INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
RECURRENCE FORMULAE FOR MULTI-POLY-BERNOULLI NUMBERS
Y. Hamahata
Department of Mathematics, Tokyo University of Science, Noda, Chiba 278-8510, Japan
hamahata@ma.noda.tus.ac.jp
H. Masubuchi
Department of Mathematics, Tokyo University of Science, Noda, Chiba 278-8510, Japan
yoshinori@ma.noda.tus.ac.jp
Received: 4/3/07, Revised: 10/10/07, Accepted: 10/15/07, Published: 11/3/07
Abstract
In this paper we establish recurrence formulae for multi-poly-Bernoulli numbers.
–Dedicated to Professor Ryuichi Tanaka on the occasion of his sixtieth birthday
1. Introduction and Background
Let us briefly recall poly-Bernoulli numbers.
For an integer k ∈ Z, put
Lik (z) =
∞
!
zn
n=1
nk
.
The formal power series Lik (z) is the k-th polylogarithm if k ≥ 1, and a rational function
if k ≤ 0. When k = 1, Li1 (z) = − log(1 − z). The formal power series Lik (z) can be used
(k)
to introduce poly-Bernoulli numbers. The rational numbers Bn (n = 0, 1, 2, . . .) are said to
be poly-Bernoulli numbers if they satisfy
∞
(1)
Lik (1 − e−x ) ! (k) xn
=
Bn
.
1 − e−x
n!
n=0
In addition, for any n ≥ 0, Bn is the classical Bernoulli number, Bn . It was shown in [1]
that special values of certain zeta functions at non-positive integers can be described in terms
of poly-Bernoulli numbers. Furthermore, Kaneko [8] presented the following recurrences for
poly-Bernoulli numbers.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
2
Theorem 1 (Kaneko) (1) For any k ∈ Z and n ≥ 0,
"
%
$
n−1 #
!
1
n
(k)
Bn(k) =
Bn(k−1) −
Bm
.
m−1
n+1
m=1
(2) For any k ≥ 1 and n ≥ 0,
Bn(k) =
n
!
(−1)m
m=0
#
n
m
$
(k−1)
Bn−m
" m
!
l=0
(−1)l
n−l+1
#
m
l
$
(1)
Bl
%
.
In this paper we consider generalized poly-Bernoulli numbers, which we refer to as multipoly-Bernoulli numbers. Kim-Kim [9] introduced these numbers and proved that special
values of certain zeta functions at non-positive integers can be described in terms of these
numbers. In [5], we established a closed formula, and a duality property for special multipoly-Bernoulli numbers. Bernoulli numbers satisfy certain recurrence relationships, which
are used in many computations involving Bernoulli numbers. Obtaining a recurrence formula
for multi-poly-Bernoulli numbers therefore seems to be a natural and important problem.
The objective of this paper is thus to establish some recurrence formulae for multi-polyBernoulli numbers. It will be apparent that these recurrence formulae (Theorems 6, 7) are
similar to those in the theorem above.
2. Multi-poly-Bernoulli Numbers
We first define a generalization of Lik (z). Let r be an integer with a value greater than one.
Definition 2 Let k1 , k2 , . . . , kr be integers. Define
!
Lik1 ,k2 ,...,kr (z) =
m1 ,m2 ,...,mr ∈Z
0<m1 <m2 <···<mr
z mr
.
mk11 · · · mkr r
Next let us establish the following fundamental result.
Lemma 3
& 1
d
Lik1 ,k2 ,...,kr−1 ,kr −1 (z) (kr %= 1)
z
Lik1 ,k2 ,...,kr (z) =
.
(1)
1
Lik1 ,k2 ,...,kr−1 (z) (kr = 1)
dz
1−z
The former case may be proven by means of a simple calculation. The latter case follows
from
∞
!
!
!
z mr −1
z mr −1
=
kr−1
kr−1
k1
k1
0<m1 <···<mr−1 <mr m1 · · · mr−1
0<m1 <···<mr−1 mr =mr−1 +1 m1 · · · mr−1
!
1
z mr−1
=
·
.
kr−1
k1
1−z
0<m1 <···<mr−1 m1 · · · mr−1
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
3
This lemma will be required later.
Let us now introduce a generalization of poly-Bernoulli numbers, making use of Lik1 ,k2 ,...,kr (z).
(k ,k2 ,...,kr )
Definition 4 Multi-poly-Bernoulli numbers Bn 1
integer k1 , k2 , . . . , kr by the generating series
(n = 0, 1, 2, . . .) are defined for each
∞
Lik1 ,k2 ,...,kr (1 − e−t ) ! (k1 ,k2 ,...,kr ) tn
=
Bn
.
(1 − e−t )r
n!
n=0
(2)
By definition, the left-hand side of (2) is
1
+
k
k
1
2
1 2 · · · rkr
!
(1 − e−t )mr −r
.
mk11 · · · mkr r
=
1
.
· · · rkr
0<m1 <···<mr
mr #=r
Hence we have,
Proposition 5
(k1 ,k2 ,...,kr )
B0
1k1 2k2
(k ,k2 )
The following tables show the values of Bn 1
\n
(1,1)
Bn
(1,0)
Bn
(0,1)
Bn
(0,0)
Bn
(0,−1)
Bn
(−1,0)
Bn
(−1,−1)
Bn
0 1
2
3
4
1
2
5
12
13
6
5
6
1
4
1
20
119
30
31
30
(k ,k2 ,k3 )
and Bn 1
5
1
− 12
5
for small n, ki .
6
7
5
84
253
42
41
42
1
12
1
2
1
2
3
2
2
3
1
2
1
2
2 4
8
16
32
64
128
6 18 54 162 486 1458 2574
3 9 27 81 243
729
1287
9 39 165 687 2829 11505 46965
1
3
29
30
43
42
7
29
30
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
\n
(1,1,1)
Bn
(1,1,0)
Bn
(1,0,1)
Bn
(0,1,1)
Bn
(1,0,0)
Bn
(0,1,0)
Bn
(0,0,1)
Bn
(0,0,0)
Bn
(0,0,−1)
Bn
(0,−1,0)
Bn
(−1,0,0)
Bn
(0,−1,−1)
Bn
(−1,0,−1)
Bn
(−1,−1,0)
Bn
(−1,−1,−1)
Bn
0
1
2
3
4
5
6
7
1
6
1
2
1
3
1
6
1
4
1
3
23
12
133
120
19
40
37
6
8
3
33
20
3
8
7
2
221
120
17
24
19
60
121
20
2383
840
53
56
1079
30
147
10
1047
140
1
8
59
6
673
168
185
168
8
63
1255
84
4321
840
303
280
8317
42
1238
21
4411
140
5
24
127
6
145
24
109
120
1
1
2
1
3
5
8
7
24
5
2
7
6
3
4
1
3
2
1
6
3
2
6
3
9
27
81
243
729
2187
12 48 192
768
3072 12288
49152
8 32 128
512
2048
8192
32768
4 16
64
256
1024
4096
16384
32 168 872 4488 22952 116808 592232
16 84 436 2244 11476 58404 296116
11 59 311 1619 8351 42779 217991
44 306 2054 13446 86414 547686 3434174
1
15
179
30
71
20
85
1177
42
433
28
4
455
3659
30
1271
20
The remainder of this paper deals with recurrences for multi-poly-Bernoulli numbers.
3. Recurrence Formulae
We present two kinds of recurrence formulae for multi-poly-Bernoulli numbers. The first
formula is as follows.
Theorem 6(1) If kr %= 1 and n ≥ 1, then
"
%
$
n−1 #
!
1
n
(k1 ,k2 ,...,kr )
Bn(k1 ,k2 ,...,kr ) =
Bn(k1 ,...,kr−1 ,kr −1) −
Bm
.
m−1
n+r
m=1
(2) If kr = 1 and n ≥ 1, then
Bn(k1 ,...,kr−1 ,1) =
'
)
& #
$ #
$(
n−1
!
1
n
n
(k1 ,...,kr−1 ,1)
Bn(k1 ,...,kr−1 ) −
(−1)n−m r
+
Bm
.
m
m−1
n+r
m=0
Proof. (1) Differentiate both sides of
Lik1 ,k2 ,...,kr (1 − e−t ) = (1 − e−t )r
∞
!
n=0
Bn(k1 ,k2 ,...,kr )
tn
.
n!
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
By (1), we have
L.H.S. =
On the other hand,
e−t
Lik1 ,...,kr−1 ,kr −1 (1 − e−t ).
1 − e−t
−t r−1 −t
R.H.S. = r(1 − e )
e
Bn(k1 ,k2 ,...,kr )
n=0
+(1 − e−t )r
Hence,
∞
!
∞
!
Bn(k1 ,k2 ,...,kr )
n=1
tn
n!
tn−1
.
(n − 1)!
Lik1 ,k2 ,...,kr −1 (1 − e−t )
(1 − e−t )r
∞
∞
!
!
tn
tn−1
=r
Bn(k1 ,k2 ,...,kr ) + (et − 1)
Bn(k1 ,k2 ,...,kr )
n!
(n − 1)!
n=0
n=1
=r
=r
∞
!
n=0
∞
!
tn
Bn(k1 ,k2 ,...,kr )
n!
Bn(k1 ,k2 ,...,kr )
n=0
+
tn
+
n!
∞ n !
∞
!
t
n!
(k1 ,k2 ,...,kr )
Bm
tm−1
(m − 1)!
n=1
m=1
∞ !
∞
n+m−1
!
(k1 ,k2 ,...,kr ) t
Bm
.
n!(m − 1)!
m=1 n=1
Here we put l = n + m − 1. The right-hand side of the last equation is then equal to
r
∞
!
n=0
tn
Bn(k1 ,k2 ,...,kr )
∞ !
∞
!
l!
tl
·
n! m=1 l=m
(l − (m − 1))!(m − 1)! l!
#
$ l
∞
∞ !
l
n
!
!
t
l
(k1 ,k2 ,...,kr ) t
(k1 ,k2 ,...,kr )
=r
Bn
+
Bm
m − 1 l!
n! l=1 m=1
n=0
*
+
$
∞
∞
n #
n
!
!
!
t
tn
n
(k1 ,k2 ,...,kr )
=r
Bn(k1 ,k2 ,...,kr ) +
Bm
.
n! n=1 m=1 m − 1
n!
n=0
+
(k1 ,k2 ,...,kr )
Bm
Comparing the coefficients of both sides, for each n ≥ 1,
$
n #
!
n
(k1 ,...,kr−1 ,kr −1)
(k1 ,k2 ,...,kr )
(k1 ,k2 ,...,kr )
Bn
= rBn
+
Bm
m−1
m=1
$
n−1 #
!
n
(k1 ,k2 ,...,kr )
(k1 ,k2 ,...,kr )
= (n + r)Bn
+
Bm
.
m−1
m=1
This implies the result.
(2) Differentiate both sides of
Lik1 ,...,kr−1 ,1 (1 − e−t ) = (1 − e−t )r
∞
!
n=0
Bn(k1 ,...,kr−1 ,1)
tn
.
n!
5
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
e−t
Lik ,...,k (1 − e−t )
1 − (1 − e−t ) 1 r−1
= Lik1 ,...,kr−1 (1 − e−t ),
∞
∞
n
!
!
tn−1
−t r−1 −t
(k1 ,...,kr−1 ,1) t
−t r
R.H.S. = r(1 − e ) e
Bn
+ (1 − e )
Bn(k1 ,...,kr−1 ,1)
.
n!
(n
−
1)!
n=0
n=1
L.H.S. =
Hence,
∞
!
Bn(k1 ,...,kr−1 )
n=0
= re−t
∞
!
tn
n!
Bn(k1 ,...,kr−1 ,1)
n=0
=r
=r
∞
∞
!
(−t)n !
n!
n=0
∞ !
∞
!
∞
!
tn
tn−1
+ (1 − e−t )
Bn(k1 ,...,kr−1 ,1)
n!
(n − 1)!
n=1
m
(k1 ,...,kr−1 ,1) t
Bm
m!
m=0
(k1 ,...,kr−1 ,1)
(−1)n Bm
m=0 n=0
−
∞
∞
!
(−t)n !
n!
n=1
∞ !
∞
!
m=0
m
(k ,··· ,kr−1 ,1) t
1
Bm+1
m!
n+m
tn+m
(k1 ,...,kr−1 ,1) t
−
(−1)n Bm+1
.
n!m! m=0 n=1
n!m!
Here we put l = n + m. The right-hand side of the last equation then becomes
∞ !
∞
!
(k1 ,...,kr−1 ,1)
r
(−1)l−m Bm
m=0 l=m
∞
∞
!
!
l!
tl
·
(l − m)!m! l!
l!
tl
·
(l − m)!m! l!
m=0 l=m+1
#
$ l
∞ !
l
!
t
l
l−m (k1 ,...,kr−1 ,1)
=r
(−1) Bm
m l!
l=0 m=0
#
$ l
∞ !
l−1
!
t
l
l−m (k1 ,...,kr−1 ,1)
−
(−1) Bm+1
m l!
l=1 m=0
(k ,...,kr−1 ,1)
1
(−1)l−m Bm+1
−
(k ,...,k
,1)
= rB0 1 r−1
" n
#
$ !
#
$% n
∞
n−1
!
!
t
n
n
(k
,...,k
,1)
1
r−1
(k1 ,...,kr−1 ,1)
+
(−1)n−m rBm
−
(−1)n−m Bm+1
.
m
m
n!
n=1
m=0
m=0
Comparing both sides, for each n ≥ 1,
Bn(k1 ,...,kr−1 )
#
$
#
$
n
n
!
!
n
n
n−m
(k1 ,...,kr−1 ,1)
n−m
(k1 ,...,kr−1 ,1)
=
(−1)
r
Bm
+
(−1)
Bm
m
m−1
m=0
m=1
& #
$ #
$(
n−1
!
n
n
(k1 ,...,kr−1 ,1)
n−m
(k1 ,...,kr−1 ,1)
= (n + r)Bn
+
(−1)
r
+
Bm
.
m
m−1
m=0
6
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
These computations imply the result.
We obtain the second formula using the integral representation of Lik1 ,k2 ,...,kr (1 − e−t ).
Theorem 7 (1) If kr %= 1, then for any n ≥ 0,
Bn(k1 ,k2 ,...,kr )
n!
= (−1)r−1
(n + r − 1)!




#
$
n
n−m

! !
! (−1)n−m+r−1−p (n − m + r − 1 − p)! n − m + r − 1
(k1 ,...,kr−1 ,kr −1)
×
Bp
p


j1 ! · · · jr−1 !

m=0  p=0 j +···+j
1
r−1
=n−m+r−1−p
×(−1)m
#
n+r−1
m
$!
m
l=0
(−1)l
n−l+r



$
 !

l!
m
(1)
(1)
Bi1 · · · Bir
.
l

i1 ! · · · ir ! 
 i +···+i

#
1
(2) If kr = 1, then for any n ≥ 0,
r
=l
Bn(k1 ,k2 ,...,kr )
n!
= (−1)r−1
(n + r − 1)!




#
$
n
n−m


! !
! (−1)n−m+r−1−p (n − m + r − 1 − p)! n − m + r − 1
(k1 ,...,kr−1 )
×
Bp
p


j1 ! · · · jr−1 !

m=0  p=0 j1 +···+jr−1
=n−m+r−1−p




#
$

!
1
m! 
n+r−1
(1)
(1)
×
Bi1 · · · Bir
.
m
n−m+r 
i1 ! · · · ir ! 
 i1 +···+ir

=m
Proof. (1) Since
we have
d
e−t
Lik1 ,k2 ,...,kr (1 − e−t ) =
Lik1 ,...,kr−1 ,kr −1 (1 − e−t ),
dt
1 − e−t
−t
Lik1 ,k2 ,...,kr (1 − e ) =
3
0
t
e−s
Lik1 ,...,kr−1 ,kr −1 (1 − e−s )ds.
−s
1−e
By this equation,
∞
!
tn
Bn(k1 ,k2 ,...,kr )
n!
n=0
3 t
1
e−s
=
Lik1 ,...,kr−1 ,kr −1 (1 − e−s )ds
−t
r
−s
(1 − e )
1−e
# t $r 30 t
−s
e
−s
−s r−1 Lik1 ,...,kr−1 ,kr −1 (1 − e )
=
e
(1
−
e
)
·
ds
et − 1
(1 − e−s )r
0
*∞
+r 3 ∞
* ∞
+r−1 ∞
t!
n−1
n
!
! (−s)n
!
t
(−s)
sn
=
Bn(1)
−
Bn(k1 ,...,kr−1 ,kr −1) ds
n!
n!
n!
n!
0 n=0
n=0
n=1
n=0
7
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
= (−1)r−1
*∞
!
tn−1
Bn(1)
n!
n=0
+r
3 t!
∞
∞
!
(−s)n !
×
n! n=r−1 j +···+j
0 n=0
1
=n
∞
r−1
(−1)n sn ! (k1 ,...,kr−1 ,kr −1) sn
B
ds.
j1 ! · · · jr−1 ! n=0 n
n!
We write I1 for the integral part of the last equation.
3 t!
∞
∞
∞
!
(−s)n ! !
(−1)n n! (k1 ,...,kr−1 ,kr −1) sn+m
I1 =
B
ds.
n! m=0 n=r−1 j +···+j j1 ! · · · jr−1 ! m
n!m!
0 n=0
1
=n
r−1
Putting l = n + m,
3 t!
∞
∞
∞
! (−1)l−m (l − m)!
(−s)n ! !
l!
sl
(k1 ,...,kr−1 ,kr −1)
I1 =
Bm
· ds
n! m=0 l=m+r−1 j +···+j
j1 ! · · · jr−1 !
(l − m)!m! l!
0 n=0
1
r−1
=l−m
#
$ l
3 t!
∞
∞ l−r+1
! (−1)l−m (l − m)!
(−s)n ! !
s
l
(k1 ,...,kr−1 ,kr −1)
=
Bm
ds
m
n! l=r−1 m=0 j +···+j
j1 ! · · · jr−1 !
l!
0 n=0
1
=
3
t
∞ !
∞ l−r+1
!
!
(−1)n
0 l=r−1 n=0 m=0
j1
r−1
=l−m
#
$ n+l
! (−1)l−m (l − m)!
s
l
(k1 ,...,kr−1 ,kr −1)
Bm
ds.
m
j1 ! · · · jr−1 !
n!l!
+···+j
r−1
=l−m
We put a = l + n. Then
# $
3 t !
∞ !
∞ l−r+1
!
! (−1)l−m (l − m)!
a!
sa
l
a−l
(k1 ,...,kr−1 ,kr −1)
I1 =
(−1)
Bm
· ds
m (a − l)!l! a!
j1 ! · · · jr−1 !
0 l=r−1 a=l m=0
j +···+j
1
=
3
t
0
r−1
=l−m
#
$# $ a
∞
a l−r+1
!
!
!
! (−1)l−m (l − m)!
l
a s
a−l
(k1 ,...,kr−1 ,kr −1)
(−1)
Bm
ds
m
l
j1 ! · · · jr−1 !
a!
j +···+j
a=r−1 l=r−1 m=0
1
r−1
=l−m
#
$ # $ a+1
∞
a l−r+1
!
!
!
! (−1)l−m (l − m)!
t
l
a
a−l
(k1 ,...,kr−1 ,kr −1)
=
(−1)
Bm
.
m
l
j1 ! · · · jr−1 !
(a + 1)!
j +···+j
a=r−1 l=r−1 m=0
1
Hence,
(−1)r−1
*∞
!
r−1
= (−1)
r−1
=l−m
tn−1
Bn(1)
n!
n=0
∞
!
!
n=0
i1 +···+ir
=n
+r
(1)
I1
(1)
Bi1 · · · Bir
tn−r
i1 ! · · · ir !
# $ # $ a+1
∞
a l−r+1
!
!
!
! (−1)l−m (l − m)!
t
l
a
a−l
(k1 ,...,kr−1 ,kr −1)
×
(−1)
Bm
m
l
j1 ! · · · jr−1 !
(a + 1)!
j +···+j
a=r−1 l=r−1 m=0
1
r−1
=l−m
8
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
∞ !
∞ !
a l−r+1
!
!
!
(1)
(1)
(−1)a−l
Bi1 · · · Bir
n!
i1 ! · · · ir !
i1 +···+ir
a=r−1 n=0 l=r−1 m=0
=n
#
$ # $ n+a+1
! (−1)l−m (l − m)!
t
l
a
(k1 ,...,kr−1 ,kr −1)
×
Bm
· t−r
m
l
j
!
·
·
·
j
!
n!(a
+
1)!
1
r−1
j +···+j
r−1
= (−1)
1
r−1
=l−m
r−1
= (−1)
∞
∞
a l−r+1
!
!
!
!
!
(1)
(1) (b − a − 1)!
(−1)a−l
Bi1 · · · Bir
i1 ! · · · ir !
i +···+ir
a=r−1 b=a+1 l=r−1 m=0
1
=b−a−1
# $# $
! (−1)l−m (l − m)!
b!
tb
l
a
(k1 ,...,kr−1 ,kr −1)
×
Bm
· · t−r
m
l (b − (a + 1))!(a + 1)! b!
j1 ! · · · jr−1 !
j +···+j
1
r−1
=l−m
(put b = n + a + 1)
r−1
= (−1)
∞ !
b−1 !
a l−r+1
!
!
!
(1)
(1) (b − a − 1)!
(−1)a−l
Bi1 · · · Bir
i1 ! · · · ir !
i +···+ir
b=r a=r−1 l=r−1 m=0
1
=b−a−1
#
$# $#
$ b−r
! (−1)l−m (l − m)!
t
l
a
b
(k1 ,...,kr−1 ,kr −1)
×
Bm
m
l
a+1
j1 ! · · · jr−1 !
b!
j +···+j
1
r−1
=l−m
= (−1)r−1
∞ n+r−1
a l−r+1
!
! !
!
(−1)a−l
n=0 a=r−1 l=r−1 m=0
!
i1 +···+ir
=n−a+r−1
(1)
(1) (n
Bi1 · · · Bir
− a + r − 1)!
i1 ! · · · ir !
#
$# $#
$
! (−1)l−m (l − m)!
tn
l
a
n+r
(k1 ,...,kr−1 ,kr −1)
×
Bm
m
l
a + 1 (n + r)!
j1 ! · · · jr−1 !
j +···+j
1
r−1
=l−m
(put n = b − r)
=
×

∞ 

!
n+r−1
!


n!
 !
(1)
(1) (n − a + r − 1)! 
(−1)r−1
Bi1 · · · Bir



(n + r)! a=r−1 i +···+ir
i1 ! · · · ir !
n=0 
1
a
!
(−1)a−l
l=r−1
#
=n−a+r−1


$ # $ l−r+1
#
$
 tn
! (−1)l−m (l − m)! l
n+r
a !
(k1 ,...,kr−1 ,kr −1)
Bm
.
a+1
l
m

j
!
·
·
·
j
!
n!
1
r−1

m=0 j +···+j
1
r−1
=l−m
9
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
10
If we compare the coefficients of both sides,
Bn(k1 ,k2 ,...,kr )
= (−1)r−1


n+r−1
!  !
n!
(1)
(1) (n − a + r − 1)! 
Bi1 · · · Bir


(n + r)! a=r−1 i +···+ir
i1 ! · · · ir !
1
=n−a+r−1
×
a
!
a−l
(−1)
l=r−1
r−1
= (−1)
#
n+r
a+1
$#
a
l
$ l−r+1
!
m=0
×
a=l
r−1
=l−m
n+r−1
! l−r+1
!
! (−1)l−m (l − m)! # l $
n!
(k1 ,...,kr−1 ,kr −1)
Bm
m
(n + r − 1)! l=r−1 m=0 j +···+j
j1 ! · · · jr−1 !
1
n+r−1
!
j1
! (−1)l−m (l − m)! # l $
(k1 ,...,kr−1 ,kr −1)
Bm
m
j
!
·
·
·
j
!
1
r−1
+···+j
r−1
=l−m


#
$# $
(−1)
n+r−1
a  !
(1)
(1) (n − a + r − 1)! 
Bi1 · · · Bir


a
l
a+1
i1 ! · · · ir !
i +···+ir
a−l
1
=n−a+r−1
*
by
= (−1)r−1
: n+r ;
a+1
+
n+r−1
a
n+r−1
! !
! n+r−1
!
n + r n+r−1
=
( a ) and
=
a+1
a=r−1 l=r−1
l=r−1 a=l
n!
(n + r − 1)!



#
$
#
$
l−r+1
!

! (−1)l−m (l − m)!
l
l n+r−1
(k1 ,...,kr−1 ,kr −1)
×
(−1)
Bm
l
m


j1 ! · · · jr−1 !
 m=0 j +···+j

l=r−1
n+r−1
!
1
r−1
=l−m




#
$
n+r−1


a
! (−1)
!
n+r−1−l
(1)
(1) (n − a + r − 1)!
×
B · · · Bir
n+r−1−a 
 i +···+ir i1

a+1
i1 ! · · · ir !

a=l
1
=n−a+r−1
(by ( n+r−1
) ( al ) = ( n+r−1
)
a
l
r−1
= (−1)
×
×


l$
!
: n+r−1−l ;
n+r−1−a )
#
$
n
!
n!
l$ +r−1 n + r − 1
(−1)
l# + r − 1
(n + r − 1)! l$ =0
! (−1)
l$ +r−1−m

m=0 j1 +···+jr−1
=l$ +r−1−m
n+r−1
!
(−1)a
a+1
a=l$ +r−1
#
(l# + r − 1 − m)!
j1 ! · · · jr−1 !
n − l#
n+r−1−a
(put l# = l − r + 1)

$
 !


i1 +···+ir
=n−a+r−1
#
l# + r − 1
m
(1)
(1)
Bi1 · · · Bir
$
(k1 ,...,kr−1 ,kr −1)
Bm







(n − a + r − 1)! 
i1 ! · · · ir !


INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
11
#
$
n
!
n!
l$ n + r − 1
=
(−1)
l# + r − 1
(n + r − 1)! l$ =0


$


#
$
l
!

! (−1)l$ +r−1−m (l# + r − 1 − m)! l# + r − 1
(k1 ,...,kr−1 ,kr −1)
×
Bm
m


j1 ! · · · jr−1 !
m=0 j +···+j

1
r−1
=l$ +r−1−m
×
$
n
!
(−1)a +r−1
a# + r
a$ =l$
#




$
 !
# 
n − l#
(1)
(1) (n − a )!
Bi1 · · · Bir
n − a# 
i1 ! · · · ir ! 
 i +···+ir

1
=n−a$
(put a# = a − r + 1)
#
$
n
!
n!
n+r−1
n−l$$
=
(−1)
n + r − 1 − l##
(n + r − 1)! l$$ =0


$$


#
$
$$
n−l
!

! (−1)n−l +r−1−m (n − l## + r − 1 − m)! n − l## + r − 1
(k1 ,...,kr−1 ,kr −1)
×
Bm
m


j1 ! · · · jr−1 !
 m=0 j1 +···+jr−1

=n−l$$ +r−1−m




#
$
$
n

# 
! (−1)a +r−1
!
l##
(1)
(1) (n − a )!
×
Bi1 · · · Bir
n − a# 
a# + r
i1 ! · · · ir ! 
 i1 +···+ir

a$ =n−l$$
=n−a$
##
#
(put l = n − l )
#
$
n
!
n!
n−l$$ n + r − 1
=
(−1)
l##
(n + r − 1)! l$$ =0


$$


#
$
$$ +r−1−m
n−l

n−l
##
!
! (−1)
(n − l + r − 1 − m)! n − l## + r − 1
(k1 ,...,kr−1 ,kr −1)
×
Bm
m


j1 ! · · · jr−1 !
 m=0 j +···+j

1
r−1
=n−l$$ +r−1−m
l$$
×
! (−1)n+r−1−a$$
n − a## + r
a$$ =0
#
##




$
 !
a## ! 
l##
(1)
(1)
Bi1 · · · Bir
a## 
i1 ! · · · ir ! 
 i +···+ir

1
#
(put a = n − a )
=a$$
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
r−1
= (−1)
×

$$

n−l
!

 m=0
#
$
n
!
n!
l$$ n + r − 1
(−1)
l##
(n + r − 1)! l$$ =0


#
$

! (−1)n−l$$ +r−1−m (n − l## + r − 1 − m)! n − l## + r − 1
(k1 ,...,kr−1 ,kr −1)
Bm
m

j1 ! · · · jr−1 !

+···+j
j1
r−1
=n−l$$ +r−1−m
l$$
×
12
!
$$
(−1)a
n − a## + r
a$$ =0




$
 !
a## ! 
l##
(1)
(1)
Bi1 · · · Bir
.
a## 
i1 ! · · · ir ! 
 i +···+ir

#
1
##
=a$$
#
(put a = n − a )
This shows the claim.
(2) Since
d
Lik ,k ,...,k ,1 (1 − e−t ) = Lik1 ,...,kr−1 (1 − e−t ),
dt 1 2 r−1
3 t
−t
Lik1 ,...,kr−1 ,1 (1 − e ) =
Lik1 ,...,kr−1 (1 − e−s )ds.
we have
0
Hence
∞
!
tn
n!
$r 3
Bn(k1 ,...,kr−1 ,1)
n=0
#
t
Lik1 ,...,kr−1 (1 − e−s )
=
(1 − e )
·
ds
(1 − e−s )r−1
0
*∞
+r 3 * ∞
+r−1 ∞
t
n−1
!
! (−s)n
!
t
sn
=
Bn(1)
−
Bn(k1 ,...,kr−1 ) ds
n!
n!
n!
0
n=0
n=1
n=0
r−1
= (−1)
et
et − 1
*∞
!
tn−1
Bn(1)
n!
n=0
−s r−1
+r 3
t
∞
!
0 n=r−1
!
∞
(−1)n sn ! (k1 ,...,kr−1 ) sm
B
ds.
j1 ! · · · jr−1 ! m=0 m
m!
j1 +···+jr−1
=n
Denote by I2 the integral part of the last equation.
3 t!
∞ !
∞
!
(−1)n n! (k1 ,...,kr−1 ) sn+m
I2 =
B
ds
j1 ! · · · jr−1 ! m
n!m!
0 m=0 n=r−1 j +···+j
1
=
3 t!
∞
∞
!
0 m=0 l=m+r−1
=
3
t
=n
!
r−1
j1 +···+jr−1
=l−m
(−1)l−m (l − m)! (k1 ,...,kr−1 )
l!
sl
Bn
· ds
j1 ! · · · jr−1 !
(l − m)!m! l!
(put l = n + m)
∞ l−r+1
!
!
! (−1)l−m (l − m)!
0 l=r−1 m=0
j1 +···+jr−1
=l−m
j1 ! · · · jr−1 !
(k1 ,...,kr−1 )
Bm
#
l
m
$
sl
ds
l!
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
∞ l−r+1
!
!
=
l=r−1 m=0
!
j1 +···+jr−1
=l−m
(−1)l−m (l − m)! (k1 ,...,kr−1 )
Bm
j1 ! · · · jr−1 !
#
l
m
$
tl+1
.
(l + 1)!
Thus,
(−1)r−1
*∞
!
tn−1
Bn(1)
n!
n=0
r−1
= (−1)
∞
!
!
n=0
×
= (−1)
i1 +···+ir
=n
∞ l−r+1
!
!
l=r−1 m=0
r−1
(1)
#
j1 +···+jr−1
=l−m
l
m
$
(1)
Bi1 · · · Bir
tn−r
i1 ! · · · ir !
(−1)l−m (l − m)! (k1 ,...,kr−1 )
Bm
j1 ! · · · jr−1 !
∞ !
∞ l−r+1
!
! !
(k1 ,...,kr−1 )
×Bm
= (−1)
I2
!
l=r−1 n=0 m=0
r−1
+r
(1)
Bi1
i1 +···+ir
=n
(1)
· · · Bir
n!
i1 ! · · · ir !
l=r−1 a=l+1 m=0
= (−1)
(1)
i1 +···+ir
=a−l−1
#
∞ !
a−1 l−r+1
!
! !
a=r l=r−1 m=0
(k1 ,...,kr−1 )
×Bm
r−1
= (−1)
#
l
m
$#
i1 +···+ir
=a−l−1
∞ n+r−1
!
! l−r+1
!
n=0 l=r−1 m=0
− l − 1)!
i1 ! · · · ir !
(1) (a
Bi1 · · · Bir
#
a
l+1
!
− l − 1)!
i1 ! · · · ir !
(1) (a
(1)
Bi1 · · · Bir
$
$
tl+1
(l + 1)!
(−1)l−m (l − m)!
j1 ! · · · jr−1 !
i1 +···+ir
=n−l+r−1
!
j1 +···+jr−1
=l−m
!
j1 +···+jr−1
=l−m
(−1)l−m (l − m)!
j1 ! · · · jr−1 !
(−1)l−m (l − m)!
j1 ! · · · jr−1 !
ta−r
a!
(1)
(1)
Bi1 · · · Bir
$#
$
tn
l
n+r
m
l+1
(n + r)!
(put n = a − r)
(k1 ,...,kr−1 )
×Bm
!
j1 +···+jr−1
=l−m
$
a!
ta
l
· · t−r
m (a − (l + 1))!(l + 1)! a!
(put a = n + l + 1)
r−1
l
m
tn+l+1
· t−r
n!(l + 1)!
∞
∞ l−r+1
!
!
! !
(k1 ,...,kr−1 )
×Bm
#
n!
i1 ! · · · ir !
!
j1 +···+jr−1
=l−m
(−1)l−m (l − m)!
j1 ! · · · jr−1 !
13
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46

∞ 

!
$ !
n+r−1 #
(−1)r−1 n! ! n + r
(1)
(1) (n − l + r − 1)!
=
Bi1 · · · Bir
l + 1 i +···+i
 (n + r)!
i1 ! · · · ir !
r
n=0 
l=r−1
1
=n−l+r−1


#
$
l−r+1
 tn
!
! (−1)l−m (l − m)!
l
(k1 ,...,kr−1 )
×
Bm
.
m j +···+j

j
!
·
·
·
j
!
n!
1
r−1

m=0
1
r−1
=l−m
If we compare the coefficients of both sides of the equation,


#
$
n+r−1
(−1)r−1 n! ! n + r  !
(1)
(1) (n − l + r − 1)! 
Bn(k1 ,k2 ,...,kr ) =
Bi1 · · · Bir


l+1
(n + r)!
i1 ! · · · ir !
i +···+i
l=r−1
×
l−r+1
!
m=0



!
j1 +···+jr−1
=l−m
r
1
=n−l+r−1
l−m

(−1) (l − m)! 

j1 ! · · · jr−1 !
#
l
m
$
(k1 ,...,kr−1 )
Bm


#
$
n+r−1
(−1)r−1 n! ! 1
n+r−1  !
(1)
(1) (n − l + r − 1)! 
=
Bi1 · · · Bir


l
(n + r − 1)!
l+1
i1 ! · · · ir !
i +···+i
l=r−1
×
l−r+1
!
m=0




!
j1 +···+jr−1
=l−m
(−1)l−m (l − m)! 

j1 ! · · · jr−1 !
r
1
=n+r−1−l
#
l
m
$
(k1 ,...,kr−1 )
Bm
#
$
: n+r ; n + r n+r−1
by l+1 =
( l )
l+1


#
$
n
(−1)r−1 n! ! 1
n+r−1  !
(1)
(1) (n − a)! 
=
Bi1 · · · Bir


(n + r − 1)!
a+r a+r−1
i1 ! · · · ir !
i +···+i
a=0

a
!

×

m=0
=
×
n−l$
!
m=0


!
!
j1 +···+jr−1
=a+r−1−m
(−1)r−1 n!
(n + r − 1)!

r
1
=n−a
(−1)a+r−1−m (a + r − 1 − m)! 

j1 ! · · · jr−1 !
(put a = l − r + 1)
n
!
l$ =0

#
$
#

a+r−1
m
$
(k1 ,...,kr−1 )
Bm

1
l# ! 
n+r−1
 !
(1)
(1)
B
·
·
·
B


i1
ir
n − l# + r n + r − 1 − l#
i1 ! · · · ir !
i +···+ir
j1 +···+jr−1
=n−l$ +r−1−m
(put l# = n − a)
=l$

#
$
(n − l + r − 1 − m)!  n − l# + r − 1 (k1 ,...,kr−1 )
Bm
.

m
j1 ! · · · jr−1 !
n−l$ +r−1−m
(−1)
1
#
14
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 7 (2007), #A46
15
This shows the claim.
(1)
Remark 8 Some recurrences exist for the original Bernoulli numbers Bn = Bn (n ≥ 0),
the case in which the upper index is fixed with the value (1). The problem of finding a
(k ,k ,...,k )
recurrence among Bn 1 2 r (n ≥ 0) for fixed (k1 , k2 , . . . , kr ), is therefore intriguing.
References
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