M
ASANOBU
K
ANEKO
Poly-Bernoulli numbers
Journal de Théorie des Nombres de Bordeaux, tome
9, n
o1 (1997),
p. 221-228
<http://www.numdam.org/item?id=JTNB_1997__9_1_221_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Poly-Bernoulli
numberspar MASANOBU KANEKO
RÉSUMÉ. Par le biais des séries
logarithmiques
multiples,
nousdéfinissons
l’analogue
enplusieurs
variables des nombres deBer-noulli. Nous démontrons une formule
explicite
ainsi qu’unthéo-rème de dualité pour ces nombres. Nous donnons aussi un
théo-rème de type von Staudt et une nouvelle preuve d’un théorème
de Vandiver.
ABSTRACT.
By using polylogarithm
series,
we define"poly-Ber-noulli numbers" which
generalize
classical Bernoulli numbers. We derive anexplicit
formula and aduality
theorem for these num-bers,together
with a vonStaudt-type
theorem for di-Bernoulli numbers and anotherproof
of a theorem of Vandiver.For every
integer k,
we define a sequence of rational numbers(n
=0,1, 2, ~ ~ ~ ),
which we refer to aspoly-Bernoulli
numbers,
by
Here,
for anyinteger k,
Lik (z)
denotes the formal power series(for
the k-thpolylogarithm
if k > 1 and a rational function if k0)
When k =
1,
is the usual Bernoulli number(with
Bi1)
=1/2):
,..-v
and
1,
the lefthand side of(1)
can be written in the form of"iterated
integrals" :
In this paper, we
give
both anexplicit
formula forB~~~
in terms of theStirling
numbers of the second kind and a sort ofduality
fornegative
indexpoly-Bernoulli
numbers. Both formulas areelementary,
and in fact almostdirect consequences of the definition and
properties
of theStirling
numbers. Asapplications,
we prove a vonStaudt-type
theorem for di-Bernoullinum-bers
(I~
=2)
andgive
an alternativeproof
of a theorem due to Vandiver on a congruence for1.
Explicit
formula andduality
An
explicit
formula forB~~~
isgiven by
thefollowing:
THEOREM 1.where
is the
Stirling
numberof
the second kind.REMARK. When k =
1,
the theorem and its many variants are classicalresults in the
study
of Bernoulli numbers(cf.
(1]).
Because the
Stirling
numbers areintegers,
we see from the formula thatfor k 0 is an
integer
(actually
positive,
as demonstrated in theremark at the end of this
section).
THEOREM 2. For anyn, k > 0,
we havePROOF OF THEOREMS 1 AND 2. One way to define the
Stirling
numbers of the second kindS(n, m) (n >
0, 0
mn)
is via the formulawhere,
for eachinteger
m >0,
we denoteby
thepolynomial
x(x
-1)(x-2)
(-m+1)
((x)o
=1).
Thenthey
satisfy
thefollowing
formulas(when n
= 0 in(3),
theidentity
00
= 1 isunderstood):
For other
definitions,
furtherproperties
andproofs,
we refer to[3].
Now Theorem 1 is
readily
derived from the definition(1)
and the formula(4).
Infact,
Hence the theorem follows.
To prove Theorem
2,
we calculate thegenerating
function ofB~ ’~~ :
The last
expression
being symmetric
in xand y
yields
Theorem 2. REMARK. Since2. Denominators of di-Bernoulli numbers
Using
Theorem1,
we cancompletely
determine the denominator of di-Bernoulli numbers as follows.THEOREM 3.
(1)
When n isodd,
B~2~ _ -~n42 Bnl~l
(n >
1). (Hence
thedescrip-tion
of
the denominator reduces to the classical Clausen-von Staudttheorem.)
(2)
When n is even(> 2),
thep-order
ord(p, n) of
for
aprime
number p is
given
asfollows.
(a) ord(p, n) >
0if p
> n + 1.(b)
we have:(i) ord(p, n) =
-2if p -
(ii)
If p -
1An,
then:(A)
0 or n - n’ modp(p - 1)
for
somme 1 n’ p - 1.(B) ord(p, n) =
-1 otherwise.(c)
ord(3, n) >
0if n -
2 mod 3 and n > 2. Otherwiseord(3, n) =
-2.(d) ord(2, n) >
0if n -
2 mod 4 and n > 2.ord(2, n) =
-1if
n - 0 mod 4.
ord(2,2) =
-2.Before
proving
thetheorem,
we establish thefollowing
lemma,
which will be needed in theproof.
LEMMA 1. Assume n > 2 is even and
p >
5 is aprime
numbers such thatm + 1
= 2p.
Thenand hence
m)/(m
+1)2
isp-integral.
PROOF.By
(3),
Since
the last sum is
equal
toNoting
thatwe see that
because p - 1 1
(recall
that n is even and p isodd).
This proves the lemma.PROOF OF THEOREM 3. 1. Let
Bn
= forn ~
1 andBl
=-1/2.
ThenE"O 0
Bn Xn /n! = x/(eX - 1).
By
(2)
in theintroduction,
we haveFrom this we see that
Since
jg"’ -
B,
= 0 for odd 1 >3,
we have for odd n2. We make use of Theorem 1. Part
(a)
is obvious because theStirling
numbers in the formula in Theorem 1 are
integers.
For the remainder ofthe
proof,
first we note that theexpression
m! / (m + 1 ) 2
in the summand of the formula is aninteger
except
when m + 1 =8, 9,
aprime
number,
or 2 xa
prime number,
as can be checked in anelementary
way.Now,
Lemma 1 tells us that anyprime
number p >
5satisfying
m + 1 =2p
does not appearOur next task is to consider the case m + 1 = p, where p is a
prime
number > 5. In this case
The
righthand
side iscongruent
modulo p to -1 if p -Iln
and to 0 ifp - 1
In.
Thusif p -
11n,
thep-order
ofm) / (m
+1)2
is -2.Since the other summands in the formula in Theorem 1 are
p-integral,
wehave shown
part
(b)-i.
Suppose
p - and calculate modulop .
Using
we see that
- , ,
It is known that
(cf.
Cor. ofProp.
15.2.2 in[2])
if n is even and p - 1Xn,
then
On the other
hand,
when we put n modp - 1 = n’,1 n’ p - 1
(since
both n and p - 1 are even, n’ is also
even),
we find=
B;~1~
mod p(see (63)
of Vandiver[4]
and Section 3below).
We therefore have
where m + 1 = p and n’ - n mod p -
1,1
n’
p - 1. Since p - 1aen,
the number is
p-integral
and mod p(Prop.
15.2.4 and Th.5following
it in[2]).
Thusr 1
This
readily gives
part
(b)-ii
of the theorem.The
only
summands in Theorem 1 which may not be3-integral
are2!S(n,
2)/32,
-5!S(n,
5)/62,
and8!S(n,
8)/92.
By
direct calculationusing
the formula(3),
we obtainpart
(c).
In a similar manner, we can determine3. A theorem of Vandiver
As an
application
of Theorems 1 and2,
we prove thefollowing
proposi-tion
originally
due to Vandiver.PROPOSITION. Let p be an odd
prime
number. For 1 i p -2,
PROOF.
By
Theorem 1 and Fermat’s littletheorem,
we see that ¿4 B.. ,- B..Theorem 2 says that the
righthand
side isequal
to whichby
Theorem 1 isequal
2,
rra)(rn,
+1)2.
LEMMA 2.
Suppose
p is an oddprime,
and 1 ~ m p - 2. ThenPROOF. The
Stirling
numberssatisfy
the recurrence formulaThus if we
put
2, m)
= bm,
weget
But
by
(3),
and we thus conclude that
This
together
with the relationbi
=S(p - 2,1)
= 1gives
the lemma and hencecompletes
theproof
of theproposition.
REMARK. If i >
1,
therighthand
side of theproposition
iscongruent
modulo p toand this
being
congruent
toBi
(even
when i =1)
is aspecial
case ofVandiver’s congruence
(63)
in[4].
Acknowledgement
The
present
paper was writtenduring
the author’sstay
in 1993 at theuniversity
ofCologne
inGermany,
as a research fellow of the Alexander von Humboldt foundation. He would like to thank the foundation and his hostprofessor
Peter Schneider for theirhospitality
andsupport.
REFERENCES
[1]
Gould,
H. W. :Explicit
formulas for Bernoullinumbers,
Amer. Math.Monthly
79
(1972), 44-51.
[2]
Ireland, K. and Rosen, M. : A Classical Introduction to Modern NumberTheory,
second edition.Springer
GTM 84(1990)
[3]
Jordan,
Charles : Calculus of FiniteDifferences,
Chelsea Publ.Co.,
NewYork,
(1950)
[4]
Vandiver,
H. S. : Ondevelopments
in an arithmetictheory
of the Bernoulliand allied
numbers,
Scripta
Math. 25(1961), 273-303
Masanobu KANEKO
Graduate School of Mathematics
Kyushu University
33Fukuoka 812-81,