Congruences for central factorial numbers modulo powers of prime

Transcription

1 DOI /s RESEARCH Open Access Congruences for central factorial numbers modulo powers of prime Haiqing Wang * and Guodong Liu *Correspondence: whqlcf@163.com Department of Mathematics, Huizhou University, Huizhou , Guangdong, People s Republic of China Abstract Central factorial numbers are more closely related to the Stirling numbers than the other well-known special numbers, and they play a major role in a variety of branches of mathematics. In the present paper we prove some interesting congruences for central factorial numbers. Keywords: Central factorial numbers of the first kind, Central factorial numbers of the second kind, Congruence, Stirling numbers Mathematics Subject Classification: 11B68, 11B73, 05A10, 11B83 Introduction and definitions Central factorial numbers are more closely related to the Stirling numbers than the other well-known special numbers, such as Bernoulli numbers, Euler numbers, trigonometric functions and their inverses. Properties of these numbers have been studied in different perspectives see Butzer et al. 1989; Comtet 1974; Liu 2011; Merca 2012; Riordan Central factorial numbers play a major role in a variety of branches of mathematics see Butzer et al. 1989; Chang and Ha 2009; Vogt 1989: to finite difference calculus, to approximation theory, to numerical analysis, to interpolation theory, in particular to Voronovskaja and Komleva-type expansions of trigonometric convolution integrals. The central factorial numbers tn, k k Z of the first kind and Tn, k k Z of the second kind are given by the following expansion formulas see Butzer et al. 1989; Liu 2011; Riordan 1968 n x [n] tn, kx k k0 1 and n x n Tn, kx [k], k0 2 respectively, where x [n] xx + n 2 1x + n 2 2 x + n 2 n + 1, n N 0 : N {0}, N being the set of positive integers, Z being the set of integers Wang and Liu. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original authors and the source, provide a link to the Creative Commons license, and indicate if changes were made.

2 Page 2 of 14 It follows from 1 that tn, k tn 2, k n 22 tn 2, k 3 with n x x x 2 n 1 2 t2n,2kx 2k 2. 4 Similarly, 2 gives Tn, k Tn 2, k k2 Tn 2, k 5 with x 2k 1 x 2 1 2x 2 1 kx 2 T2n,2kx 2n 6 and k!tn, k k k n 1 i i 2 i. i0 7 Several papers obtain useful results on congruences of Stirling numbers, Bernoulli numbers and Euler numbers see Chan and Manna 2010; Lengyel 2009; Sun 2005; Zhao et al But only a few of congruences on central factorial numbers for odd prime moduli which can be found in Riordan 1968, p For example, let t n x n k0 tn, kx k, then t p x x p x mod p, 8 t p+k x t p x t k x 9 Conclusions In the present paper we prove some other interesting congruences for central factorial numbers. In Congruences for Tap m 1 p 1 + r, k modulo powers of prime p section, some congruence relations for Tap m 1 p 1 + r, k modulo powers of prime p are established. For a is odd, m, k N and k 2 m 1 a, we prove that { k!t2 m 1 2 k 1 mod 2 a, k m, k 0 mod 4, 2 k 1 mod 2 m, k 2 mod 4. For p is odd prime, m, a, k N, r N 0, k p 1 and r < p m 1 p 1, in Congruences for Tap m 1 p 1 + r, k modulo powers of prime p section we also show that Tap m 1 p 1 + r, k Tr, k mod p m, 1 r < p m 1 p 1

3 Page 3 of 14 and k!tap m 1 p 1, k 1 k 2 +1 k k 2 mod p m, k is even. In Congruences for t2ap m,2k and T2n,2ap m modulo powers of p section, congruences on t2ap m,2k and T2n,2ap m modulo powers of p are derived. Moreover, the following results are obtained: 1 for a, k, m N, b N 0 and 2 m 1 a k 2 m a, we prove a congruence for t2 m+1 a + 2b,2k mod 2 m ; 2 for a, n, m N, b N 0 and n 2 m a, we prove a congruence for T2n,2 m+1 a + 2b mod 2 m ; 3 for p is a odd prime number and a, k, m N, b N 0, we deduce a congruence for t2ap m + 2b,2k mod p m ; 4 for p is a odd prime number, a, n, m N, b N 0, we deduce a congruence for T2n,2ap m + 2b mod p m. Congruences for Tap m 1 p 1 + r, k modulo powers of prime p Theorem 1 For a is odd, m, k N and k 2 m 1 a, we have k!t2 m 1 a, k { 2 k 1 mod 2 m, k 0 mod 4, 2 k 1 mod 2 m, k 2 mod Proof Using Euler s Theorem, ϕ2 m 2 m 1. Therefore, by Fermat s Little Theorem, we get c ϕ2m c 2m 1 1 mod 2 m if c is odd. Observe that, when c is even, c 2m 1 0 mod 2 m. Then by 7, if k 0 mod 4, we yield k!t2 m 1 a, k i0 If k 2 mod 4, we have k k 2 m 1 1 i a i 2 i This completes the proof of Theorem 1. k!t2 m 1 a, k, i odd 1 i k i 2 k 1 mod 2 m. k k 2 m 1 1 i a i 2 i i0 i0, i even 1 i k i 2 k 1 mod 2 m.

4 Page 4 of 14 Remark By Theorem 1 and 5, we readily get k!t2 m 1 a + 2, k { k 2 k 3 mod 2 m, k 0 mod 4, k 2 k 3 mod 2 m, k 2 mod Theorem 2 For p is odd prime, m, a, k N, r N 0, k p 1 and r < p m 1 p 1, we have T ap m 1 p 1 + r, k Tr, k mod p m, 1 r < p m 1 p 1, 12 k!t ap m 1 p 1, k 1 k k 2 +1 k2 mod p m, k is even. 13 Proof By Euler s Theorem and Fermat s Little Theorem, we get a ϕpm a pm 1 1 mod p m if a, p 1, where a, p is the greatest common factor of a and p. Then by 7 and noting that k 2i, p 1, we get k!t ap m 1 p 1 + r, k k k ap m 1 1 i +r i 2 i i0 k k r 1 i i 2 i Observe that k!, p 1. Hence, T ap m 1 p 1 + r, k Tr, k mod p m. i0 k!tr, k mod p m. The proof of 12 is complete. If r 0, then k is even. Therefore, k!t ap m 1 p 1, k k k ap m 1 1 i i 2 i i0 k 1 i 1 k k 2 k i 2 i0 1 k 2 +1 k k 2 mod p m. The proof of 13 is complete. This completes the proof of Theorem 2. As a direct consequence of Theorem 2, we have the following corollary. Corollary 3 For p is odd prime, a, k N and r N 0, we have Tap 1 + r, p { 0 mod p, 3 r p 2, 1 mod p, r 1. 14

5 Page 5 of 14 Tap 1 + r, p 1 { 0 mod p, 1 r p 1, 1 mod p, r Tp + 2, k + 2 Tp, k 0 mod p, 3 k p 1. T2p + 2, k + 2 T2p, k 0 mod p, 4 k p 1. Proof By setting m 1 in 12 and using 5, we have Tap 1 + r, p Tap 1 + r 2, p 2 Tr 2, p 2 0 mod p, 3 r p 2, Tap 1 + 1, p Tap 1 1, p 2 Tp 2, p 2 1 The proof of 14 is complete. Setting m 1 and k p 1 in 12, we can readily get Tap 1 + r, p 1 0 Setting m 1 and k p 1 in 13, and noting that 1 j j 1 mod p j 0, 1, 2,..., p 1, p 1! 1 mod p, we have Tap 1, p 1 1 The proof of 15 is complete. If m 1 and a r in 12, then Trp, k Tr, k 18 Taking r 1, 2 in 18 and using 5, we immediately get 16 and 17. This completes the proof of Corollary 3. Congruences for t2ap m,2k and T2n,2ap m modulo powers of p To establish the main results in this section, we need to introduce the following lemmas. Lemma 4 If m N, then 2 m 1 1 2i 1x 2 1 x 2 2m 1 mod 2 m, 2 m 1 1 2ix 2 1 mod 2 m Proof We prove this lemma by induction on m. We see that 19 is true for m 1. Assume that it is true for m 1, 2,..., j 1. Then

6 Page 6 of 14 2 j 1 2i 1x 2 2 j 1 1 2i 1x j + 2i 1x 2 2 j 1 [ 1 2i 1x j+1 x 2 2 j 1 + 2i 1 1 2i 1x 2] 2 j 1 1 2i 1x 2 2 mod 2 j+1. For any polynomials Ax, Bx, we have Ax Bx mod 2 m Ax 2 Bx 2 mod 2 m+1, so we obtain the desired result. That is, 2 j 2 j 1 1 2i 1x 2 1 2i 1x 2 The proof of 19 is complete. Similarly, we can prove 20 as follows. 2 1 x 2 2j mod 2 j+1. 2 j 1 2ix 2 2 j 1 1 2ix j + 2ix 2 2 j 1 [ 1 2ix j+1 x 2 2 j 1 + 2i1 2ix 2] 2 j 1 1 2ix 2 1 mod 2 j+1. 2 mod 2 j+1 This completes the proof of Lemma 4. Similarly, we can get the following results. Lemma 5 If m N, then 2 m 1 2 m 1 x 2 2i 1 2 x 2 1 2m 1 mod 2 m, x 2 2i 2 x 2m mod 2 m. We are now ready to state the following theorems

7 Page 7 of 14 Theorem 6 Let a, b, k, m N and 2 m 1 a k 2 m a, then t2 m+1 a,2k 1 k 2m 1 a 2 m 1 a k 2 m 1 a mod 2 m, 23 t2 m+1 a + 2b,2k t2 m+1 a,2jt2b,2k 2j mod 2 m. j1 24 Proof By 4 and Lemma 5, we find that 2 m a t2 m+1 a,2kx 2k 2 x 2 2 m a 2 x x 2 2 m a 1 2 x 2 2 m a 2. Thus 2 m a 2 m t 2 m+1 a,2k x 2k x 2 i 2 2 m 1 a x 2 2i 1 2 x 2 m 1 a 1 x 2 m a 2 m 1 a k0 2 m a k2 m 1 a 1 k 2 m 1 a k x 2 2i 2 2 2m 1 x 2m a+2k 2 1 k m 1 a k 2 m 1 x 2k mod 2 m. a a This completes the proof of 23. For 24, we can prove this as follows. 2 m a+b t 2 m+1 a + 2b,2k x 2k 2 x x 2 2 m a 2 x 2 2 m a x 2 2 m a + b 1 2 x x 2 2 m a 2 x x 2 b m a b t 2 m+1 a,2k x 2k t2b,2kx 2k 2 2 m a+b k2 j1 t 2 m+1 a,2j t 2b,2k 2j x 2k 2 mod 2 m. This completes the proof of Theorem 6.

8 Page 8 of 14 Remark t Taking a 1 and k 2 m 1,2 m 1 + 1, 2 m in 23, we readily get t 2 m+1,2 m 1 mod 2 m, 2 m+1,2 m m 1 mod 2 m, t 2 m+1,2 m m 2 mod 2 m, m 3. Theorem 7 Let a, b, n, m N and n 2 m a, then n 2 T 2n,2 m+1 m 1 a 1 a n 2 m a mod 2 m, 25 T 2n,2 m+1 a + 2b n j0 T 2j,2 m+1 a T2n 2j,2b mod 2 m. 26 Proof By 6 and Lemma 4, we have 2 m a T 2n,2 m+1 a x 2n x 2 1 ix 2 2 m a x 2 1 ix 2 x 2m+1 a 1 2 m 1 1 2i 1x2 2 m 1 1 2i2 x 2m+1 a 1 1 x 2 2m 1 a n + 2 m 1 a 1 n n 2 m 1 a 1 n 2 m a n2 m a x 2m+1 a+2n x 2n mod 2 m. This completes the proof of 25. For 26, we can prove this as follows. 2 m a+b T 2n,2 m+1 a + 2b x 2n x 2 1 ix 2 2 m a x 2 b 1 ix 2 x 2 1 ix 2 T 2n,2 m+1 a x 2n T2n,2bx 2n n j0 a T 2j,2 m+1 a T2n 2j,2bx 2n mod 2 m.

9 Page 9 of 14 This completes the proof of Theorem 7. Remark Taking a 1 and n 2 m + 1, 2 m + 2 in 25, we readily get T 2 m+1 + 2, 2 m+1 2 m 1 mod 2 m, T 2 m+1 + 4, 2 m+1 2 m 2 mod 2 m, m 3. Lemma 8 If p is a odd prime number and m N, then p m 1 ix 2 1 x 2p m 1 mod p m. 27 Proof Apparently, by Lagrange congruence, we have 1 x1 2x 1 p 1x1 px 1 x mod p and 1 + x1 + 2x 1 + p 1x1 + px 1 x Thus 1 x 2 1 2x 2 1 px 2 1 x 2 Hence 27 is true for the case m 1. Suppose that 27 is true for some m 1. Then for the case m + 1, p m+1 1 ix 2 p m p m 1 ix 2 1 p m + ix 2 1 2p m + ix 2 1 p 1p m + ix 2 1 ix 2 p 1 ix 2 jp m 2 + 2jp m ix j1 + terms involving powers of p 2m and higher ]. For any prime p and polynomials Ax, Bx, we have Ax Bx mod p m. This implies that Ax p Bx p mod p m+1. With j1 jpm 2 + 2jp m ix 0 mod p m+1, we obtain the desired result. That is, p m+1 1 ix 2 p m 1 ix 2 This completes the proof of Lemma 8. p 1 x 2p m mod p m+1.

10 Page 10 of 14 Similarly, we get the following results. Lemma 9 If p is a odd prime number and m N, then p m Theorem 10 x 2 i 2 x p x 2pm 1 mod p m. Let p is a odd prime number and a, b, k, m N, then t2ap m,2k 1 2k 2apm 1 2ap m 1 2k 2ap m 1 mod p m, t2ap m + 2b,2k t2ap m,2jt2b,2k 2j mod p m, j1 30 where 2k 2ap m 1 mod p 1. Proof By 4 and Lemma 9, we find that ap m t2ap m,2kx 2k 2 x 2 ap m 2 x x 2 ap m 1 2 x 2 ap m 2. Thus ap m t2ap m,2kx 2k p m x 2 i 2 x p x 2apm 1 2ap m 1 k0 ap m where 2k 2ap m 1 mod p 1. This completes the proof of 29. By 4 we get ap m +b a 1 k 2ap m 1 kap m 1 1 t2ap m + 2b,2kx 2k 2 k x 2apm 1 +k 2k 2ap m 1 2ap m 1 2k 2ap m 1 x 2k mod p m, x x 2 ap m 2 x 2 ap m x 2 ap m + b 1 2 x x 2 ap m 2 x x 2 b 1 2 ap m t2ap m,2kx 2k ap m +b k2 b t2b,2kx 2k 2 t2ap m,2jt2b,2k 2jx 2k 2 mod p m. j1

11 Page 11 of 14 This completes the proof of Theorem 10. Remark Taking a 1 and 2k 2p m 1,2p m 1 + p 1,2p m 1 + 2p 1 in 29, we readily get t2p m,2p m 1 1 mod p m, t2p m,2p m 1 + p 1 2p m 1 mod p m, t2p m,2p m 1 + 2p 1 2p 2m 2 p m 1 mod p m Obviously, t2p,2 1 mod p, 34 t2p, p If u, v N,1 u < 2p v. By setting m 1, a p v,2k 2p v + up 1 in 29, and noting that p j 1 mod p j 0, 1, 2,..., p 1 with ip+r jp+s i r j s mod p i j, we have 2p t2p v+1,2p v + up 1 1 u v 0 36 u The following corollary is a direct consequence of Theorem 10. Corollary 11 Let p be a odd prime and α be a positive integer. Then for any 1 k p α p 1, we have tp α p 1,2k { 1 mod p, 2k 0 mod p α 1 p 1, 0 mod p, otherwise. 37 Proof Let m 1, 2a p α 1 p 1 in 29 of Theorem 10. Then we have p α 2 tp α p 1,2kx 2k p α 1 j0 p 1 j α 1 p 1 j x pα 1 +j By the Lucas congruence, we obtain p α 1 p 1 p 1 j mod p, j 0 mod p α 1, j p α 1 0 mod p, otherwise. With p 1 1 j 1 j mod p j 0, 1, 2,..., p 1,

12 Page 12 of 14 we can deduce p α 2 tp α p 1,2kx 2k j0 p j1 p j1 1 j p α 1 p 1 p α 1 j p 1 1 j 1 j 1 x pα 1 j which is obviously equivalent to 37. The following theorem includes the congruence relations for T2n,2ap m and T2n,2ap m + 2b. x pα 1 j+1 x pα 1 j mod p, Theorem 12 If p is a odd prime number, a, b, n, m N, then T2n,2ap m 2n 2ap m 1 1 2n 2ap m mod p m 38 and n T2n,2ap m + 2b T2j,2ap m T2n 2j,2b mod p m, j0 39 where 2n 2ap m mod p 1. Proof By 6 and Lemma 8, we have T2n,2ap m x 2n ap m p m x 2 1 ix 2 x 2 1 ix 2 a x 1 2apm 1 x 2apm 1 n + 2ap m 1 1 x 2apm +n n 2n 2ap m 1 1 x 2n mod p m. 2n 2ap m nap m This completes the proof of 38. For 39, we can prove this as follows.

13 Page 13 of 14 ap m +b T2n,2ap m + 2bx 2n x 2 1 ix 2 ap m x 2 b x 2 1 ix 2 1 ix 2 T2n,2ap m x 2n T2n,2bx 2n This completes the proof of Theorem 12. j0 n T2j,2ap m T2n 2j,2mx 2n mod p m. Remark Taking a 1 and 2n 2p m + p 1,2p m + 2p 1 in 38, we have T2p m + p 1,2p m 2p m 1 mod p m, T2p m + 2p 1,2p m 2p 2m 2 + p m 1 mod p m Obviously, T3p 1, 2p 2 T4p 2, 2p 3 mod p, If u N 0, v N. By setting m 1, a p v 1,2n 2p u+v in 38, and noting that ip+r jp+s i r j s mod p i j, we have 2p T2p u+v,2p v v 1 u i0 p i 1 2p v p v 1 u i0 p i ui0 p i That is, 2p v ui0 p i ui0 p i 2 1 T2p u+v,2p v 1 44 Authors contributions Both authors read and approved the final manuscript. Competing interests The authors declare that they have no competing interests. Received: 8 October 2015 Accepted: 18 March 2016

14 Page 14 of 14 References Butzer PL, Schmidt M, Stark EL, Vogt L 1989 Central factorial numbers; their main properties and some applications. Numer Funct Anal Optim 10: Chang CH, Ha CW 2009 Central factorial numbers and values of Bernoulli and Euler polynomials at rationals. Numer Funct Anal Optim 303 4: Comtet L 1974 Advanced combinatorics: the art of finite and infinite expansions translated from the French by J. W. Nienhuys. Reidel, Dordrecht and Boston Chan O-Y, Manna DV 2010 Congruences for Stirling numbers of the second kind. Contemp Math 517: Liu GD 2011 The D numbers and the central factorial numbers. Publ Math Debr 791 2:41 53 Lengyel T 2009 On the 2-adic order of Stirling numbers of the second kind and their differences. In: DMTCS Proc AK, pp Merca Mircea 2012 A special case of the generalized Girard Waring formula. J Integer Seq, Article , 15:7 Riordan J 1968 Combinatorial identities. Wiley, New York Sun ZW 2005 On Euler numbers modulo powers of two. J Number Theory 115: Vogt L 1989 Approximation by linear combinations of positive convolution integrals. J Approx Theory 572: Zhao J, Hong S, Zhao W 2014 Divisibility by 2 of Stirling numbers of the second kind and their differences. J Number Theory 140: