A note on the denominators of Bernoulli numbers - Project Euclid

No. 5]

Proc. Japan Acad., 90, Ser. A (2014)

71

A note on the denominators of Bernoulli numbers By Takao K OMATSU,Þ Florian LUCAÞ,Þ and Claudio (Communicated by Shigefumi MORI,

Abstract:

M.J.A.,

DE

J. P ITA RUIZ V. Þ

April 14, 2014)

We show that

gcdð2!Sð2n þ 1; 2Þ; . . . ; ð2n þ 1Þ!Sð2n þ 1; 2n þ 1ÞÞ ¼ denominator of B2n ; where Sðn; kÞ is the Stirling number of the second kind and Bn is the Bernoulli number. Key words: Bernoulli numbers; Stirling numbers.

1. Introduction. Let Sðn; kÞ be the Stirling number of the second kind, which counts the number of partitions of a set with n elements in k e kÞ ¼ k!Sðn; kÞ. disjoint nonempty subsets. Put Sðn; Let Bm be the mth Bernoulli number. Theorem 1. The formula e e gcdðSð2n þ 1; 2Þ; . . . ; Sð2n þ 1; 2n þ 1ÞÞ ¼ denominator of B2n ; holds. It is interesting to note that there are already classical formulas expressing the Bernoulli number in terms of Stirling numbers such as B2n ¼

2nþ1 X k¼1

¼1

ð1Þk1 ðk  1Þ!Sð2n þ 1; kÞ k 2nþ1 X k¼2

e ð1Þk Sð2n þ 2; kÞ 2 k

(see, for example, Chapter 1 in [2]). We shall not use this formula in our argument. Proof. We use the von Staudt–Clausen theorem ([1,3]) which states that Y denominator of B2n ¼ p: ðp1Þj2n

We first show that the right–hand side divides each e of the numbers Sð2n þ 1; kÞ. Let p be a prime such that p  1 j 2n. To proceed, we recall that 2010 Mathematics Subject Classification. Primary 11B68, 11B73. Þ Graduate School of Science and Technology, Hirosaki University, Hirosaki, Aomori 036-8561, Japan. Þ Mathematical Institute, UNAM Juriquilla, 76230 Santiago de Quere´taro, Me´xico. Þ School of Mathematics, University of the Witwatersrand, P. O. Box Wits 2050, South Africa. Þ Universidad Panamericana, Me´xico DF, Me´xico.

doi: 10.3792/pjaa.90.71 #2014 The Japan Academy

Sðn; kÞ ¼

  k 1X k n j ; ð1Þkj k! j¼0 j

to derive that ð1Þ

e Sð2n þ 1; kÞ ¼

k X j¼0

ð1Þkj

  k 2nþ1 : j j

Let j 2 f1; . . . ; 2ng. By Fermat’s Little Theorem and since p  1 j 2n, we get ð2Þ

j2nþ1  j (mod pÞ:

Hence, inserting the above congruence (2) for j ¼ 1; 2; . . . ; k into (1), we get   k X k e Sð2n þ 1; kÞ  j (mod pÞ: ð1Þkj j j¼0 However, the last sum above   k X kj k ð3Þ j¼0 ð1Þ j j¼0 for k  2 as it can be seen by putting x ¼ 1 into the identity d kðx  1Þk1 ¼ ðx  1Þk dx   ! k d X kj k xj ¼ ð1Þ dx j¼0 j   k X kj k ¼ jxj1 : ð1Þ j j¼0 This shows that the right-hand side divides the lefthand side. Now we show that the left-hand side divides the right–hand side. Note that in the left-hand side, e the last term inside the gcd is Sð2n þ 1; 2n þ 1Þ ¼ ð2n þ 1Þ!, which implies that every prime p dividing the left-hand side satisfies p  2n þ 1. Let pt be the

72

T. KOMATSU, F. LUCA and C.

exact power of p appearing in the left-hand side. We show that t ¼ 1 and that ðp  1Þ j 2n, statements which together imply the desired conclusion. We then have   k X kj k j2nþ1  0 (mod pt Þ; ð4Þ ð1Þ j j¼0 for k ¼ 2; . . . ; 2n þ 1. Making k ¼ 2 in (4) above we get ð5Þ

 2 þ 22nþ1  0 (mod pt Þ:

Making k ¼ 3 in (4) above we get ð6Þ

3  3  22nþ1 þ 32nþ1  0 (mod pt Þ;

and inserting also (5) into (6), we get the congruence 32nþ1  3 (mod pt Þ. So, let us show by induction on k ¼ 1; 2; 3; . . . ; 2n þ 1 that k2nþ1  k (mod pt Þ. Assume that k  4 and that the above congruences are satisfied for 1; 2; . . . ; k  1. Formula (4) together with the induction hypothesis implies that   k1 X kj k j þ k2nþ1  0 (mod pt Þ ð1Þ j j¼0 which together with the identity (3) gives that

DE

J. PITA RUIZ V.

[Vol. 90(A),

k2nþ1  k (mod pt Þ. Hence, it is indeed the case that k2nþ1  k (mod pt Þ; for k ¼ 1; . . . ; 2n þ 1. Making k ¼ p, we get that p2nþ1  p (mod pt Þ, showing that t ¼ 1. Finally, since p  2n þ 1, it follows that 1; 2; . . . ; 2n þ 1 cover all residue classes modulo p, therefore we have a2nþ1  a (mod pÞ for all integers a. In particular, a2n  1 (mod pÞ for all integers a coprime to p, implying that ðp  1Þ j 2n because the multiplicative group modulo p is cyclic of order p  1. This concludes the proof of the theorem.  Acknowledgements. The authors thank the referee for suggestions which improved the quality of this paper. Work by T. K. was done during his visit to Mexico in Fall, 2013. F. L. was supported in part by Project PAPIIT IN104512. References [ 1 ] T. Clausen, Theorem, Astron. Nachr. 17 (1840), 351–352. [ 2 ] H. Rademacher, Topics in analytic number theory, Springer, New York, 1973. [ 3 ] K. G. C. von Staudt, Beweis eines Lehrsatzes, die Bernoullischen Zahlen betreffend, J. Reine Angew. Math. 21 (1840), 372–374.