Southest Asin Bulletin of Mthemtics (2013 37: 795 800 Southest Asin Bulletin of Mthemtics c SEAMS. 2013 Congruences for Stirling Numbers of the Second Kind Modulo 5 Jinrong Zho School of Economic Mthemtics, Southwestern University of Finnce nd Economics, Chengdu 61007, P.R. Chin Emil: mthzjr@fomil.com Received 10 Februry 2013 Accepted 28 July 2013 Communicted by K.P. Shum AMS Mthemtics Subject Clssifiction(2000: 11B73, 11A07 Abstrct. Let n nd k be positive integers nd S(n,k be the Stirling numbers of the second kind. In this pper, the uthor estblishes the congruences for S(n, k modulo 5 in terms of binomil coefficients. Keywords: Stirling number of the second kind; Congruence; Binomil coefficient 1. Introduction The Stirling numbers re common topics in number theory nd combintorics. Let N denote the set of nturl numbers. The Stirling numbers of the first kind, denoted by s(n, k (with lower-cse s, count the number of permuttions of n elements with k disjoint cycles. The Stirling numbers of the second kind S(n,k (with cpitl S is defined forn N nd positiveinteger k n s the numberofwystoprtitionset ofnelementsintoectlyk non-emptysubsets. One cn chrcterize the Stirling numbers of the first nd the second kind by ( n k0 s(n,kk nd n k0 S(n,k( k, respectively, where ( n is the flling fctoril ( n ( 1...( n+1. The Stirling numbers of the first nd second kind cn be considered to inverse of one nother: m(j,k l0 s(l,js(k,l δ jk nd m(j,k l0 S(l,js(k,l δ jk,
796 J.R. Zho where δ jk is the Kronecker delt. Divisibility is n importnt nd interesting topic in number theory. See [5], [6] nd [12] for some interesting results on this topic. For emple, the elements of n rbitrry gcd-closed set of the form pqr were considered nd studied in [6]. Divisibility properties of Stirling numbers hve been studied from number of different perspectives. Given prime p nd positive integer m, there eist unique integers nd m, with p nd n 0, such tht m p n. The number n is clled the p-dic vlution of m, denoted by n v p (m. The numbers min{v p (k!s(n,k : m k n} re importnt in lgebric topology, see [2]. Lengyel [9] studied the 2-dic vlutions of S(n, k nd conjectured, proved by Wnnemcker [11], tht v 2 (S(2 n,k s 2 (k 1, where s 2 (k mens the bse 2 digitl sum of k. Lengyel [10] showed tht if 1 k 2 n, then v 2 (S(c2 n,k s 2 (k 1 for ny positive integer c. Amdeberhn, Mnn nd Moll [1] suggested tht v 2 (S(2 n + 1,k + 1 s 2 (k 1, which confirmed by Hong, Zho nd Zho [7]. Congruence is nother centrl topic in the field of Stirling numbers of the second kind. It is known tht for ech fied k, the sequence {S(n,k,n k} is periodic modulo prime powers. The length of this period hs been studied by Crlitz [3] nd Kwong [8]. Chn nd Mnn [] chrcterized S(n, k modulo prime powers in terms of binomil coefficients when k is multiple modulus. In this pper, we estblishes the congruences for S(n,k modulo 5 in terms of binomil coefficients. For ny rel number, we denote by the biggest integer no more thn. In fct, we hve the following min results. Theorem 1.1. Let n nd be positive integers with n 5. Then the following congruences re true modulo 5: { (n if n (mod, (i S(n, 5 1 (ii S(n,5+1 ( n 1. 2 (n if n (mod, (iii S(n,5+2 3 (n 3 if n 1 (mod, (n 2 if n 2 (mod, (n if n (mod, (n 3 (iv S(n,5+3 if n 1 (mod, {(n (v S(n,5+ if n (mod, In the net section, we will show Theorem 1.1.
Congruences for Stirling Numbers of the Second Kind Modulo 5 797 2. Proof of Theorem 1.1 At first, we will use the elementry property of S(n,k s follows. For ech fied k, S(n, k hs generting function (i. By (1, we hve S(n,k n S(n,5 n 5 k 1 i. (1 ( 5 1 i 1 i 5( 1 5 1 (1 ( 5 ( j j j0 ( 1+j j+5 (mod 5. (2 j j0 Collecting powers nd reindeing in (2, we derive tht if n (mod, then ( 1+ n 5 ( n S(n,5 n 5 (mod 5, 1 nd if n (mod, then S(n,5 0 (mod 5. So prt (i is proved. (ii. From (1, we cn deduce tht S(n,5+1 n m0 5+1 r1 ( 5 1 i 1 i 1 ( S(m,5 m( r n 1 S(t,5 n (mod 5. (3 t0 Then by (3 nd the definition of Stirling numbers of the second kind, we get n 1 S(n,5+1 S(t,5 t0 5 t n 1 j ( mod n 1 t5 It then follows from ( nd prt (i tht ( t S(n,5+1 1 n 1 s ( s 1 1 S(t,5 (mod 5. ( 5 t n 1 t+s n 1 5 j0 ( t 1 1 ( j + 1 1 (mod 5. (5
798 J.R. Zho Thus by (5 nd pplying the identity c ( ( b+j b+1+c b b+1 j0 with b 1 nd c n 1 5, we obtin ( n 1 S(n,5+1 (mod 5 s required. Prt (ii is true. (iii. By (1, we get tht (5+5 S(n,5+2 n (1 3(1 (1 5 1 i 3 It follows tht ( S(m,5(+1 m 1+3+2 2 m0 S(n+3,5(+1 n +3 S(n+2,5(+1 n +2 3 S(n+1,5(+1 n (mod 5. S(n,5+2 S(n+3,5(+1+3S(n+2,5(+1 +2S(n+1,5(+1 (mod 5. (6 On the other hnd, by prt (i we derive tht S(n+3,5(+1 S(n+2,5(+1 S(n+1,5(+1 if n+3 +1 mod if n+2 +1 mod if n+1 +1 mod {(n 2 {(n 3 {(n It then follows from (6, (7, (8 nd (9 tht 2 (n if n (mod, S(n,5+2 3 (n 3 if n 1 (mod, (n 2 if n 2 (mod, (mod 5, (7 (mod 5, (8 (mod 5. (9 (mod 5
Congruences for Stirling Numbers of the Second Kind Modulo 5 799 s desired. Prt (iii is proved. (iv. From (1, we derive tht S(n,5+3 n ( m0 Thus by (10 we know tht (5+5 S(m,5(+1 m 1+ 2 (1 (1 5 1 i 2 ( S(n+2,5(+1+S(n+1,5(+1 n (mod 5. (10 S(n,5+3 S(n+2,5(+1+S(n+1,5(+1 (mod 5. (11 It then follows from (8, (9 nd (11 tht if n +3 (mod, then ( n 3 S(n,5+3 S(n+2,5(+1 (mod 5; if n +2 (mod, then S(n,5+3 S(n+1,5(+1 ( n (mod 5; if ether n +1 (mod or n (mod, then S(n,5+3 0 (mod 5. This implies tht prt (iv is true. (v. Using (1, we cn obtin tht (5+5 S(n,5+ n (1 5 1 i ( S(m,5(+1 m 1 S(n+1,5(+1 n (mod 5. (12 m0 It then follows from (9 nd (12 tht S(n,5+ S(n+1,5(+1 {(n if n (mod, 0, otherwise (mod 5 s required. Prt (v is proved. The proof of Theorem 1.1 is complete. References [1] T. Amdeberhn, D. Mnn, V. Moll, The 2-dic vlution of Stirling numbers, Eperimentl Mth. 17 (2008 69 82.
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