∑_{n=0}^{2K−2s} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{n + 2s − 2}{n}2^{2s−2s′+n−2}B_{n}.

Next we compute (P C)_{s,s′}. Using the second expression in (17) for P_{s,r}, we have

\begin{aligned}(P C)_{s,s′} & = −\frac{2}{2s − 1} ∑_{r=s′}^{K−1} \binom{2K − 2s′}{2K − 2r} ∑_{n=max(0,2r−2s+1)}^{2K−2s} \binom{2K − 2r}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}. \\ & (22) \end{aligned}

Using the binomial identity

\binom{a}{b}\binom{b}{c} = \binom{a}{a − c}\binom{a − c}{a − b},

we have

\binom{2K − 2s′}{2K − 2r}\binom{2K − 2r}{2K − 2s − n + 1} = \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{2s − 2s′ + n − 1}{2r − 2s′}.

Now we discuss the cases s < s′, s = s′ and s > s′ separately. When s = s′, we have

(P C)_{s,s} = −\frac{2}{2s − 1} ∑_{r=s}^{K−1} \binom{2K − 2s}{2K − 2r} ∑_{n=2r−2s+1}^{2K−2s} \binom{2K − 2r}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}.

In this case, we have an n = 1 term when r = s. When r > s, the summation over n ≥ 2r − 2s + 1 can be replaced by the summation over n ≥ 2r − 2s + 2. The term with r = s and n = 1 contributes the term 1. Therefore,

(P C)_{s,s} = 1 − \frac{2}{2s − 1} ∑_{r=s}^{K−1} \binom{2K − 2s}{2K − 2r} ∑_{n=2r−2s+2}^{2K−2s} \binom{2K − 2r}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}

= 1 − \frac{2}{2s − 1} ∑_{n=2}^{2K−2s} ∑_{r=s}^{s+n/2−1} \binom{2K − 2s}{n − 1}\binom{n − 1}{2r − 2s}\binom{n + 2s − 2}{n}B_{n}

= 1 − \frac{2}{2s − 1} ∑_{n=2}^{2K−2s} \binom{2K − 2s}{n − 1}\binom{n + 2s − 2}{n}2^{n−2}B_{n}.

The last equality follows from the first equation in (19). Comparing to the s = s′ case in (21) shows that

(P B + P C)_{s,s} = 1.

Next we consider the case s < s′. Since r ≥ s′ in the summation (22), we find that 2r − 2s + 1 ≥ 3. Therefore,

(P C)_{s,s′} = −\frac{2}{2s − 1} ∑_{r=s′}^{K−1} ∑_{n=2r−2s+2}^{2K−2s} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{2s − 2s′ + n − 1}{2r − 2s′}\binom{n + 2s − 2}{n}B_{n}

= −\frac{2}{2s − 1} ∑_{n=2s′−2s+2}^{2K−2s} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{n + 2s − 2}{n}B_{n} ∑_{r=s′}^{s+n/2−1} \binom{2s − 2s′ + n − 1}{2r − 2s′}

= −\frac{2}{2s − 1} ∑_{n=2s′−2s+2}^{2K−2s} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{n + 2s − 2}{n}2^{2s−2s′+n−2}B_{n}

= −(P B)_{s,s′}.

Finally, we consider the case s > s′. In this case

(P C)_{s,s′} = −\frac{2}{2s − 1} ∑_{n=0}^{2K−2s} ∑_{s′≤r≤s+n/2−1} \binom{2K − 2s′}{2K − 2r}\binom{2K − 2r}{2K − 2s − n + 1}\binom{n + 2s − 2}{n}B_{n}.

Splitting out the n = 1 term, we have

(P C)_{s,s′} = −\frac{2}{2s − 1} ∑_{\begin{matrix}0≤n≤2K−2s \\ n is even \end{matrix}} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{n + 2s − 2}{n}B_{n} ∑_{r=s′}^{s+n/2−1} \binom{2s − 2s′ + n − 1}{2r − 2s′}

− \frac{2}{2s − 1}\binom{2K − 2s′}{2s − 2s′}\binom{2s − 1}{1}B_{1} ∑_{r=s′}^{s} \binom{2s − 2s′}{2r − 2s′}

= −\frac{2}{2s − 1} ∑_{n=0}^{2K−2s} \binom{2K − 2s′}{2s − 2s′ + n − 1}\binom{n + 2s − 2}{n}2^{2s−2s′+n−2}B_{n}

= −(P B)_{s,s′}.

This completes the proof of (18), and so the assertion of the theorem is proved.

4 Acknowledgment

This work is supported by the XMUM Research Fund XMUMRF/2018-C2/IMAT/0003. We would like to thank the anonymous referee and the editor for their suggestions and comments.

References

[1] L. Carlitz, Problem 795, Math. Mag. 44 (1971), 107.

[2] G. E. Andrews, R. Askey, and R. Roy, Special Functions, Cambridge University Press, 1999.

[3] D. Ma, Inverse of some matrix related to double zeta values of odd weight, J. Number Theory 166 (2016), 166–180.

[4] H. Prodinger, Carlitz’s symmetric Bernoulli number identity, J. Integer Sequences 17 (2014), Article 14.4.1.

[5] D. Zagier, Evaluation of the multiple zeta values ζ(2, …, 2, 3, 2, …, 2), Ann. of Math. 175 (2012), 977–1000.

2020 Mathematics Subject Classification: Primary 11B65; Secondary 11B68, 11M32. Keywords: double zeta value, Bernoulli number, generating function.

Received March 1 2022; revised versions received March 2 2022; June 28 2022; June 29 2022. Published in Journal of Integer Sequences, June 29 2022.

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