Algorithms for Bernoulli and Related Polynomials

Transcription

1 Journal of Integer Sequences, Vol. 10 (2007, Article Algoriths for Bernoulli Related Polynoials Ayhan Dil, Veli Kurt Mehet Cenci Departent of Matheatics Adeniz University Antalya, Turey adil@adeniz.edu.tr vurt@adeniz.edu.tr cenci@adeniz.edu.tr Abstract We investigate soe algoriths that produce Bernoulli, Euler Genocchi polynoials. We also give closed forulas for Bernoulli, Euler Genocchi polynoials in ters of weighted Stirling nubers of the second ind, which are extensions of nown forulas for Bernoulli, Euler Genocchi nubers involving Stirling nubers of the second ind. 1 Introduction Priary Concepts Bernoulli nubers B n, n Z, n 0, originally arise in the study of finite sus of a given power of consecutive integers. They are given by B 0 1, B 1 1/2, B 2 1/6, B 4 1/30,..., with B n 0 for odd n > 1, B n 1 n 1 n+1 +1 B for all even n 2. In the sybolic notation, Bernoulli nubers are given recursively by { (B +1 n 1, if n 1; B n 0, if n > 1. 1

2 with the usual convention about replacing B j by B j after expansion. The Bernoulli polynoials B n (x, n Z, n 0, can be expressed in the for ( n B n (x (B +x n B x n. (1 The generating functions of Bernoulli nubers polynoials are respectively given by t e t 1 t n B n n!, te xt e t 1 B n (x tn n!, for t < 2π ([2]. The Euler polynoials E n (x, n Z, n 0, ay be defined by the generating function [23, 24] 2e xt e t +1 E n (x tn n!, (2 for t < π. The Euler nubers E n are defined by ( 1 E n 2 n E n, n 0. 2 The Genocchi nubers polynoials, G n G n (x, n Z, n 0, are defined respectively as follows [8, p.49]: 2t e t +1 t n G n n!, (3 2te xt e t +1 G n (x tn n!, for t < π, which have several cobinatorial interpretations in ters of certain surjective aps of finite sets [10, 12, 13]. The well-nown identity G n 2(1 2 n B n, n 0, shows the relation between Genocchi Bernoulli nubers. Fro (2 (3, it is easy to see that G n ne n 1 (0, n 0, with E 1 : 0. Several arithetical properties of Bernoulli, Euler Genocchi polynoials can be obtained fro the generating functions. We list here three of the, which will be used in the next section. B n (1+x B n (x+nx n 1, n 0, (4 E n (1+x E n (x+2x n, n 0, (5 G n (1+x G n (x+2nx n 1, n 0. (6 2

3 Bernoulli, Euler Genocchi nubers can be directly coputed fro generating functions by dividing nuerator to denoinator (after exping e t coparing the coefficients on either sides. Besides this ethod, any recurrence forulas for Bernoulli, Euler Genocchi nubers are obtained [3, 6, 8, 9, 16, 18, 19, 22]. Alternative ethods for coputing these nubers are developed as well. In this paper, we concern two of the, the Euler-Seidel atrices the Aiyaa-Tanigawa algorith. Definition 1.1. ([11] Let a 0, a 1, a 2,...,a n,... be an initial sequence. The Euler-Seidel atrix corresponding to the sequence {a n } is deterined by the sequence { a n}, whose eleents are recursively given by a 0 n a n (n 0, a n a 1 n +a 1 n+1 (n 0, 1. (7 The sequence {a 0 n} is the first row the sequence { a 0} is the first colun of the atrix. Fro the recurrence relation (7, it can be seen that a n i0 ( a 0 i n+i, (n 0, 1. (8 The first row colun can be deterined fro (8 as follows: a n 0 a 0 n i i i0 i0 a 0 i, n 0, ( 1 n i a i 0, n 0. (9 Thus, as the first row is given, the first colun can be found vice versa. Next two propositions show the relationship between the generating functions of a 0 n a n 0. Proposition 1.1. Let a(t a 0 nt n be the generating function of the initial sequence {a 0 n}. Then the generating function of the sequence {a n 0} is given by a(t a n 0t n 1 ( t 1 t a. 1 t The proof of this proposition is given by Euler [14]. Proposition 1.2. Let A(t a 0 t n n n! 3

4 be the exponential generating function of the initial sequence {a 0 n}. Then the exponential generating function of the sequence {a n 0} is given by A(t a n 0 t n n! et A(t. The proof of this proposition is given by Seidel [25]. Of notable interest of Euler-Seidel atrices is the exaples for Bernoulli related nubers. Let A(t t e t 1. Then A(t tet e t 1 t+ t e t 1, thus A(t t+a(t. Therefore, a 0 0 B 0 1, a , a0 n a n 0 for n 2, the corresponding Euler-Seidel atrix is For the initial sequence is A(t 2t e t +1, a 0 0 0, a 0 1 1, a 0 2n+1 0 a 0 2n G 2n for n 1. Since it is seen that A(t 2tet e t +1 2t 2t e t +1 2t A(t, a 0 0 0, a 1 0 1, a 2n a 2n 0 G 2n for n 1. 4

5 Thus, the corresponding Euler-Seidel atrix is In their study of values at nonpositive integer arguents of ultiple zeta functions, Aiyaa Tanigawa [1] found an algorith for coputing Bernoulli nubers in a anner siilar to Pascal s triangle for binoial coefficients. The algorith is as follows: Starting with the sequence {1/n}, n 1, as the 0-th row, the -th nuber in the (n+1-st row a n+1 is deterined recursively by a n+1 (+1 ( a n a n +1, for 0,1,2,... n 0,1,2,... Then, a n 0 of each row is the n-th Bernoulli nuber B n with B 1 1/2. Kaneo [20] reforulated the Aiyaa-Tanigawa algorith as follows: Proposition 1.3. Given an initial sequence a 0 with 0,1,2,... define sequence a n for n 1 recursively by a n (+1 ( a n 1 a+1 n 1. Then the leading eleents are given by a n 0 ( 1!S(n+1,+1a 0. Here, S(n, denotes the Stirling nubers of the second ind which are defined as follows [8, Chap. 5]: (e t 1 S(n, tn! n!. (10 n For 1 n, S(n, > 0 for 1 n <, S(n, 0. The Stirling nubers of the second ind S(n, satisfy the recurrence relation S(n, S(n 1, 1+S(n 1, (11 for n, 1 with S(n,0 S(0, 0, except S(0,0 1. If the initial sequence is a 0 1/(+1, 0, in Proposition 1.3, then the leading eleents are Bernoulli nubers with B 1 1/2. By replacing the initial sequence a 0 5

6 1/(+1 by a 0 1/(+1, 0, applying sae algorith, Kaneo also derived the resulting sequence as poly-bernoulli nubers [20]. Chen [7] changed the recursive step in Proposition 1.3 to proved the following: a n a n 1 (+1a n 1 +1 (n 1, 0, Proposition 1.4. Given an initial sequence a 0 with 0,1,2,... define sequence a n for n 1 recursively by a n a n 1 (+1a+1. n 1 Then a n 0 ( 1!S(n,a 0. Given initial sequences 1/(+1, 0, 1/2, 0, ( 1 [/4] 2 [/2] (1 δ 4,+1, 0, Chen obtained the leading eleents respectively as Bernoulli, Euler tangent nubers. Here, [x] is the greatest integer x δ 4,i 1 if 4 i δ 4,i 0 otherwise. Tangent nubers are defined as follows (cf. [24, p. 35]: tgt ( 1 n+1 T 2n+1 t 2n+1, (2n+1! with T 0 1. At this point, we note that by choosing the initial sequence a 0 1/2, 0, we obtain the leading eleents a0 n 1 as G n /n for n 1, which is not obtained in [7]. For other studies on the Aiyaa-Tanigawa algorith, see [17, 21]. In this paper, we give the Euler-Seidel atrices the Aiyaa-Tanigawa algoriths for Bernoulli, Euler Genocchi polynoials. These atrices algoriths are polynoial extensions of the corresponding atrices algoriths for Bernoulli, Euler Genocchi nubers. In particular, the Aiyaa-Tanigawa algoriths for these polynoials lead new closed forulas for Bernoulli, Euler Genocchi polynoials in ters of weighted Stirling nubers of the second ind. 2 Euler-Seidel Matrices for Bernoulli, Euler Genocchi Polynoials We start with the polynoial extension of Definition 1.1. Definition 2.1. Let a 0 (x, a 1 (x, a 2 (x,...,a n (x,... be initial sequence of polynoials in x. The Euler-Seidel atrix corresponding to the sequence {a n (x} is deterined by the sequence { a n(x } for n 0 1, whose eleents are recursively given by a 0 n(x a n (x, a n(x an 1 (x+an+1(x. 1 6

7 Fro the recurrence relation above, we have a n(x a n 0 (x a 0 n(x ( i i i i0 i0 i0 a 0 n+i(x, n 0, a 0 i (x, n 0, (12 ( 1 n i a i 0(x, n 0. Let A(x,t A(x,t be the exponential generating functions of a 0 n(x a n 0 (x, respectively. Then, we have A(x,t e t A(x,t. Note that for x 0, the above equalities reduce to Duont s. Now, we construct the Euler-Seidel atrices corresponding to Bernoulli, Euler Genocchi polynoials. Let a 0 n(x B n (x, n 0. Then Using (4, we obtain A(x,t B n (x tn n! text e t 1, A(x,t e t A(x,t te(x+1t e t 1 B n (x+1 tn n!. (13 A(x,t A(x,t+te xt, a n 0 (x B n (x+nx n 1, n 0. Therefore, given Bernoulli polynoials as the first row, then the first colun of the corresponding Euler-Seidel atrix is again Bernoulli polynoials except the opposite sign for the coefficient of the second highest power. This is because a n 0 (x B n (x+nx n 1 ( n B x n +nx n 1 B 0 x n +B 1 x n 1 +nx n 1 + B 0 x n n 2 xn 1 +nx n 1 + B 0 x n + n 2 xn ( n B x n 2 ( n B x n 2 B x n,

8 by using (1 B 1 1/2. The Euler-Seidel atrix corresponding to Bernoulli polynoials is as follows: 1 x 1 x 2 x+ 1 x x2 + 1x 2 x+ 1 x 2 1 x x2 1x x 2 +x+ 1 x x2 1x x x2 + 1x 2 Now let a 0 n(x E n (x, n 0. Then Using (5, we get A(x,t E n (x tn n! 2ext e t +1, A(x,t e t A(x,t 2e(x+1t e t +1 E n (x+1 tn n!. (14 A(x,t A(x,t+2e xt, a n 0 (x E n (x+2x n, n 0. Thus, if Euler polynoials are taen as the first row, then the first colun of the corresponding Euler-Seidel atrix is negative Euler polynoials except the coefficient of the highest power. The atrix is 1 x 1 2 x 2 x x x x+ 1 2 x x x2 x+ 1 4 x 2 +x x x2 x 1 4 x x2 1 4 Finally, let a 0 n(x G n (x, n 0. Then A(x,t G n (x tn n! 2text e t +1, Using (6 yields A(x,t e t A(x,t 2te(x+1t e t +1 G n (x+1 tn n!. (15 A(x,t A(x,t+2te xt, a n 0 (x G n (x+2nx n 1, n 0. 8

9 The corresponding atrix is 0 1 2x 1 3x 2 3x 1 2x 3x 2 x 1 2x+1 3x 2 +x 1 3x 2 +3x We conclude this section by giving alternative proofs of the equations Fro (13, we see that B n (x+1 E n (x+1 G n (x+1 ( n B (x, n 0, ( n E (x, n 0, ( n G (x, n 0. a n 0 (x tn n! A(x,t B n (x+1 tn n!. Thus, by coparing the coefficients of power series on both sides, we get a n 0 (x B n (x+1. By (12, we have B n (x+1 a n 0 (x ( n a 0 (x ( n B (x. Other two equalities can be proved in the sae way by using appropriate relations (14 or (15. 3 The Aiyaa-Tanigawa Algoriths for Bernoulli, Euler Genocchi Polynoials In this section, we derive the Aiyaa-Tanigawa algoriths for Bernoulli, Euler Genocchi polynoials give closed forulas for these polynoials in ters of weighted Stirling nubers of the second ind. The weighted Stirling nubers of the second ind, S(n,, x, are defined as follows [4, 5]: e xt (e t 1! n S(n,,x tn n!. (16 9

10 When x 0, S(n,,0 S(n, are the Stirling nubers of the second ind. For n 0, S(0,0,x 1, for 1 < n, S(n,,x 0 for other values of n, S(n,, x are deterined by the following recurrence forula: S(n+1,,x (+xs(n,,x+s(n, 1,x. (17 The relationship between S(n,,x S(n, can be given by the following equation: Lea 3.1. We have S(n,,x n 0 ( n x S(n,. Proof. By (16 (10, we have n S(n,,x tn n! ext (e t 1 e xt S(n, tn! n! n x ntn t n+ S(n+, n! (n+! ( x S(n +, t n+.! (n +! 0 Coparing the coefficients of power series yields the result. Given an initial sequence a 0 with 0,1,2,..., let the sequence a n for n 1 be given recursively by a n a n 1 (+1a+1 n 1 (18 as in Proposition 1.4. Let g n (t be the generating function of a n, g n (t We define a n (x g n (x,t by eans of a n t. (19 a n (x (x+a n 0 g n (x,t (x+g(t n 0 x a n, n 0, (20 x g n (t, n 0, (21 where we use the usual convention about replacing (a j by a j (g(t j by g j (t in the binoial expansions. Note that for x 0, (20 reduces to (18 (21 reduces to (19. 10

11 Proposition 3.1. Given an initial sequence a 0, with 0,1,2,..., let the sequence a n for n 1 be given as (18 a n (x be given as (20. Then a n 0 (x ( 1!S(n,,xa 0, (22 where S(n,,x are the weighted Stirling nubers of the second ind. Proof. Fro (21 (19, we have ( n g n (x,t x g n (t 0 By the recurrence forula (18, we get g n (x,t Applying the forula (cf. [15, p. 310] we have x x 0 { a n 1 (+1a n 1 +1 t +1 x (+1a n 1 +1 t x (t 1 ( (t 1 dt d n n g n (x,t 0 ( n x a n t. } (+1a n 1 +1 t (+1a n 1 +1 t x (t 1 d dt (g n 1(t x ( (t 1 d dt n g 0 (t. S(n,(t 1 ( d dt, n ( d x S(n,(t 1 g 0 (t. dt Setting t 0 interchanging suations, we obtain a n 0 (x ( 1!a 0 Using Lea 3.1, the result follows. 11 n 0 ( n x S(n,.

12 Theore 3.1. Let a 0 1/(+1, 0, in (22. Then the leading polynoials a n 0 (x, n 0, are given by a n 0 (x ( 1! S(n,,x B n (x. +1 Proof. We have ( ( 1! S(n,,x +1 ( 1! +1 This proves the theore. n t n n! S(n,,x tn n! ( 1! e xt (e t 1 +1! e xt (1 e t 1 e t 1 e xt 1 e t ( log ( 1 ( 1 e t text e t 1 B n (x tn n!. Theore 3.2. Let a 0 1/2, 0, in (22. Then the leading polynoials a n 0 (x, n 0, are given by a n 0 (x ( 1! S(n,,x E 2 n (x. Proof. We have ( ( 1! S(n,,x t n 2 n! ( 1! S(n,,x tn 2 n! n ( 1! e xt (e t 1 2! ( 1 e e xt t 2ext 2 e t +1 E n (x tn n!. Coparing the coefficients of power series yields the result. Theore 3.3. Let a 0 1/2, 0, as in Theore 3.2. Then the polynoials a n 0 (x, n 0, are given by (n+1a n 0 (x ( 1 +1 (+1! S(n+1,+1,x G 2 +1 n (x. 12

13 Proof. The proof follows fro Theore 3.2 the relations G n (x ne n 1 (x, n 0, S(n,0,x 0. 4 Acnowledgent The authors would lie to than the anonyous referee for valuable coents. This wor was supported by Adeniz University Scientific Research Project Unit. References [1] S. Aiyaa Y. Tanigawa, Multiple zeta values at non-positive integers, Raanujan J. 5 (2001, [2] T. M. Apostol, Introduction to Analytic Nuber Theory, Undergraduate Texts in Matheatics, Springer-Verlag, [3] M. Can, M. Cenci, V. Kurt, A recurrence relation for Bernoulli nubers, Bull. Korean Math. Soc. 42 (2005, [4] L. Carlitz, Weighted Stirling nubers of the first second inds I, Fibonacci Quart. 18 (1980, [5] L. Carlitz, Weighted Stirling nubers of the first second inds II, Fibonacci Quart. 18 (1980, [6] C.-H. Chang C.-W. Ha, On recurrence relations for Bernoulli Euler nubers, Bull. Austral. Math. Soc. 64 (2001, [7] K.-W. Chen, Algoriths for Bernoulli nubers Euler nubers, J. Integer Seq. 4 (2001, Article [8] L. Cotet, Advanced Cobinatorics. The Art of Finite Infinite Expansions, Revised enlarged edition, D. Riedel Publishing Co., Dordrecht, [9] E. Y. Deeba D. M. Rodriguez, Stirling s series Bernoulli nubers, Aer. Math. Monthly 98 (1991, [10] D. Duont, Interprétations cobinatoires des nobres de Genocchi, Due Math. J. 41 (1974, [11] D. Duont, Matrices d Euler-Seidel, Séinaire Lotharingien de Cobinatoire, 1981, Paper B05c. (Forerly Publ. I.R.M.A. Strasbourg, 1982, 182/S-04, pp [12] D. Duont G. Viennot, A cobinatorial interpretation of Seidel generation of Genocchi nubers, Ann. Discrete Math. 6 (1980,

14 [13] D. Duont A. Rrianarivony, Dérangeents et nobres de Genocchi, Discrete Math. 132 (1994, [14] L. Euler, De transforatione serieru, Opera Onia, series pria, Vol. X, Teubner, [15] R. Graha, D. Knuth, O. Patashni, Concrete Matheatics, Addison-Wesley, [16] F. T. Howard, Applications of a recurrence for Bernoulli nubers, J. Nuber Theory 52 (1995, [17] Y. Inaba, Hyper-sus of powers of integers the Aiyaa-Tanigawa atrix, J. Integer Seq. 8 (2005, Article [18] C. Jordan, Calculus of Finite Differences, Chelsea, New Yor, [19] M. Kaneo, A recurrence forula for the Bernoulli nubers, Proc. Japan Acad. Math. Sci. Ser. A 71 (1995, [20] M. Kaneo, The Aiyaa-Tanigawa algorith for Bernoulli nubers, J. Integer Seq. 3 (2000, Article [21] D. Merlini, R. Sprugnoli, M. C. Verri, The Aiyaa-Tanigawa transforation, Integers: Electronic Journal of Cobinatorial Nuber Theory 5 (1 (2005, Paper A5. [22] H. Moiyaa, A new recurrence forula for Bernoulli nubers, Fibonacci Quart. 39 (2001, [23] N. E. Nörlund, Méoire sur les polynôes de Bernoulli, Acta Math. 43 (1922, [24] N. E. Nörlund, Vorlesungen Über Differenzenrechnung, Chelsea, New Yor, [25] P. L. von Seidel, Über eine einfache Entstehungsweise der Bernoullischen Zahlen und einiger verwten Reihen, Sitzungsberichte der Münch. Aad. Math. Phys. Classe 7 (1877, Matheatics Subject Classification: Priary 11B68. Keywords: Bernoulli polynoials, Euler polynoials, Genocchi polynoials, weighted Stirling nubers of the second ind, Euler-Seidel atrices, Aiyaa-Tanigawa algorith. (Concerned with sequences A000110, A000182, A000364, A001898, A027641, A027642, A Received Septeber ; revised version received May Published in Journal of Integer Sequences, May Return to Journal of Integer Sequences hoe page. 14