Proc Amer Math Soc 1382010, no 1, 37 46 SOME CONGRUENCES FOR THE SECOND-ORDER CATALAN NUMBERS Li-Lu Zhao, Hao Pan Zhi-Wei Sun Department of Mathematics, Naning University Naning 210093, People s Republic of China zhaolilu@gmailcom, haopan79@yahoocomcn, zwsun@nueducn Abstract Let p be any odd prime We mainly show that 1 2 3 0 mod p where C 2 3 1 2 1 C 2 1 /2 1 mod p, /2 + 1 is the th Catalan number of order 2 1 Introduction The well-nown Catalan numbers are those integers C n 1 n + 1 2n n 2n 2n n n 1 n 0, 1, 2, As usual we regard x as 0 for 1, 2, There are many combinatorial interpretations for these important numbers see, eg, [St, pp 219-229] With the help of a sophisticated binomial identity, H Pan Z W Sun [PS] obtained some congruences on sums of Catalan numbers; in particular, by [PS, 116 18], for any prime p > 3 we have 0 C 3 p 3 1 2 mod p 1 C 3 p 1 2 3 mod p, 10 2000 Mathematics Subect Classification Primary 11B65; Secondary 05A10, 11A07 The third author is the corresponding author He was supported by the National Natural Science Foundation grant 10871087 the Overseas Cooperation Fund of China 1
2 LI-LU ZHAO, HAO PAN AND ZHI-WEI SUN where the Legendre symbol a 3 {0, ±1} satisfies the congruence a a 3 mod 3 Recently Z W Sun R Tauraso [ST1, ST2] obtained some further congruences concerning sums involving Catalan numbers For m, n N {0, 1, 2, }, we define C n m 1 mn + n mn + n mn + n m mn + 1 n n n 1 call it the nth Catalan number of order m Clearly C n 1 C n C n 2 1 3n 2n + 1 n In contrast with 10, we have the following result involving the secondorder Catalan numbers Theorem 11 Let p be an odd prime Then 2 C 2 2 1 /2 1 mod p 11 1 1 2 C 2 4 1 1 /2 mod p 12 Actually Theorem 11 follows from our next two theorems Theorem 12 Let p > 5 be a prime Then 3 2 6 1/2 1 mod p, 13 5 0 0 2 3 + 1 1 4 1/2 + 1 5 mod p, 14 Theorem 13 For any prime p we have 2 3 0 mod p 15 For any odd prime p we can also prove the following congruences: 3 + 2 5 2 1 /2 1 mod p, 1 1 1 2 1 2 1 3 + 1 1 /2 1 mod p, 3 + 2 3 2 1 /2 1 mod p
SOME CONGRUENCES FOR THE SECOND-ORDER CATALAN NUMBERS 3 We omit their proofs which are similar to those of Theorems 12-13 With the help of Theorems 12 13, we can easily deduce Theorem 11 Proof of Theorem 11 via Theorems 12 13 Clearly 11 12 hold for p 3, 5 Assume p > 5 By 13 14, 0 2 2 + 1 3 3 2 3 0 2 0 2 1 /2 1 mod p This proves 11 For 12 it suffices to note that 1 2 2 + 1 3 1 2 3 2 1 3 + 1 2 2 2 + 1 3 This concludes the proof We are going to provide two lemmas in the next section Theorems 12 13 will be proved in Sections 3 4 respectively Lemma 21 For m, n N we have 2 Some Lemmas m/3 2 n 0 1 m n 3 m + n 2 m 3 n n m n 2 2 m 0 0 21 Proof Let P x 2 + 2x 4x 3 n, denote by [x ]P x the coefficient of x in the expansion of P x Then 2 n [x m ]P x [x m ]1 + x 2x 3 n m/3 0 m/3 0 m/3 0 n 2 [x m 3 ]1 + x n 2 n n m 3 n 3 m + n 2 m 3
4 LI-LU ZHAO, HAO PAN AND ZHI-WEI SUN Since P x 1 x n 2x + 1 2 + 1 n we also have [x m ]P x 0 n 0 n 1 x n 2x + 1 2, n n m 2 n 2 1 m m 0 Therefore 21 is valid For any prime p, if n, N s, t {0, 1,, p 1} then we have the following well-nown Lucas congruence cf [Gr] or [HS]: pn+s p+t mod p This will be used in the proof of the following lemma n s t Lemma 22 Let p > 5 be a prime Then we have 1 s s0 1 s s0 t0 Proof Observe that 1 s s0 /2 s0 /2 s0 /2 s0 t0 t0 1 s 2 t 3 1/2 + 2 t 5 mod p 22 2 t 3 1 1 /2 mod p 23 p + t 10 2 t t t0 1 s 3 + 1 s 3 + 1 s 3 s0 9 s s0 2 t + t sp+1/2 sp+1/2 sp+1/2 sp+1/2 sp+1/2 1 s t0 1 s t0 2 t t 1 s 2 t t tp t0 1 s 2 t t p 1 s r0 2 p+r r + p 2 t t tp 2 t t
SOME CONGRUENCES FOR THE SECOND-ORDER CATALAN NUMBERS 5 For s p + 1/2,, p 1, by Lucas congruence we have p p p + p p 2 r 2 r 3 p mod p p + r r r0 Thus, with the help of Fermat s little theorem, we get 1 s 2 t 1 9p 1 s 2 t 10 3 9s s0 t0 r0 sp+1/2 1 2 p+1 1 9 /2 9 2 3 10 3 1/2 + 2 mod p 5 This proves 22 In view of Lucas congruence Fermat s little theorem, we also have 1 s 2 t p + t s0 sp+1/2 t0 1 s t0 p 2 t t 3 p p+1/2 1 9/2 9 10 3 10 1 1 /2 mod p sp+1/2 1 s 3 p 1 p+1/2 3 10 1 + 1 p+1/2 3 So 23 is also valid We are done 3 Proof of Theorem 12 In order to prove Theorem 12, we first present an auxiliary result Theorem 31 Let p > 5 be a prime, let d, δ {0, 1} Then 1 d+δ 3 + d 2 δ 2 4 δ 10 δp d 3 δp+ d 3δ 25d 3 + 1 /2 mod p 10 Proof Applying 21 with n p 1 m δp + p 1 d, we get δp+ d/3 2 2 p 1 3 + d δp δp + p 1 d 3 0 1 δp+ d p 1 0 δp+ d 0 31 2 p 1 2 δp + p 1 d
6 LI-LU ZHAO, HAO PAN AND ZHI-WEI SUN Observe that δp+ d/3 0 δp d 3 δp+ d δp d 3 δp+ d 1 d+δ 1 δp+ d 1 0 2 p 1 3 + d δp δp + p 1 d 3 2 p 1 3 + d δp p + δp 1 d 3 3 + d 2 1 δp+ d 3 3 + d 2 mod p δp d 3 δp+ d p 1 0 δp d <δp+p d 2 δ d 1 s s0 Therefore t0 δp+ d 2 2 0 2 p 1 2 δp + p 1 d 0 2 t mod p δp d + t δp d 3 δp+ d 2 δ d 1 s s0 3 + d 2 t0 Recall that d {0, 1} We have 1 s s0 s0 2 t δp d + t t0 1 s d t d 1 s s0 2 d s0 t0 t0 2 d+t δp + t 1 s 2 t δp + t 1 2 δp d+t 2 δp d + t t0 2 t mod p δp d + t 2 d+t + d 2 p δp + t δp 1 + d s0 δp + p 1 1 s 2 p δp 1 δp + p 1
SOME CONGRUENCES FOR THE SECOND-ORDER CATALAN NUMBERS 7 hence 1 d+δ d2 δ d s0 δp d 3 δp+ d 1 s δp 1 2 3 + d 2 d2 δ 1 3δ 2 1 s p 1 Since s0 1 s t0 s0 by Lemma 22, we finally get 1 d+δ 2 δ 2 δ 1 s 2 t δp + t δp + p 1 s0 t0 d3δ 22 δ 1 1 /2 mod p 2 t 4 δ δp + t 10 + 3 10 2 3δ 1/2 mod p δp d 3 δp+ d 3 + d 2 4 δ 10 + 3 10 2 3δ 1/2 + d 3δ 2 1/2 2 4 δ 3δ 25d 3 + 1 /2 mod p 10 10 This proves 31 Proof of Theorem 12 Let d {0, 1} If 2p d/3 p 1, then 2 + d + 1 2 + 2 2p 3 + d hence 3 + d 3 + d 2 + d + 1 0 mod p! Therefore 0 2p d 3 3p 3 d 3 2 d 3 + d 2 0 mod p With the help of Theorem 31, we have 3 + d 3 + d 2 2 1 δ0 δp d 3 δp+ d 1 4 δ 1 d 2 δ 10 δ0 1d 1 5 3 + d 2 3δ 25d 3 + 1 /2 10 1 + 10d 6 1 /2 mod p
8 LI-LU ZHAO, HAO PAN AND ZHI-WEI SUN This yields 13 14 We are done 4 Proof of Theorem 13 Proof of Theorem 13 Obviously 15 holds for p 2, 3 Below we assume p > 3 Let δ {0, 1} Applying 21 with m p + δp n p we get p 2 p 2 p 3 δp p + δp 3 0 1 δ+1 p p+δp p 2 p 2 p + δp 0 0 41 Observe that p 2 p 3 δp p + δp 3 0 p 3 δp 2 3 δp δp 3 p+δ p 3 δp 1 δ + 2 3 δp δp<3<p+δp For 1,, p 1 clearly p p p 1 p 1 1 1 mod p 2 Thus δp<3<p+δp δp<3<p+δp 1 δ+1 1 δ+1 p 3 p 3 δp 2 3 δp 2 p 13 δ 3 δp 3 δp 2 p 1 3 δp + δp 3 δp<3<p+δp δp<3<p+δp 2 3 mod p 2 by Lucas congruence
SOME CONGRUENCES FOR THE SECOND-ORDER CATALAN NUMBERS 9 Notice that Clearly δp<2<2p δp<2<2p δp<2<2p + 1 δ p 0 p+δp p 2 δp 2 2p p p p δp<2<2p + 0 p+δp p p 2 {δp,2p} p+δp δp p+δp p 2 p + δp p 2 2 δp p 2 2 δp δp p+δp p p 2 2 δp δp p 2 2 δp δp p 2 p 2 2 δp + 2 p+δp 0 δp p p + δp 2 δp 2 δp 0 p<2<2p p 2 p 2 p+δp p p 2 δ 2 1 δ 2 1 + 1 δ 2 1+δ 2 1 1 δ δ2 p 2 δ2 p 2 mod p 2 Note that δ {0, 1} 2 p p/2<<p p+δp p p 2 2 1 δ δp 2δp δp by Lucas congruence 1 p 2 p 2 Also, p p 0 2 1 δ 0 p+δp p p 2 2 δp 22p δp p 2p 2 δp {δp,p+δp} 2 2 2 δp + 2 p+δp 4 δp 2 p+1 mod p 2 δ 1 + δ
10 LI-LU ZHAO, HAO PAN AND ZHI-WEI SUN 2 1 mod p 3 by the Wolstenholme congruence Recall that 2 1 2p p cf [Gr] or [HT] Combining the above with 41, we have 2 p 1 δ + 1 δ+1 p 3 δp<3<p+δp 2 3 1 δ+1 δ2 2 p + 1 δ + 4 δp 2 p+1 mod p 2 Setting δ 0 δ 1 respectively, we obtain 2 p 2 p p 2 3 2 p+1 2 mod p 2 3 2 p p 3 It follows that 2 3 p 0<3<2p p<3<2p 2 0<3<p 2 If 2p 3 < 3p, then 3 Therefore 1 2 3 0<3<2p 3 2 2 p + 4 p 2 p+1 mod p 2 3 4 p 4 2 p + 4 2 p 2 2 0 mod p 2 3 2 + 1! 2 3 + 2p 3<3p This completes the proof of Theorem 13 References 0 mod p 2 3 0 mod p [Gr] A Granville, Arithmetic properties of binomial coefficients I Binomial coefficients modulo prime powers, in: Organic Mathematics Burnady, BC, 1995, 253 276, CMS Conf Proc, 20, Amer Math Soc, Providence, RI, 1997 [HT] C Helou G Teranian, On Wolstenholme s theorem its converse, J Number Theory 128 2008, 475 499 [HS] H Hu Z W Sun, An extension of Lucas theorem, Proc Amer Math Soc 129 2001, 3471 3478 [PS] H Pan Z W Sun, A combinatorial identity with application to Catalan numbers, Discrete Math 306 2006, 1921 1940 [St] R P Stanley, Enumerative Combinatorics, Vol 2, Cambridge Univ Press, Cambridge, 1999 [ST1] Z W Sun R Tauraso, On some new congruences for binomial coefficients, preprint, arxiv:07091665 [ST2] Z W Sun R Tauraso, New congruences for central binomial coefficients, Adv in Appl Math, to appear http://arxivorg/abs/08050563