Multi-Poly-Bernoulli Numbers and Finite Multiple Zeta Values

Kohtaro Imatomi, Masanobu Kaneko,^{1} and Erika Takeda Graduate School of Mathematics Kyushu University Motooka 744, Nishi-ku Fukuoka 819-0395 Japan

             k-imatomi@math.kyushu-u.ac.jp
              mkaneko@math.kyushu-u.ac.jp
                eririntake@yahoo.co.jp

Abstract

We define the multi-poly-Bernoulli numbers slightly differently from the similar numbers given in earlier papers by Bayad, Hamahata, and Masubuchi, and study their basic properties. Our motivation for the new definition is the connection to “finite multiple zeta values”, which have been studied by Hoffman and Zhao, among others, and are recast in a recent work by Zagier and the second author. We write the finite multiple zeta value in terms of our new multi-poly-Bernoulli numbers.

1The second author is supported by the Japan Society for the Promotion of Science, Grant-in-Aid for Scientific Research (B) 23340010.

1 Introduction

For any index set (k_{1}, …, k_{r}) with k_{i} ∈ ℤ and r ≥ 1, we define two kinds of multi-poly-Bernoulli numbers B_{n}^{(k_{1},...,k_{r})} and C_{n}^{(k_{1},...,k_{r})} by the following generating series:

\frac{Li_{k_{1},...,k_{r}}(1 − e^{−t})}{1 − e^{−t}} = ∑_{n=0}^{∞} B_{n}^{(k_{1},...,k_{r})} \frac{t^{n}}{n!},  (1)

\frac{Li_{k_{1},...,k_{r}}(1 − e^{−t})}{e^{t} − 1} = ∑_{n=0}^{∞} C_{n}^{(k_{1},...,k_{r})} \frac{t^{n}}{n!},  (2)

where Li_{k_{1},...,k_{r}}(z) is the multiple polylogarithm series given by

Li_{k_{1},...,k_{r}}(z) = ∑_{m_{1}>···>m_{r}>0} \frac{z^{m_{1}}}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} .  (3)

When r = 1, the numbers B_{n}^{(k)} and C_{n}^{(k)} are “poly-Bernoulli numbers” defined and studied in our previous work [7, 1]. When r = 1 and k_{1} = 1, the multiple polylogarithm is just the usual logarithm − log(1−z), and thus both B_{n}^{(1)} and C_{n}^{(1)} are the classical Bernoulli numbers (with B_{1}^{(1)} = 1/2 and C_{1}^{(1)} = −1/2).

The finite multiple zeta value is the collection of truncated sums

∑_{p>m_{1}>···>m_{r}>0} \frac{1}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}}

modulo p for all primes p, considered in the quotient ring ∏_{p} ℤ/pℤ/ ⨁_{p} ℤ/pℤ. (For more details on finite multiple zeta values, we refer the reader to papers by Hoffman [6], Zhao [10], and Kaneko-Zagier [8], among others.) In §4 we show the congruence

∑_{p>m_{1}>···>m_{r}>0} \frac{1}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} ≡ −C_{p−2}^{(k_{1}−1,k_{2},...,k_{r})} (mod p),

which reveals that each component of a finite multiple zeta value is a multi-poly-Bernoulli number.

2 Recurrences and explicit formulas for the multi-poly-Bernoulli numbers

In this section, we first prove a simple relation between B_{n}^{(k_{1},...,k_{r})} and C_{n}^{(k_{1},...,k_{r})} and proceed to give recursion relations and explicit formulas for the multi-poly-Bernoulli numbers. Since the two generating series (1) and (2) differ only by the factor e^{t}, two obvious relations between B_{n}^{(k_{1},...,k_{r})} and C_{n}^{(k_{1},...,k_{r})} are

B_{n}^{(k_{1},...,k_{r})} = ∑_{i=0}^{n} \binom{n}{i}C_{i}^{(k_{1},...,k_{r})} and C_{n}^{(k_{1},...,k_{r})} = ∑_{i=0}^{n}(−1)^{n−i}\binom{n}{i}B_{i}^{(k_{1},...,k_{r})}.

A simpler relation expressing B_{n}^{(k_{1},...,k_{r})} in terms of C_{n}^{(k_{1},...,k_{r})} is the following. Proposition 1. For any r ≥ 1, k_{i} ∈ ℤ and n ≥ 1, we have

B_{n}^{(k_{1},...,k_{r})} = C_{n}^{(k_{1},...,k_{r})} + C_{n−1}^{(k_{1}−1,k_{2},··· ,k_{r})}.

Proof. From the obvious identity

\frac{d}{dz} Li_{k_{1},...,k_{r}}(z) = \frac{1}{z} Li_{k_{1}−1,k_{2},...,k_{r}}(z)

(note that we allow the first index to be 0 or negative), we have

\frac{d}{dt} Li_{k_{1},...,k_{r}}(1 − e^{−t}) = \frac{Li_{k_{1}−1,k_{2},...,k_{r}}(1 − e^{−t})}{1 − e^{−t}} · e^{−t} = \frac{Li_{k_{1}−1,k_{2},...,k_{r}}(1 − e^{−t})}{e^{t} − 1},

and hence

∑_{n=1}^{∞} C_{n−1}^{(k_{1}−1,k_{2},...,k_{r})} \frac{t^{n}}{n!} = ∫_{0}^{t} ∑_{n=0}^{∞} C_{n}^{(k_{1}−1,k_{2},...,k_{r})} \frac{t^{n}}{n!} dt = ∫_{0}^{t} \frac{Li_{k_{1}−1,k_{2},...,k_{r}}(1 − e^{−t})}{e^{t} − 1} dt
= Li_{k_{1},...,k_{r}}(1 − e^{−t}) = Li_{k_{1},...,k_{r}}(1 − e^{−t}) (\frac{e^{t}}{e^{t} − 1} − \frac{1}{e^{t} − 1})
= \frac{Li_{k_{1},...,k_{r}}(1 − e^{−t})}{1 − e^{−t}} − \frac{Li_{k_{1},...,k_{r}}(1 − e^{−t})}{e^{t} − 1}
= ∑_{n=0}^{∞} B_{n}^{(k_{1},...,k_{r})} \frac{t^{n}}{n!} − ∑_{n=0}^{∞} C_{n}^{(k_{1},...,k_{r})} \frac{t^{n}}{n!}.

This proves the proposition. Proposition 2. For any k_{1}, …, k_{r} ∈ ℤ and n ≥ 0, we have

B_{n}^{(k_{1},...,k_{r})} = \frac{1}{n + 1} (B_{n}^{(k_{1}−1,k_{2},...k_{r})} − ∑_{m=1}^{n−1} \binom{n}{m − 1}B_{m}^{(k_{1},...,k_{r})})

and

C_{n}^{(k_{1},...,k_{r})} = \frac{(−1)^{n}}{n + 1} (∑_{m=0}^{n}(−1)^{m}\binom{n}{m}C_{m}^{(k_{1}−1,k_{2},...,k_{r})} − ∑_{m=1}^{n−1}(−1)^{m}\binom{n}{m − 1}C_{m}^{(k_{1},...,k_{r})}) ,

where an empty sum is understood to be 0.

Proof. For B_{n}^{(k_{1},...,k_{r})}, we multiply the defining equation (1) by 1 − e^{−t} and differentiate with respect to t to obtain

\frac{e^{−t}}{1 − e^{−t}} Li_{k_{1}−1,k_{2},...,k_{r}}(1 − e^{−t}) = e^{−t} ∑_{n=0}^{∞} B_{n}^{(k_{1},...,k_{r})} \frac{t^{n}}{n!} + (1 − e^{−t}) ∑_{n=1}^{∞} B_{n}^{(k_{1},...,k_{r})} \frac{t^{n−1}}{(n − 1)!},

and from this we have

∑_{n=0}^{∞} B_{n}^{(k_{1},...,k_{r})} \frac{t^{n}}{n!} = ∑_{n=0}^{∞} B_{n}^{(k_{1}−1,k_{2},...,k_{r})} \frac{t^{n}}{n!} − (e^{t} − 1) ∑_{n=1}^{∞} B_{n}^{(k_{1},...,k_{r})} \frac{t^{n−1}}{(n − 1)!}.

Comparing the coefficients of t^{n}/n! on both sides, we obtain the desired relation for B_{n}^{(k_{1},...,k_{r})}. The relation for C_{n}^{(k_{1},...,k_{r})} is obtained similarly.

We make a remark on the following point: since we allow k_{1} to be positive or negative, the recursions above need never ‘stop’. However, by the identity

Li_{0,k_{2},...,k_{r}}(z) = \frac{z}{1 − z} Li_{k_{2},...,k_{r}}(z),

the B_{n}^{(0,k_{2},...,k_{r})} or C_{n}^{(0,k_{2},...,k_{r})} can be written as simple linear combinations of B_{m}^{(k_{2},...,k_{r})} or C_{m}^{(k_{2},...,k_{r})} (0 ≤ m < n). The proposition therefore gives a way to express B_{n}^{(k_{1},...,k_{r})} or C_{n}^{(k_{1},...,k_{r})} in terms of numbers with smaller n and lower “depth” (length r of the index set).

Next, we write our multi-poly-Bernoulli numbers as finite sums involving Stirling numbers of the second kind. Recall that the Stirling numbers of the first kind \left[{n \atop m}\right] and the second

kind \left\{{n \atop m}\right\} are the integers uniquely determined by the following recursions for all integers

m, n (see Knuth [9]):

\left[{0 \atop 0}\right] = 1, \left[{n \atop 0}\right] = \left[{0 \atop m}\right] = 0 (n, m ≠ 0), \left[{n + 1 \atop m}\right] = \left[{n \atop m − 1}\right] + n\left[{n \atop m}\right] (∀n, m ∈ ℤ),

\left\{{0 \atop 0}\right\} = 1, \left\{{n \atop 0}\right\} = \left\{{0 \atop m}\right\} = 0, (n, m ≠ 0), \left\{{n + 1 \atop m}\right\} = \left\{{n \atop m − 1}\right\} + m\left\{{n \atop m}\right\} (∀n, m ∈ ℤ).

Theorem 3. For any r ≥ 1, k_{1}, k_{2}, ⋯, k_{r} ∈ ℤ, and n ≥ 0, we have

B_{n}^{(k_{1},...,k_{r})} = (−1)^{n} ∑_{n+1≥m_{1}>m_{2}>···>m_{r}>0} \frac{(−1)^{m_{1}−1}(m_{1} − 1)!\left\{{n \atop m_{1}−1}\right\}}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}}

and

C_{n}^{(k_{1},...,k_{r})} = (−1)^{n} ∑_{n+1≥m_{1}>m_{2}>···>m_{r}>0} \frac{(−1)^{m_{1}−1}(m_{1} − 1)!\left\{{n+1 \atop m_{1}}\right\}}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}}.

Proof. Using the well-known generating series (cf. Graham et al. [3, §7.4])

\frac{(e^{t} − 1)^{m}}{m!} = ∑_{n=m}^{∞} \left\{{n \atop m}\right\}\frac{t^{n}}{n!} (m ≥ 0),  (4)

we have

∑_{n=0}^{∞} B_{n}^{(k_{1},...,k_{r})} \frac{t^{n}}{n!} = \frac{Li_{k_{1},··· ,k_{r}}(1 − e^{−t})}{1 − e^{−t}}

= ∑_{m_{1}>m_{2}>···>m_{r}>0} \frac{(1 − e^{−t})^{m_{1}−1}}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}}

= ∑_{m_{1}>m_{2}>···>m_{r}>0} \frac{(−1)^{m_{1}−1}(m_{1} − 1)!}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} ∑_{n=m_{1}−1}^{∞} \left\{{n \atop m_{1} − 1}\right\}\frac{(−t)^{n}}{n!}

= ∑_{n=r−1}^{∞}(−1)^{n} ∑_{n+1≥m_{1}>m_{2}>···>m_{r}>0} \frac{(−1)^{m_{1}−1}(m_{1} − 1)!\left\{{n \atop m_{1}−1}\right\}}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} \frac{t^{n}}{n!}.

Comparing the coefficients of t^{n}, we obtain the formula for B_{n}^{(k_{1},...,k_{r})}. To obtain the formula for C_{n}^{(k_{1},...,k_{r})}, we proceed similarly by using the equation

\frac{e^{t}(e^{t} − 1)^{m−1}}{(m − 1)!} = ∑_{n=m−1}^{∞} \left\{{n + 1 \atop m}\right\}\frac{t^{n}}{n!} (m ≥ 1),

which follows from equation (4) by differentiation:

∑_{n=0}^{∞} C_{n}^{(k_{1},...,k_{r})} \frac{t^{n}}{n!} = \frac{Li_{k_{1},...,k_{r}}(1 − e^{−t})}{e^{t} − 1}

= ∑_{m_{1}>m_{2}>···>m_{r}>0} \frac{e^{−t}(1 − e^{−t})^{m_{1}−1}}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}}

= ∑_{m_{1}>m_{2}>···>m_{r}>0} \frac{(−1)^{m_{1}−1}(m_{1} − 1)!}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} ∑_{n=m_{1}−1}^{∞} \left\{{n + 1 \atop m_{1}}\right\} \frac{(−t)^{n}}{n!}

= ∑_{n=r−1}^{∞}(−1)^{n} ∑_{n+1≥m_{1}>m_{2}>···>m_{r}>0} \frac{(−1)^{m_{1}−1}(m_{1} − 1)!\left\{{n+1 \atop m_{1}}\right\}}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} \frac{t^{n}}{n!}.

We end this section by giving formulas for the special case k_{1} = ⋯ = k_{r} = 1.

Proposition 4. For any r ≥ 1 and n ≥ r − 1, we have

B_{n}^{(_{1,1,...,1}^{r})} = \frac{1}{n + 1}\binom{n + 1}{r}B_{n−r+1}^{(1)}, C_{n}^{(_{1,1,...,1}^{r})} = \frac{1}{n + 1}\binom{n + 1}{r}C_{n−r+1}^{(1)}.

Proof. Use the identity Li^{1,...,1}_{r}(z) = (− log(1 − z))^{r}/r! to obtain

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

and

∑_{n=0}^{∞} C_{n}^{(_{1,...,1}^{r})} \frac{t^{n}}{n!} = \frac{t^{r−1}}{r!} \frac{t}{e^{t} − 1} = \frac{t^{r−1}}{r!} ∑_{n=0}^{∞} C_{n}^{(1)} \frac{t^{n}}{n!}.

The proposition follows.

3 Multi-poly-Bernoulli numbers with negative indices

Proposition 5. ^{2} We have the following generating series identities for multi-poly-Bernoulli numbers with non-positive upper indices:

1)

∑_{k_{1},...,k_{r}=0}^{∞} ∑_{n=0}^{∞} B_{n}^{(−k_{1},...,−k_{r})} \frac{t^{n}}{n!} \frac{x_{1}^{k_{1}}}{k_{1}!} ⋯ \frac{x_{r}^{k_{r}}}{k_{r}!}

= \frac{(1 − e^{−t})^{r−1}}{(e^{−x_{1}} + e^{−t} − 1)(e^{−x_{1}−x_{2}} + e^{−t} − 1) ⋯ (e^{−x_{1}−···−x_{r}} + e^{−t} − 1)},

2)

∑_{k_{1},...,k_{r}=0}^{∞} ∑_{n=0}^{∞} C_{n}^{(−k_{1},...,−k_{r})} \frac{t^{n}}{n!} \frac{x_{1}^{k_{1}}}{k_{1}!} ⋯ \frac{x_{r}^{k_{r}}}{k_{r}!}

= \frac{e^{−t}(1 − e^{−t})^{r−1}}{(e^{−x_{1}} + e^{−t} − 1)(e^{−x_{1}−x_{2}} + e^{−t} − 1) ⋯ (e^{−x_{1}−···−x_{r}} + e^{−t} − 1)}.

Proof. Use the defining identity

∑_{n=0}^{∞} B_{n}^{(−k_{1},...,−k_{r})} \frac{t^{n}}{n!} = (1 − e^{−t})^{−1} ∑_{m_{1}>···>m_{r}>0} m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}(1 − e^{−t})^{m_{1}}

2This proposition and the corollary below are also pointed out by Genki Shibukawa.

to obtain

∑_{k_{1},...,k_{r}=0}^{∞} ∑_{n=0}^{∞} B_{n}^{(−k_{1},...,−k_{r})} \frac{t^{n}}{n!} \frac{x_{1}^{k_{1}}}{k_{1}!} ⋯ \frac{x_{r}^{k_{r}}}{k_{r}!} = (1 − e^{−t})^{−1} ∑_{m_{1}>···>m_{r}>0}(1 − e^{−t})^{m_{1}}e^{m_{1}x_{1}+···+m_{r}x_{r}}

= (1 − e^{−t})^{−1} ∑_{n_{1},...,n_{r}=1}^{∞}(1 − e^{−t})^{n_{1}+···+n_{r}}e^{(n_{1}+···+n_{r})x_{1}+(n_{2}+···+n_{r})x_{2}+···+n_{r}x_{r}}

= (1 − e^{−t})^{−1} ∑_{n_{1},...,n_{r}=1}^{∞}((1 − e^{−t})e^{x_{1}})^{n_{1}}((1 − e^{−t})e^{x_{1}+x_{2}})^{n_{2}} ⋯ ((1 − e^{−t})e^{x_{1}+···+x_{r}})^{n_{r}}

= (1 − e^{−t})^{r−1} \frac{e^{x_{1}}}{1 − ((1 − e^{−t})e^{x_{1}})} \frac{e^{x_{1}+x_{2}}}{1 − ((1 − e^{−t})e^{x_{1}+x_{2}})} ⋯ \frac{e^{x_{1}+···+x_{r}}}{1 − ((1 − e^{−t})e^{x_{1}+···+x_{r}})}
= \frac{(1 − e^{−t})^{r−1}}{(e^{−x_{1}} + e^{−t} − 1)(e^{−x_{1}−x_{2}} + e^{−t} − 1) ⋯ (e^{−x_{1}−···−x_{r}} + e^{−t} − 1)}.

The identity for C_{n}^{(−k_{1},...,−k_{r})} readily follows from this because the defining generating series differ only by the factor e^{−t}.

Corollary 6. For any integers k_{1}, …, k_{r} ≥ 0 and n ≥ 0, the multi-poly-Bernoulli numbers B_{n}^{(−k_{1},...,−k_{r})} and C_{n}^{(−k_{1},...,−k_{r})} are positive integers.

Proof. The right-hand sides of 1) and 2) of the proposition can be rewritten as

e^{t}(e^{t} − 1)^{r−1} · \frac{e^{x_{1}}}{1 − (e^{t} − 1)(e^{x_{1}} − 1)} · \frac{e^{x_{1}+x_{2}}}{1 − (e^{t} − 1)(e^{x_{1}+x_{2}} − 1)} ⋯ \frac{e^{x_{1}+···+x_{r}}}{1 − (e^{t} − 1)(e^{x_{1}+···+x_{r}} − 1)}

and

(e^{t} − 1)^{r−1} · \frac{e^{x_{1}}}{1 − (e^{t} − 1)(e^{x_{1}} − 1)} · \frac{e^{x_{1}+x_{2}}}{1 − (e^{t} − 1)(e^{x_{1}+x_{2}} − 1)} ⋯ \frac{e^{x_{1}+···+x_{r}}}{1 − (e^{t} − 1)(e^{x_{1}+···+x_{r}} − 1)}

respectively, from which the positivity of B_{n}^{(−k_{1},...,−k_{r})} and C_{n}^{(−k_{1},...,−k_{r})} is obvious. That both are integers follows from the explicit formulas in Theorem 3.

We can give some explicit formulas different from Theorem 3 for special indices.

Proposition 7. For r ≥ 1, k ≥ 0 and n ≥ 0, we have 1)

B_{n}^{(−k,_{0,...,0}^{r−1})} = (−1)^{n} ∑_{j=0}^{n}(−1)^{j}(j + 1)^{k}j!\binom{j}{r − 1}\left\{{n \atop j}\right\},

C_{n}^{(−k,_{0,...,0}^{r−1})} = (−1)^{n} ∑_{j=0}^{n}(−1)^{j}(j + 1)^{k}j!\binom{j}{r − 1}\left\{{n + 1 \atop j + 1}\right\}.

B_{n}^{(_{0,...,0}^{r−1},−k)} = ∑_{j=0}^{min\{n+1−r,k\}}(r + j − 1)!j!\left\{{n + 1 \atop r + j}\right\}\left\{{k + 1 \atop j + 1}\right\},
C_{n}^{(_{0,...,0}^{r−1},−k)} = ∑_{j=0}^{min\{n+1−r,k\}}(r + j − 1)!j!\left\{{n \atop r + j − 1}\right\}\left\{{k + 1 \atop j + 1}\right\}.

Proof. Set x_{2} = ⋯ = x_{r} = 0 in 1) of Proposition 5 to obtain

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

From this we have

∑_{r=1}^{∞} ∑_{k,n=0}^{∞} B_{n}^{(−k,_{0,...,0}^{r−1})} \frac{t^{n}}{n!} \frac{x^{k}}{k!} y^{r−1} = \frac{1}{(e^{−x} + e^{−t} − 1)} ∑_{r=1}^{∞} (\frac{1 − e^{−t}}{e^{−x} + e^{−t} − 1})^{r−1} y^{r−1}
= \frac{1}{(e^{−x} + e^{−t} − 1)} \frac{1}{1 − \frac{1−e^{−t}}{e^{−x}+e^{−t}−1}y}
= \frac{1}{e^{−x} + e^{−t} − 1 − (1 − e^{−t})y}
= \frac{e^{x}}{1 − e^{x}(1 + y)(1 − e^{−t})}
= ∑_{j=0}^{∞} e^{(j+1)x}(1 + y)^{j}(1 − e^{−t})^{j}
= ∑_{j=0}^{∞} e^{(j+1)x}(1 + y)^{j}(−1)^{j}j! ∑_{n=j}^{∞}(−1)^{n}\left\{{n \atop j}\right\}\frac{t^{n}}{n!}
= ∑_{n=0}^{∞} ∑_{j=0}^{n} e^{(j+1)x}(1 + y)^{j}(−1)^{n+j}j!\left\{{n \atop j}\right\}\frac{t^{n}}{n!}.

Comparing the coefficients of \frac{t^{n}}{n!} \frac{x^{k}}{k!} y^{r−1} on both sides, we obtain 1) for B_{n}^{(−k,0,...,0)}.

Similarly, we compute

∑_{r=1}^{∞} ∑_{k,n=0}^{∞} C_{n}^{(−k,_{0,...,0}^{r−1})} \frac{t^{n}}{n!} \frac{x^{k}}{k!} y^{r−1} = e^{−t} · \frac{e^{x}}{1 − e^{x}(1 + y)(1 − e^{−t})}

= ∑_{j=0}^{∞} e^{(j+1)x}(1 + y)^{j}e^{−t}(1 − e^{−t})^{j}

= ∑_{j=0}^{∞} e^{(j+1)x}(1 + y)^{j}(−1)^{j}j! ∑_{n=j}^{∞}(−1)^{n}\left\{{n + 1 \atop j + 1}\right\}\frac{t^{n}}{n!}

= ∑_{n=0}^{∞} ∑_{j=0}^{n} e^{(j+1)x}(1 + y)^{j}(−1)^{n+j}j!\left\{{n + 1 \atop j + 1}\right\}\frac{t^{n}}{n!}

and obtain the formula for C_{n}^{(−k,0,...,0)}.

For 2), we set x_{1} = ⋯ = x_{r−1} = 0 in the formulas in Proposition 5 and have

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

= (e^{t} − 1)^{r−1} · \frac{e^{t+x}}{1 − (e^{t} − 1)(e^{x} − 1)}
= ∑_{j=0}^{∞} e^{t}(e^{t} − 1)^{r+j−1}e^{x}(e^{x} − 1)^{j}

= ∑_{j=0}^{∞}(r + j − 1)! ∑_{n=r+j−1}^{∞} \left\{{n + 1 \atop r + j}\right\}\frac{t^{n}}{n!} · j! ∑_{k=j}^{∞} \left\{{k + 1 \atop j + 1}\right\}\frac{x^{k}}{k!}

and

∑_{k=0}^{∞} ∑_{n=0}^{∞} C_{n}^{(_{0,...,0}^{r−1},−k)} \frac{t^{n}}{n!} \frac{x^{k}}{k!} = ∑_{j=0}^{∞}(r + j − 1)! ∑_{n=r+j−1}^{∞} \left\{{n \atop r + j − 1}\right\}\frac{t^{n}}{n!} · j! ∑_{k=j}^{∞} \left\{{k + 1 \atop j + 1}\right\}\frac{x^{k}}{k!} .

Comparing the coefficients of \frac{t^{n}}{n!} \frac{x^{k}}{k!} on both sides, we obtain 2).

4 Connection to the finite multiple zeta values

As recalled in the introduction, the finite multiple zeta value ζ_{A}(k_{1}, …, k_{r}) is an element in the ring A := ∏_{p} ℤ/pℤ/ ⨁_{p} ℤ/pℤ represented by (ζ_{A}(k_{1}, …, k_{r})_{(p)})_{p} ∈ ∏_{p} ℤ/pℤ where

ζ_{A}(k_{1}, …, k_{r})_{(p)} := ∑_{p>m_{1}>···>m_{r}>0} \frac{1}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} mod p.

Theorem 8. For any r ≥ 1 and k_{i} ∈ ℤ, we have

ζ_{A}(k_{1}, …, k_{r})_{(p)} = −C_{p−2}^{(k_{1}−1,k_{2},...,k_{r})} mod p.

More generally, we have for any r ≥ 1, j ≥ 0, and k_{i} ∈ ℤ

ζ_{A}(\underbrace{1, …, 1}_{j}, k_{1}, …, k_{r})_{(p)} = −C_{p−j−2}^{(k_{1}−1,k_{2},...,k_{r})} mod p.

Proof. By the explicit formula in Theorem 3, we have

C_{p−2}^{(k_{1}−1,k_{2},...,k_{r})} = − ∑_{p−1≥m_{1}>m_{2}>···>m_{r}>0} \frac{(−1)^{m_{1}−1}(m_{1} − 1)!\left\{{p−1 \atop m_{1}}\right\}}{m_{1}^{k_{1}−1}m_{2}^{k_{2}} ⋯ m_{r}^{k_{r}}}

= ∑_{p−1≥m_{1}>m_{2}>···>m_{r}>0} \frac{(−1)^{m_{1}}m_{1}!\left\{{p−1 \atop m_{1}}\right\}}{m_{1}^{k_{1}}m_{2}^{k_{2}} ⋯ m_{r}^{k_{r}}}.

By the well-known formula (cf. Graham et al. [3, §6.1])

(−1)^{m_{1}}m_{1}!\left\{{p − 1 \atop m_{1}}\right\} = ∑_{l=0}^{m_{1}}(−1)^{l}\binom{m_{1}}{l}l^{p−1},

we see

(−1)^{m_{1}}m_{1}!\left\{{p − 1 \atop m_{1}}\right\} ≡ ∑_{l=1}^{m_{1}}(−1)^{l}\binom{m_{1}}{l} (mod p),

the sum on the right being equal to (1 − 1)^{m_{1}} − 1 = −1. This proves that

C_{p−2}^{(k_{1}−1,k_{2},...,k_{r})} mod p = −ζ_{A}(k_{1}, …, k_{r})_{(p)}.

For the second identity, we proceed as follows. First note

∑_{p−1≥i_{1}>···>i_{j}>m_{1}>···>m_{r}≥1} \frac{1}{i_{1} ⋯ i_{j}m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}}

= ∑_{p−j>m_{1}>···>m_{r}≥1} \frac{1}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} ∑_{p−1≥i_{1}>···>i_{j}>m_{1}} \frac{1}{i_{1} ⋯ i_{j}} .

By changing i_{l} → p − i_{l}, we see that

∑_{p−1≥i_{1}>···>i_{j}>m_{1}} \frac{1}{i_{1} ⋯ i_{j}} ≡ (−1)^{j} ∑_{p−m_{1}>i_{j}>···>i_{1}≥1} \frac{1}{i_{1} ⋯ i_{j}} (mod p).

Using the formula

∑_{p−m_{1}>i_{j}>···>i_{1}≥1} \frac{1}{i_{1} ⋯ i_{j}} = \frac{1}{(p − m_{1} − 1)!} \left[{p − m_{1} \atop j + 1}\right] ,

and the congruences

\left[{n \atop m}\right] ≡ \left\{{p − m \atop p − n}\right\} (mod p) (1 ≤ m ≤ n ≤ p − 1)

(see Hoffman [6, §5]) and

\frac{1}{(p − m_{1} − 1)!} ≡ (−1)^{m_{1}+1}m_{1}! (mod p),

we obtain (for odd p)

ζ_{A}(\underbrace{1, …, 1}_{j}, k_{1}, …, k_{r})_{(p)} = ∑_{p−j>m_{1}>···>m_{r}≥1} \frac{(−1)^{j+m_{1}+1}m_{1}!\left\{{p−j−1 \atop m_{1}}\right\}}{m_{1}^{k_{1}} ⋯ m_{r}^{k_{r}}} mod p
= −C_{p−j−2}^{(k_{1}−1,k_{2},...,k_{r})} mod p.

This concludes the proof of the theorem.

Combining the theorem with Proposition 4, we see that

ζ_{A}(2, \overbrace{1, …, 1}^{k−2}) = (−C_{p−2}^{(_{1,...,1}^{k−1})} mod p)_{p} = (−\frac{1}{p − 1}\binom{p − 1}{k − 1}B_{p−k} mod p)_{p}
= (B_{p−k} mod p)_{p}.

(We used \binom{p−1}{k−1} ≡ (−1)^{k−1} (mod p) and (−1)^{p−k}B_{p−k} = B_{p−k} for large enough p.) As is discussed in Kaneko-Zagier [8], the element (B_{p−k} mod p)_{p} on the right is regarded as an analogue of kζ(k). (Note that the obvious analogue ζ_{A}(k) of ζ(k) is 0.)

References

[1] T. Arakawa and M. Kaneko, Multiple zeta values, poly-Bernoulli numbers, and related zeta functions, Nagoya Math. J. 153 (1999), 189–209.

[2] A. Bayad and Y. Hamahata, Multiple polylogarithms and multi-poly-Bernoulli polynomials, Funct. Approx. Comment. Math. 46 (2012), 45–61.

[3] R. Graham, D. Knuth, and O. Patashnik, Concrete Mathematics, Addison-Wesley, 1989.

[4] Y. Hamahata and H. Masubuchi, Special multi-poly-Bernoulli numbers, J. Integer Seq. 10 (2007), Article 07.4.1.

[5] Y. Hamahata and H. Masubuchi, Recurrence formulae for multi-poly-Bernoulli numbers, Integers 7 (2007), #A46.

[6] M. Hoffman, Quasi-symmetric functions and mod p multiple harmonic sums, preprint,

   http://arxiv.org/abs/math/0401319.

[7] M. Kaneko, Poly-Bernoulli numbers, J. Théor. Nombres Bordeaux 9 (1997), 199–206.

[8] M. Kaneko and D. Zagier, Finite multiple zeta values, in preparation.

[9] D. Knuth, Two notes on notation, Amer. Math. Monthly 99 (1992), 403–422.

[10] J. Zhao, Wolstenholme type theorem for multiple harmonic sums, Int. J. Number Theory 4 (2008), 73–106.

2010 Mathematics Subject Classification: Primary 11B68; Secondary 11M32. Keywords: multi-poly-Bernoulli number, Stirling number, finite multiple zeta value.

(Concerned with sequences \underline{A027643}, \underline{A027644}, \underline{A027645}, \underline{A027646}, \underline{A027647}, \underline{A027648}, \underline{A027649}, \underline{A027650}, and \underline{A027651}.)

Received October 24 2013; revised version received February 17 2014. Published in Journal of Integer Sequences, February 17 2014.

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