Department of Mathematics, Nanjing University Nanjing , People s Republic of China

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1 Proc Amer Math Soc , no 1, SOME CONGRUENCES FOR THE SECOND-ORDER CATALAN NUMBERS Li-Lu Zhao, Hao Pan Zhi-Wei Sun Department of Mathematics, Naning University Naning , People s Republic of China zhaolilu@gmailcom, haopan79@yahoocomcn, zwsun@nueducn Abstract Let p be any odd prime We mainly show that mod p where C 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 ] 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, ], for any prime p > 3 we have 0 C 3 p mod p 1 C 3 p mod p, 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 the Overseas Cooperation Fund of China 1

2 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 1 mod p C /2 mod p 12 Actually Theorem 11 follows from our next two theorems Theorem 12 Let p > 5 be a prime Then /2 1 mod p, / mod p, 14 Theorem 13 For any prime p we have mod p 15 For any odd prime p we can also prove the following congruences: /2 1 mod p, /2 1 mod p, /2 1 mod p

3 SOME CONGRUENCES FOR THE SECOND-ORDER CATALAN NUMBERS 3 We omit their proofs which are similar to those of Theorems With the help of Theorems 12 13, we can easily deduce Theorem 11 Proof of Theorem 11 via Theorems Clearly hold for p 3, 5 Assume p > 5 By 13 14, /2 1 mod p This proves 11 For 12 it suffices to note that This concludes the proof We are going to provide two lemmas in the next section Theorems 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 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 4 LI-LU ZHAO, HAO PAN AND ZHI-WEI SUN Since P x 1 x n 2x 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 /2 mod p 23 p + t 10 2 t t t0 1 s s 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

5 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 s s0 t0 r0 sp+1/2 1 2 p / /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 mod p sp+1/2 1 s 3 p 1 p+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 /2 mod p 10 Proof Applying 21 with n p 1 m δp + p 1 d, we get δp+ d/3 2 2 p d δp δp + p 1 d δp+ d p 1 0 δp+ d p 1 2 δp + p 1 d

6 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 p d δp δp + p 1 d 3 2 p d δp p + δp 1 d d 2 1 δp+ d 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 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

7 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 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 δ 1/2 mod p δp d 3 δp+ d 3 + d 2 4 δ δ 1/2 + d 3δ 2 1/2 2 4 δ 3δ 25d /2 mod p This proves 31 Proof of Theorem 12 Let d {0, 1} If 2p d/3 p 1, then 2 + d p 3 + d hence 3 + d 3 + d 2 + d 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 δ0 δp d 3 δp+ d 1 4 δ 1 d 2 δ 10 δ0 1d d 2 3δ 25d / d 6 1 /2 mod p

8 8 LI-LU ZHAO, HAO PAN AND ZHI-WEI SUN This yields 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 δ+1 p p+δp p 2 p 2 p + δp 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 δp δp<3<p+δp For 1,, p 1 clearly p p p 1 p 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

9 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 δ 2 1+δ δ δ2 p 2 δ2 p 2 mod p 2 Note that δ {0, 1} 2 p p/2<<p p+δp p p δ δp 2δp δp by Lucas congruence 1 p 2 p 2 Also, p p δ 0 p+δp p p 2 2 δp 22p δp p 2p 2 δp {δp,p+δp} δp + 2 p+δp 4 δp 2 p+1 mod p 2 δ 1 + δ

10 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 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 p+1 2 mod p 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 <3<2p p + 4 p 2 p+1 mod p p 4 2 p p mod p ! p 3<3p This completes the proof of Theorem 13 References 0 mod p mod p [Gr] A Granville, Arithmetic properties of binomial coefficients I Binomial coefficients modulo prime powers, in: Organic Mathematics Burnady, BC, 1995, , CMS Conf Proc, 20, Amer Math Soc, Providence, RI, 1997 [HT] C Helou G Teranian, On Wolstenholme s theorem its converse, J Number Theory , [HS] H Hu Z W Sun, An extension of Lucas theorem, Proc Amer Math Soc , [PS] H Pan Z W Sun, A combinatorial identity with application to Catalan numbers, Discrete Math , [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: [ST2] Z W Sun R Tauraso, New congruences for central binomial coefficients, Adv in Appl Math, to appear