A summation on Bernoulli numbers
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1 Journal of Number Theory 111 ( A summaton on Bernoull numbers Kwang-Wu Chen Department of Mathematcs and Computer Scence Educaton, Tape Muncpal Teachers College, No. 1, A-Kuo West Road, Tape, Tawan 100, Republc of Chna Receved December 003; revsed 5 Aprl 004 Communcated by A. Granvlle Avalable onlne 7 December 004 Abstract In ths paper, our am s to nvestgate the summaton form of Bernoull numbers B n, such as n k=0 ( nk Bk+m. We derve some basc denttes among them. These numbers can form a Sedel matrx. The upper dagonal elements of ths Sedel matrx are called the medan Bernoull numbers. We determne the prme dvsors of ther numerators and denomnators. And we characterze ther ordnary generatng functon as the unque soluton of some functonal equaton. At last, we also obtan the contnued fracton representaton of ther ordnary generatng functon and ther value of Hankel determnant. 004 Elsever Inc. All rghts reserved. MSC: prmary 11B68; 15A15 Keywords: Medan Bernoull numbers; Sedel matrx; Hankel determnant 1. Introducton The Bernoull numbers B n are defned by the recurrence relaton m =1 ( m B m = 0 (m, B 0 = 1. E-mal address: kwchen@tmtc.edu.tw X/$ - see front matter 004 Elsever Inc. All rghts reserved. do: /j.jnt
2 K.-W. Chen / Journal of Number Theory 111 ( Thus we have B 0 = 1, B 1 = 1, B = 1 6, B 3 = 0, and so on. The Bernoull numbers have extensve applcatons on many areas. One of remarkable propertes s related to the values at ntegers of the Remann zeta functon ζ(s, such as ζ(n = ( 1 n+1 (πn B n, ζ(1 n = B n (n! n for postve ntegers n. Recently, Akyama and Tangawa [1] expressed the values ζ k (0,...,0, n and ζ k ( n, 0,...,0 of the Euler Zager s multple zeta functon, whch s defned by ζ k (s 1,...,s k = 1 n s 1 0<n 1 <n < <n k 1 n s n s k k as a lnear combnaton of B n+ for 1 k (See [1, Eqs. (6, (10]. However, the summaton formulae for these knds are very few n the lterature. We lst two man denttes of these knds (Refs. [6,9] n+1 ( n + 1 (k + n + 1B k+n = 0 (1 k k=0 and m k=0 ( m B k+n = ( 1 m+n k k=0 ( n B k+m. ( k In order to nvestgate the summatons of the form m k=0 ( m k Bk+n, the notaton of a Sedel matrx wll be llustrated n the followng: The Sedel matrx (a n,k n,k 0 assocated wth the ntal sequence a 0,n s defned by [ 5] a n,k = a n 1,k + a n 1,k+1 (3 or, equvalently, by a n,k = ( n a 0,k+. (4
3 374 K.-W. Chen / Journal of Number Theory 111 ( We usually denote the exponental generatng functon of the ntal sequence a 0,n as A(x = a 0,n x n. n! The ordnary generatng functon of the ntal sequence a 0,n s denoted as a(x = a 0,n x n+1. The ordnary generatng functon a(x s, n a formal sense, the Laplace transform of A(x, that s, 0 A(te t/x dt = a(x. The Bernoull polynomals B n (x are defned by te xt e t 1 = B n (xt n. n! It s easly to get B n = B n (0. The Sedel matrx (a n,k n,k 0 obtaned by takng the sequence of Bernoull numbers as the ntal sequence s represented as follows (Ref. Matrx (ES1 n [3]. 1 1/ 1/6 0 1/30 0 1/ / 1/3 1/6 1/30 1/30 1/4 1/4... 1/6 1/6 /15 1/15 1/105 1/ /30 1/15 8/105 4/ /30 1/30 1/105 4/ /4 1/1... 1/4 1/ Therefore we take a 0,n = B n n Eq. (4. Then a n,k = ( n B k+. We denote ths Sedel matrx as the BS-matrx. Some basc propertes of the BS-matrx gve Eqs. (1 and (. We wll provde more detal later n Secton.
4 K.-W. Chen / Journal of Number Theory 111 ( Three mportant sequences n the Sedel matrx are the ntal sequence (a 0,n, the fnal sequence (a n,0, and the upper dagonal sequence (a n,n+1. The medan Genocch numbers H n+1 (Refs. [3 5] could be defned by the upper dagonal elements of the Sedel matrx assocated wth the ntal sequence as the Genocch numbers G n, whch s defned by t e t + 1 = n=1 G n t n. n! Dumont [4] defned the medan Euler numbers R n, L n as the upper dagonal elements of the Sedel matrx assocated wth the ntal sequence as the Euler numbers E n, ( 1 n+1 E n, respectvely, whch s defned by sech (x + tanh(x = E n x n. n! Snce the ntal sequence B n and the fnal sequence B n (1 = ( 1 n B n of the BSmatrx are both well-known, n ths paper, we focus our attenton on the upper dagonal sequence, whch we call the medan Bernoull numbers, K n, n accord wth the above nomenclature n [4,5]. It can be seen that the man dagonal elements are just negatve twce of the upper dagonal elements. And ths wll be proved n Secton. Let p be a prme number. We call a ratonal number p-ntegral f ts denomnator s not dvded by p. Ifa and b are ratonal numbers, then by a b(mod p r we mean that (a b/p r s p-ntegral. For example, 1 5 (mod. We defne ord p(a to be the largest nteger for whch a/p ord p(a s p-ntegral; so ord ( 1 = 1 and ord ( 4 3 =. In Secton 3, we determne the prme dvsors of the denomnators D n of the medan Bernoull numbers K n and the order of of the numerators N n of K n. Tables 1 and gve the prme factorzatons of N n and D n for 0 n 30. Theorem 1.1. (1 The denomnators D n are square-free numbers. ( For a postve nteger n, the set of the all odd prme dvsors of D n s { p : odd prme n m (3 If s a dvsor of D n, then n = 0 or 1. (4 For n 1, ord (N n = [ n 1 p 1 n } m 1, m N+. ], where [x] means the largest nteger n x. In Secton 4, we characterze the ordnary generatng functon of K n as the unque soluton of some functonal equaton.
5 376 K.-W. Chen / Journal of Number Theory 111 ( Table 1 Medan Bernoull numerators N n factored n Prme factorzaton of N n Theorem 1.. Let m(x = K n x n+1 = 1 x x 1 15 x x x x6 (5 be the ordnary generatng functon of K n. Then m(x s the unque soluton of the functonal equaton x x ( x m + x + ( x m = x. (6 1 x x 1 + x
6 K.-W. Chen / Journal of Number Theory 111 ( Table Medan Bernoull denomnators D n factored n Prme factorzaton of D n In the last secton, we obtan the contnued fracton representaton of m(x, and we also obtan the Hankel determnant det 0,j n ( K+j. Theorem 1.3. For any non-negatve nteger n, the Hankel determnant of the medan Bernoull numbers s ( ( det K+j = 1 n+1 n 0,j n =1 ( ( n +1 (4 3(4 1. (7 (4 + 1
7 378 K.-W. Chen / Journal of Number Theory 111 ( Some basc relatons Consder the Sedel matrx (a n,k n,k 0 wth the ntal sequence B n. From Eq. (4 we have ( n a n,k = B k+. (8 Proposton.1. For k, n 0, a k,n = ( 1 k+n a n,k. (9 Proof. Ths result s followed on usng nducton on the frst ndex k n a k,n and the recursve relaton for a n,k. Substtute a n,k by Eq. (8 n Eq. (9, then we get Eq. (. Corollary.. For k, n 0, k ( k B n+ = ( 1 k+n From Proposton.1, t s clearly that a n,n+1 = a n+1,n. Hence ( n B k+. (10 a n,n = a n+1,n a n,n+1 = a n,n+1 = K n. (11 Therefore the man dagonal elements are just negatve twce of the upper dagonal elements. In the followng we gve the proof of Eq. (1. Corollary.3. For any non-negatve nteger n, we have n+1 ( n + 1 ( + n + 1B +n = 0. (1 Proof. Eq. (11 s equvalent to 1 n + 1 ( n B n+ = n+1 =1 ( n + 1 (n B n+ = B n+1. ( n B n+, 1 Multply (n + 1 on both sdes, then we get the desred dentty.
8 K.-W. Chen / Journal of Number Theory 111 ( Let k = n n Eq. (3 and we apply Eq. (11, then we have a recursve relaton wth K n. n 1 ( n 1 K n = K n 1 + B n++1. (13 Solvng ths recursve relaton we get, for n 1, n ( n 1 K n = B + j=0 j+1 ( j ( j + 1 B +j+3. (14 In the end of ths secton we lst two expressons of K n whch can be easly got from Eqs. (8 and (11. For n 0, K n = ( n B n+1+ (15 and K n = 1 ( n B n+. (16 3. Prme dvsors of the D n and N n We frst nvestgate the order of of K n. Proposton 3.1. For n and m n + 1, [ ] n 1 ord (a n,m =. (17 And ord (a 1,m = 1, for m. Proof. We use nducton on n. For n = 1 and m, a 1,m = B m + B m+1. Hence a 1,k = a 1,k 1 = B k. As a result ths mples ord (a 1,m = 1. For n = and m 3, we have a,k = B k + B k+ and a,k 1 = B k for k. Snce B t 1 (mod 4 for t (Ref. [8, p. 47], t could get B k + B k+ 1 (mod. Thus
9 380 K.-W. Chen / Journal of Number Theory 111 ( ord (a,k = 0. On the other hand, a,k 1 = B k 1 (mod 4, for k. Therefore ord (a,k 1 = 0. Assume that for n 3 and k<n, ord (a k,m = [ k 1 ] wth m k + 1. Then for m n + 1 a n,m = a n 1,m + a n 1,m+1 = a n,m + a n,m+ + a n,m+1. Now by the nducton hypothess we have [ ] n 1 ord (a n,m + a n,m+ [ ] n 1 and ord (a n,m+1 =. Ths completes our proof. Corollary 3.. For n we have [ ] n 1 ord (K n =. (18 The nformaton of Eqs. (9, (11, and Proposton 3.1 have been collected, the value of ord (a n,m for all n, m can be easly predcted. To descrbe the denomnators of K n we need the von Staudt Claussen Theorem: Theorem 3.3 (Ireland and Rosen [8, p. 33]. For m 1, B m = A m 1 p, (19 p 1 m where A m Z and the sum s over all prmes p such that p 1 m. Proposton 3.4. Let q be a prme number. Then for any non-negatve nteger n, qk n s q-ntegral and the denomnator D n of K n s a square-free number. Proof. Gven the fact that qb k s q-ntegral for any k and from the result of Eq. (15, we know that qk n s q-ntegral and therefore D n s square-free.
10 K.-W. Chen / Journal of Number Theory 111 ( Combne Eqs. (15 and (19, we can rewrte K n as K n = ( n B n+1+ m ( m A m+ 1 =1 = m 1 ( m 1 A m+ p 1 m+ p 1 m+ 1 f n = m, p 1 f n = m 1. p (0 It can be seen that the largest possble prme dvsor of D n s n + 1. Now we need a result whch s proved by Glasher n Lemma 3.5 (Granvlle [7]. For any gven prme q and ntegers 1 j,k q 1, we have 1 m n m j (mod q 1 ( n m for all postve ntegers n k(mod q 1. ( k (mod q (1 j Let q be a fxed odd prme number. Now we need only to know whether q s a dvsor of D n or not, for q 1 3q 5 n. Proposton 3.6. Gven a fxed odd prme q, f q 1 n q 1, then q s a dvsor of D n ; f q n 3q 5, then q s not a dvsor of D n. Proof. From Eq. (0 we need to consder whether the followng two factors: m =1 p 1 m+ ( m 1 p for n = m and m 1 p 1 m+ ( m 1 p for n = m 1
11 38 K.-W. Chen / Journal of Number Theory 111 ( are q-ntegral or not. Here, we frst deal wth the case for n = m. The method for the remanng case, n = m 1, s smlar, so we omt t. n q 1, then ths mples If q 1 So q cannot be a dvsor of ( m 1. Let n<n+ 1 q n + 1. C = { 1 m q 1 s a dvsor of m + }. Snce n q 1 n, there exsts an unque nteger 0 q m wth q 1 = m + q. If 1 q m, then q C. If q = 0, then q 1 = m s a dvsor of 4m. And = m C. Therefore the set C s not empty. If C contans at least two dstnct elements, then ths wll force the number q 1 s less than n. Ths contradcton ensures a fact that the number of the elements n C s one. The prme number q only occur once n the summaton m =1 p 1 m+ ( m 1 p, therefore q s a dvsor of D n. Now we assume that q n 3q 5. We can rewrte n = q + j, for 0 j q 5 If 1 q, then we have n + q + 1 m + n = q + j, q + 1 m + 3q 5. Dvdng (q 1 from m +, we have a remander r wth 3 r q 3. Thus q 1 s not a dvsor of m +. On the other hand, f m + 1 q, then we use the same trck and we also can prove that q 1 s not a dvsor of m +. Therefore f q 1 s a dvsor of m +, for some. Then both 1 and m + 1 must be less than q, and ths mples q ( m 1. Ths completes our proof. Now we come to the results and the followng theorem can be derved from them. Theorem 3.7. For a postve nteger n, the set of the all odd prme dvsors of D n s { p : odd prme n p 1 n },m N+. ( m m 1.
12 K.-W. Chen / Journal of Number Theory 111 ( Proof. For a fxed odd prme q, q s a dvsor of D m+k(q 1, for q 1 m q 1 and k 0. Hence for a fxed postve nteger n, wehave That s for k 0, q 1 + k(q 1 n = m + k(q 1 q 1 + k(q 1. n k + 1 q 1 n k Ordnary generatng functon and functonal equaton Proposton 4.1 (Dumont [4, Proposton 6]. Gven a Sedel matrx (a n,k n,k 0, then the ordnary generatng functon of ts ntal sequence, of ts man dagonal, and of ts upper dagonal, respectvely, denoted by a(x = a 0,n x n+1, d 0 (x = a n,n x n, d 1 (x = a n,n+1 x n+1 satsfy the dentty and Let m(x = b(x = a(x = x d 0 ( x 1 + x + d 1 ( x 1 + x. (3 B n x n+1 = x 1 x x x x7 K n x n+1 = 1 x x 1 15 x x x5. That s, b(x s the ordnary generatng functon of the ntal sequence B n, and m(x s the ordnary generatng functon of the upper dagonal sequence K n. Now we combne Eq. (11 and Proposton 4.1 and we get the relaton between these two functons n the followng theorem.
13 384 K.-W. Chen / Journal of Number Theory 111 ( Theorem 4.. We have ( b(x = 1 + ( x m. (4 x 1 + x Corollary 4.3. For k 1, we have k ( k n k n k n K n = B k, (5 k 1 ( k 1 n k 1 n k 1 n K n = B k 1. (6 Proof. Ths follows by extractng the coeffcent of x n n Eq. (4. The two equatons n the above corollary can be rewrtten as [ n ] ( n m n m n m K m = (δ 1n 1B n n 1. (7 m=0 Theorem 4.4. Let (x be a formal power seres. Then the followng three assertons are equvalent: t (1 (x = 0 snh(x e t/x dt; ( x ( (x and + x are both odd; ( 1 + ( x x x (3 = x. 1 x 1 + x Proof. ((1 (: Let f(x= 0 e t/x F(tdt. Recall that 0 e t/x t n dt = n!x n+1, so f(x s even (resp. odd, f and only f F(x s odd (resp. even. Note that Snce t snh t ( x + x = 1 + x 0 t coth te t/x dt. and t cosh t are both even, asserton ( follows mmedately.
14 K.-W. Chen / Journal of Number Theory 111 ( ( x (( (3: Snce + x s odd, we have 1 + x ( ( x x + x = x. 1 x 1 + x Asserton (3 follows from the fact that (x s odd. ((3 (1: Let (x = 0 e t/x F(tdt. Then ( x x = = 0 1 x ( x 1 + x e t/x F (t(e t e t dt. But x = 0 e t/x dt; hence F (t(e t e t = t and F(t = t snh t. t e t e t = Theorem 4.5. The followng denttes hold: ( x b(x =, (8 x + (x = 1 ( 4x x m 1 x. (9 Proof. Accordng to Theorem 4., t suffces to prove Eq. (8. Note that So te t snh t = t e t 1 = n B n t n. n! ( x = 1 x + 1 n+1 B n x n+1, whch clearly mples Eq. (8. The next corollary s an mmedate consequence of Theorem 4.4 and Theorem 4.5.
15 386 K.-W. Chen / Journal of Number Theory 111 ( Corollary 4.6. The formal power seres b(x and m(x are the unque solutons of the functonal equatons ( x b b(x = x (30 1 x and x x ( x m + x + ( x m = x, (31 1 x x 1 + x respectvely. The medan Bernoull numbers K n have certan connectons wth the number B n ( 1. Corollary 4.7. We have n+1 ( 1 n+1 K n = m=0 ( n ( 1 m m B m ( 1 m (3 and n B n ( 1 = m=0 ( n m+1 K m. (33 m Proof. Recall that t snh t = tet e t 1 = n B n ( 1 tn (n! for B n+1 ( 1 = 0. So, by applyng the formal Laplace transform, we get (x = n B n ( 1 xn+1. (34 Upon substtutng ths n Eq. (9, and replacng 4x /(1 x by y, weget ( y n+1 ( K n y n+1 = n B n ( y n 1 4 4
16 K.-W. Chen / Journal of Number Theory 111 ( = = y n+1 ( 1 n+1 n 1 y n+1 m=0 m=0 ( n ( 1 m m B m ( 1 m ( n ( 1 n+1+m m n 1 B m ( 1 m. Comparng the coeffcents of y n on the two sdes then yelds Eq. (3. Eq. (33 s followed mmedately usng the bnomal transform on Eq. (3. 5. Contnued fractons and Hankel determnants Rogers [11] found the contnued fracton expresson of (x as (x = x x 16x x = x 16x x x n 4 x... (35 + n Now usng Eq. (9 and replacng 4x /(1 x by y, we get the contnued fracton expanson for m(x. Proposton 5.1. The ordnary generatng functon m(x of the medan Bernoull numbers satsfes the contnued fracton representaton m(x = x x 16x x (x x (n 1 4 x + 4n 1 +. (36 (n 4 x (4n + 1(x Values of Hankel determnants of a sequence are known to be related to coeffcents of contnued fracton expansons of ts ordnary generatng functon. Proposton 5. (Krattenthaler [10, Theorem 11]. Let (μ n n 0 be a sequence of numbers wth generatng functon μ n t n+1 wrtten n the form μ n t n+1 = μ 0 t b 1 t b t.... ( a 0 t 1 + a 1 t 1 + a t Then the Hankel determnant det 0,j n (μ +j equals μ n+1 0 b n 1 bn 1 b n 1 b n.
17 388 K.-W. Chen / Journal of Number Theory 111 ( We need to change the representaton form of m(x n Proposton 5.1 as the form as Eq. (37. The followng two lemmas are needed. Lemma 5.3 (Dumont [4]. The followng representatons of a seres f(x are equvalent: f(x = x c 1 x c x c 3 x = x c 1 c x c 3 c 4 x 1 + c 1 x 1 + (c + c 3 x 1 + (c 4 + c 5 x.... Lemma 5.4. Let {b n } n 1 and {c n } n 1 be two sequences of complex numbers. If then, for n 1, we have 1 + θx + b 1x b x b 3 x b 4 x θx θx +... = 1 + c 1x c x c 3 x c 4 x..., ( c 1 = b 1 + θ, c 1 c = b 1 b, c n + c n+1 = b n + b n+1 + θ, c n+1 c n+ = b n+1 b n+. (39 Proof. Denote by A 0 (x and B 0 (x the left- and rght-hand sde of Eq. (38 and set and A 0 (x = 1 + θx + B 0 (x = 1 + b 1 x 1 + b x A (x c 1 x 1 + c. x B (x From the equalty A 0 (x = B 0 (x, we derve mmedately the followng: c 1 = b 1 + θ, c 1 c = b 1 b
18 K.-W. Chen / Journal of Number Theory 111 ( and A (x + b x = B (x + c x. The proof can then be readly completed by nducton. ( Now we can obtan the Hankel determnant det 0,j n K+j. Theorem 5.5. The Hankel determnant of the medan Bernoull numbers s ( ( det K+j = 1 n+1 n 0,j n =1 ( ( n +1 (4 3(4 1. (40 (4 + 1 Proof. We frst rewrte m(x as the form as the left-hand sde of Eq. (38 m(x = x b 1 x 1 + θx b x b 3 x b 4 x 1 + θx θx +..., where θ = 1 4 and for n 1, b n 1 = b n = Then from Lemmas 5.3 and 5.4 we have and for n 1, (n 1 4 4(4n 3(4n 1, (n 4 4(4n 1(4n + 1. m(x = x c 1 x c x c 3 x = x c 1 c x c 3 c 4 x 1 + c 1 x 1 + (c + c 3 x 1 + (c 4 + c 5 x... c 1 = b 1 + θ = 1 3, c 1 c = b 1 b = 1 40, c n + c n+1 = b n + b n+1 + θ = 8n4 + 8n 3 + 6n + n 1, (4n + 3(4n 1
19 390 K.-W. Chen / Journal of Number Theory 111 ( c n+1 c n+ = b n+1 b n+ = Applyng Proposton 5., we have ( det (K +j = 1 n+1 n (c 1 c n+1 0,j n =1 n+1 n ( = ( 1 =1 (n (n (4n + 1(4n + 3 (4n + 5. ( (4 3(4 1 (4 + 1 n +1. In vew of Eqs. (3 and (33, t seems that the Hankel determnant of B n ( 1 shall have the smlar formula. Corollary 5.6. det (B +j ( 1 0,j n = n ( ( n +1 (4 3(4 1. (41 (4 + 1 =1 Proof. Upon usng Eqs. (34 and (35, we have n B n ( 1 x(n+1 = x x 16x 3 4 x n 4 x n Thus n B n ( 1 xn+1 = x x 16x n 4 x n Now we use the smlar method as the proof of Theorem 5.5 and let 4x be y, then we get our concluson. Remark 5.7. It s worth notng that ( n+1 det (B +j ( 1 0,j n = det 0,j n (K +j. (4
20 Acknowledgments K.-W. Chen / Journal of Number Theory 111 ( The author would lke to thank the referee for some useful comments and suggestons. Ths research was supported by the Natonal Scence Foundaton of Tawan, Republc of Chna, Grant NSC M References [1] S. Akyama, Y. Tangawa, Multple zeta values at non-postve ntegers, Ramanujan J. 5 (4 ( [] K.-W. Chen, Applcatons related to the generalzed Sedel matrx, Ars Combn. (003 1pp, accepted for publcaton. [3] D. Dumont, Matrces d Euler-Sedel, n: Sém. Lothar. Combn., 5-éme Sesson, 1981, Publ. IRMA Strasbourg, vol. 18/s-04, 198, pp [4] D. Dumont, Further trangles of Sedel Arnold type and contnued fractons related to Euler and Sprnger numbers, Adv. Appl. Math. 16 (3 ( (do: /aama [5] D. Dumont, J. Zeng, Further results on the Euler and Genocch numbers, Aequatones Math. 47 ( [6] I.M. Gessel, Applcatons of the classcal umbra calculus, Algebra Unversals 49 ( [7] A. Granvlle, Arthmetc propertes of bnomal coeffcents, Canad. Math. Soc. Conf. Proc. 0 ( [8] K. Ireland, M. Rosen, A Classcal Introducton to Modern Number Theory, Sprnger, New York, 198. [9] M. Kaneko, A recurrence formula for the Bernoull numbers, Proc. Japan Acad. Ser. A Math. Sc. 71 ( [10] C. Krattenthaler, Advanced determnant calculus, Sém. Lothar. Combn. vol. 4 (The Andrews Festschrft, 1999, artcle B4q, 67pp. [11] L.J. Rogers, On the representaton of certan asymptotc seres as convergent contnued fractons, Proc. London Math. Soc. 4 ( (
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