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1. Introduction
For m, n N = 0, 1, 2,... , the Delannoy number ∈ { }
m n k Dm,n := 2 (1.1) N k k kX∈ in combinatorics counts lattice paths from (0, 0) to (m, n) in which only east (1, 0), north (0, 1), and northeast (1, 1) steps are allowed (cf. R. P. Stanley [St99, p. 185]). The n-th central Delannoy number Dn = Dn,n has another well-known expression:
n n n + k n n + k 2k D = = . (1.2) n k k 2k k kX=0 kX=0 For n Z+ = 1, 2, 3,... , the n-th little Schr¨oder number is given by ∈ { } n k−1 sn := N(n,k)2 (1.3) kX=1 with the Narayana number N(n,k) defined by
1 n n N(n,k) := Z. n k k 1 ∈ − (See [Gr, pp.268–281] for certain combinatorial interpretations of the Narayana number N(n,k) = N(n,n +1 k).) For n N, the n-th large Schr¨oder number is given by − ∈ n n n + k 1 n n + k S := = C , (1.4) n k k k +1 2k k Xk=0 kX=0 where C denotes the Catalan number 2k /(k +1) = 2k 2k . It is well known k k k − k+1 that S = 2s for every n = 1, 2, 3,... . Both little Schr¨oder numbers and large n n Schr¨oder numbers have many combinatorial interpretations (cf. [St97] and [St99, pp. 178, 239-240]); for example, sn is the number of ways to insert parentheses into an expression of n + 1 terms with two or more items within a parenthesis, and Sn is the number of lattice paths from the point (0, 0) to (n,n) with steps (1, 0), (0, 1) and (1, 1) which never rise above the line y = x. Surprisingly, the central Delannoy numbers and Schr¨oder numbers arising nat- urally in enumerative combinatorics, have nice arithmetic properties. In 2011 the author [S11b] showed that
p−1 p−1 D S k ( 1)(p−1)/22E (mod p) and k 0 (mod p) k2 ≡ − p−3 6k ≡ Xk=1 Xk=1 PROPERTIES OF DELANNOY NUMBERS AND SCHRODER¨ NUMBERS 3 for any prime p > 3, where E0,E1,E2,... are the Euler numbers. In 2014 the author [S14a] proved that 1 n−1 (2k + 1)D2 Z for all n =1, 2, 3,..., n2 k ∈ Xk=0 and that p−1 2 D2 (mod p) for any odd prime p, k ≡ p k=0 · X where ( p ) denotes the Legendre symbol. Definition 1.1. We define n n 2 D (x) := xk(x + 1)n−k for n N, (1.5) n k ∈ Xk=0 and n s (x) := N(n,k)xk−1(x + 1)n−k for n Z+. (1.6) n ∈ k=1 X + Obviously Dn(1) = Dn for n N, and sn(1) = sn for n Z . In this paper we obtain somewhat∈ curious results involving∈ the polynomials Dn(x) and sn(x). Our first theorem is as follows. Theorem 1.1. (i) For any n Z+, we have ∈ 1 n−1 D (x)s (x) = W (x(x + 1)), (1.7) n k k+1 n Xk=0 where n W (x) = w(n,k)C xk−1 Z[x] (1.8) n k−1 ∈ kX=1 with 1 n 1 n + k w(n,k) = − Z. (1.9) k k 1 k 1 ∈ − − (ii) Let p be an odd prime. For any p-adic integer x, we have p−1
Dk(x)sk+1(x) Xk=0 p(1 x(x +1)) (mod p3) if x 0, 1 (mod p), − ≡ − ≡ 2p2 + 2x+1 p2(x2q (x) (x + 1)2q (x +1)) (mod p3) otherwise, ( x(x+1) p − p (1.10) p−1 where qp(z) denotes the Fermat quotient (z 1)/p for any p-adic integer z 0 (mod p). − 6≡
Remark 1.1. It is interesting to compare the new numbers w(n,k) with the Narayana numbers N(n,k). Clearly, Theorem 1.1 in the case x = 1 yields the following consequence. 4 ZHI-WEI SUN
Corollary 1.1. For any positive integer n, we have 1 n−1 n D s = w(n,k)C 2k−1 1 (mod2). (1.11) n k k+1 k−1 ≡ kX=0 Xk=1 Also, for any odd prime p we have p−1 D s 2p2(1 3q (2)) (mod p3). (1.12) k k+1 ≡ − p Xk=0
Remark 1.2. For the prime p = 588811, we have qp(2) 1/3 (mod p) and hence p−1 3 ≡ k=0 Dksk+1 0 (mod p ). In 2016 J.-C. Liu [L] confirmed the author’s conjec- ture (cf. [S11b,≡ Conjecture 1.1]) that P p−1 p−1 ( 1)k +3 D S 2p − (mod p4) for any prime p > 3. k k ≡ − k kX=1 kX=1 From Theorem 1.1 we can deduce a novel combinatorial identity. Corollary 1.2. For any n Z+, we have ∈ n n n + k C (n + 1)/2 n 2 k−1 = ⌊ ⌋ . (1.13) k k 1 ( 4)k−1 4n−1 n/2 Xk=1 − − ⌊ ⌋
Remark 1.3. If we let an denote the left-hand side of (1.13), then the Zeilberger algorithm (cf. [PWZ, pp.101-119]) cannot find a closed form for an, and it only yields the following second-order recurrence relation: (n + 1)2a + (2n + 3)a (n + 1)(n + 3)a =0 for n =1, 2, 3,.... n n+1 − n+2 Now we give our second theorem which can be viewed as a supplement to The- orem 1.1. Theorem 1.2. Let p be any odd prime. Then p−1 kD (x)s (x) 2(x(x + 1))(p−1)/2 (mod p). (1.14) k k+1 ≡ kX=0 In particular, p−1 2 kD s 2 (mod p). (1.15) k k+1 ≡ p kX=0 In the next section we are going to show Theorems 1.1-1.2 and Corollary 1.2. In Section 3 we will give applications of Theorems 1.1-1.2 to central trinomial coefficients, Motzkin numbers, and their generalizations. Section 4 contains two related conjectures. Throughout this paper, for any polynomial P (x) and n N, we use [xn]P (x) to denote the coefficient of xn in P (x). ∈ PROPERTIES OF DELANNOY NUMBERS AND SCHRODER¨ NUMBERS 5
2. Proofs of Theorems 1.1-1.2 and Corollary 1.2
Recall the following definition given in [S12a] motivated by the large Schr¨oder numbers.
Definition 2.1. For n N we set ∈
n n n + k xk n n + k S (x) = = C xk. (2.1) n k k k +1 2k k kX=0 Xk=0 Lemma 2.1. We have
n n n + k D (x) = xk for n N. (2.2) n k k ∈ Xk=0 Also, for any n Z+ we have ∈ D (x) D (x)=2x(2n + 1)S (x) (2.3) n+1 − n−1 n and
(x + 1)sn(x) = Sn(x). (2.4)
Proof. For k,n N, we obviously have ∈
n n 2 k n 2 n j [xk]D (x) =[xk] xj(x + 1)n−j = − n j j k j j=0 j=0 X X − n k n k n n + k = = k j k j k k j=0 X − with the help of the Chu-Vandermonde identity (cf. [G, (3.1)]). This proves (2.2). Now fix n Z+. For k N, by (2.2) we clearly have ∈ ∈ [xk+1](D (x) D (x)) n+1 − n−1 n +1+(k + 1) 2(k + 1) n 1+(k + 1) 2(k + 1) = − 2(k + 1) k +1 − 2(k + 1) k +1 2n +1 n + k 2k +2 2(2n + 1) n + k 2k = = 2k +1 2k k +1 k +1 2k k k+1 =[x ]2x(2n + 1)Sn(x).
So (2.3) follows. 6 ZHI-WEI SUN
For each k N, it is apparent that ∈ n k k j−1 n−j [x ](x + 1)sn(x) =[x ](x + 1) N(n,j)x (x + 1) j=1 X n j k+1 n j = N(n,j) − + N(n,j) − k j k j +1 6 j=1 0 n n n + k x 1 k P (x) = − . n k k 2 kX=0 Lemma 2.2. Let n Z+. Then ∈ n n + k 2k 2k n(n + 1)S (x)2 = xk−1(x + 1)k+1 (2.5) n 2k k k +1 Xk=1 and D (x) + D (x) n n + k 2k 2 2k +1 n−1 n+1 S (x) = xk(x + 1)k+1. (2.6) 2 n 2k k (k + 1)2 kX=0 Remark 2.2. The identities (2.5) and (2.6) are (2.1) and (3.6) of the author’s paper [S12a] respectively. Lemma 2.3. For any m, n Z+ with m 6 n, we have the identity ∈ n k + m 2k +1 2m +1 m(m + 1) 2m − k(k + 1) kX=m (2.7) (n m)(n + m + 1) n + m = − . n +1 2m PROPERTIES OF DELANNOY NUMBERS AND SCHRODER¨ NUMBERS 7 Proof. When n = m, both sides of (2.7) vanish. Let m, n Z+ with n > m. If (2.7) holds, then ∈ n+1 k + m 2k +1 2m +1 m(m + 1) 2m − k(k + 1) kX=m (n m)(n + m + 1) n + m = − n +1 2m n + 1 + m 2(n +1)+1 + 2m +1 m(m + 1) 2m − (n + 1)(n + 2) n + m +1 (n m)(n m + 1) m(m + 1)(2n + 3) = − − +2m +1 2m n +1 − (n + 1)(n + 2) (n +1)+ m (n +1 m)((n +1)+ m + 1) = − . 2m (n +1)+1 In view of the above, we have proved Lemma 2.3 by induction. Let A and B be integers. The Lucas sequence un(A,B)(n = 0, 1, 2,...) is defined by u0(A,B)=0, u1(A,B)= 1, and u (A,B) = Au (A,B) Bu (A,B) for n =1, 2, 3,.... n+1 n − n−1 It is well known that if ∆ = A2 4B = 0 then − 6 αn βn u (A,B) = − for all n N, n α β ∈ − where α and β are the two roots of the quadratic equation x2 Ax + B = 0 (so that α + β = A and αβ = B). It is also known (see, e.g., [S10,− Lemma 2.3]) that for any odd prime p we have ∆ up(A,B) (mod p), and up−( ∆ )(A,B) 0 (mod p) if p ∤ B. ≡ p p ≡ 5 In particular, Fp ( ) (mod p) and p Fp−( 5 ) for any odd prime p, where Fn = ≡ p | p u (1, 1) with n N is the n-th Fibonacci number. n − ∈ Lemma 2.4. Let p be an odd prime, and let x be any integer not divisible by p. Then 4x +1 4x +1 W (x) 1 (mod p). (2.8) p ≡ 2x p − Moreover, if x 1/4 (mod p) then ≡ − W (x) 2p (mod p2), (2.9) p ≡ 8 ZHI-WEI SUN otherwise we have 4x +1 4x +1 W (x) 2p + 1 xp−1 +(p + 1) 1 p ≡ 2x − p − 4x +1 4x +1 2 2 4x+1 2x + up−( 4x+1 ) 2x +1,x (mod p ). − 2−( p ) p p 4x (2.10) Proof. Clearly, p−1 (p−1)/2 2p 1 p 1 1 − = 1 + 1 + p + 1 (mod p2). p 1 k ≡ k p k ≡ − kY=1 kX=1 − (J. Wolstenholme [W] even showed that 2p−1 1 (mod p3) if p > 3.) Thus p−1 ≡ 1 2p 2 2p 1 w(p, p) = = − 2(1 p) (mod p2), p p 1 p +1 p 1 ≡ − − − and 1 2p 2 1 2p 1 C = − = − (2p +1) (mod p2). p−1 p p 1 2p 1 p 1 ≡ − − − − For each k =1,...,p 1, clearly − 1 p 1 p + k 1 p + k w(p, k) = − − k k 1 k 1 p +1 − − p 1 p j p + j = 1 + − k p +1 j · j 0 Therefore p−1 p−1 k−1 Wp(x) =w(p, p)Cp−1x + w(p, k)Ck−1x kX=1 p−1 p 2(1 p)C xp−1 + 1 p + C ( x)k−1 ≡ − p−1 − k k−1 − k=1 X p (2 2p)C xp−1 +(p + 1) C ( x)k−1 +1 C ( x)p−1 ≡ − p−1 k−1 − − p−1 − kX=2 p−1 1 + p 2 C ( x)k−1 k − k−1 − Xk=1 p−1 p +1 (1 3p)(1+2p)xp−1 +(p + 1) C ( x)k ≡ − − k − Xk=1 p−1 p 2k k ( x)k−1 − 2 k − k=1 X p−1 p−1 C p 2k 2p +1 xp−1 +(p + 1) k m k (mod p2), ≡ − mk − 2 kmk k=1 k=1 X X where m is an integer with m 1/x (mod p2). By [S10, Theorem 1.1], we have ≡ − p−1 Ck p−1 m 4 ∆ 2 m 1 − 1 + up−( ∆ )(m 2, 1) (mod p ) mk ≡ − − 2 p − p − Xk=1 m 4 ∆ − 1 (mod p), ≡ − 2 p − where 1 1 4x +1 ∆ := m(m 4) 4 = (mod p2). − ≡ −x −x − x2 So (2.8) follows. If m 4 (mod p), then by [S12b, Lemma 3.5] we have 6≡ p−1 2k ( m)p−1 1 m 4 ∆ up−( ∆ )(2 m, 1) k − − + − p − kmk ≡ p 2 p p k=1 X mp−1 1 m 4 ∆ up−( ∆ )(m 2, 1) = − − p − (mod p) p − 2 p p 10 ZHI-WEI SUN and hence from the above we deduce that W (x) 2p +1 xp−1 p ≡ − p−1 m 4 ∆ +(p + 1) m 1 − 1 + up−( ∆ )(m 2, 1) − − 2 p − p − p mp−1 1 m 4 ∆ up−( ∆ )(m 2, 1) m − − p − − 2 p − 2 p p ! m m 4 4x +1 2p +1 xp−1 + 1 (mp−1 1) (p + 1) − 1 ≡ − − 2 − − 2 p − m 4 m 4x +1 − 1 up−( 4x+1 )(m 2, 1) − 2 − 2 p p − 2x +1 4x +1 4x +1 2p +1 xp−1 + (1 xp−1)+(p + 1) 1 ≡ − 2x − 2x p − 4x +1 1 4x +1 2 + 1 + up−( 4x+1 )(m 2, 1) (mod p ) 2x 2x p p − which is equivalent to (2.10) since ( x)k−1u (m 2, 1) = u ( x(m 2), ( x)2) u (2x +1,x2) (mod p2) − k − k − − − ≡ k for all k N. ∈ When m 4 (mod p) (i.e., x 1/4 (mod p)), we have ≡ ≡ − p−1 2k p−1 2k k k 2q (2) (mod p) kmk ≡ k4k ≡ p k=1 k=1 X X by [ST, (1.12)], and hence m 4 p W (x) 2p +1 xp−1 +(p + 1) mp−1 1 + − m2q (2) p ≡ − − 2 − 2 p m 4 2p +1 xp−1 + mp−1 1 + − 4(2p−1 1) ≡ − − 2 − − 4x +1 2p +1 xp−1 +1 xp−1 2(22(p−1) 1) ≡ − − − 2x − − 2 p + (4x +1)+(1 xp−1)+(1 4p−1) ≡ − − 2 p + (4x +1)+1 (4x +1 1)p−1 ≡ − − 2(p + (4x +1)+(p 1)(4x + 1)) 2p (mod p2). ≡ − ≡ Therefore (2.9) is valid. PROPERTIES OF DELANNOY NUMBERS AND SCHRODER¨ NUMBERS 11 Proof of Theorem 1.1. (i) Fix n Z+. For each k Z+, by (2.3) and Lemma 2.2 we have ∈ ∈ S (x) D (x) + D (x) S (x) x D (x) k = k−1 k+1 k (2k + 1) S (x)2 k−1 x +1 2 · x +1 − x +1 k k k + j 2j 2 2j +1 = (x(x + 1))j 2j j (j + 1)2 j=0 X 2k +1 k k + j 2j 2j (x(x + 1))j − k(k + 1) 2j j j +1 j=0 X k k + j 2j 2 2j +1 2k +1 j = (x(x + 1))j 2j j (j + 1)2 − k(k + 1) · j +1 j=0 X k k + j 2k +1 = C2 2j +1 j(j + 1) (x(x + 1))j. 2j j − k(k + 1) Xj=0 Combining this with (2.4) we obtain k k + j 2k +1 D (x)s (x) = C2 2j +1 j(j + 1) (x(x + 1))j (2.11) k−1 k 2j j − k(k + 1) j=0 X for any k Z+. Therefore, ∈ n n k k + j 2k +1 D (x)s (x) = C2 2j +1 j(j + 1) (x(x + 1))j k−1 k 2j j − k(k + 1) kX=1 kX=1 Xj=0 n n k + j 2k +1 =n + C2(x(x + 1))j 2j +1 j(j + 1) j 2j − k(k + 1) j=1 X Xk=j n−1 (n j)(n + j + 1) n + j = C2(x(x + 1))j − j n +1 2j j=0 X with the help of Lemma 2.3. It follows that n−1 n−1 n n 1 n + j +1 D (x)s (x) = C (x(x+1))j − = nW (x(x+1)). k k+1 j j +1 j j n kX=0 Xj=0 For any k Z+, we have ∈ 1 n n + k 1 n 1 n + k w(n,k) = = − n k k 1 n +1 k 1 k − − and hence w(n,k)=(n +1)w(n,k) nw(n,k) Z. So Wn(x) Z[x]. This proves part (i) of Theorem 1.1. − ∈ ∈ 12 ZHI-WEI SUN (ii) Let x be any p-adic integer, and let y be an integer with y x(x + 1) (mod p2). By (1.7) we have ≡ p−1 D (x)s (x) = pW (x(x + 1)) pW (y) (mod p3). (2.12) k k+1 p ≡ p kX=0 If y 0 (mod p), then ≡ W (y) w(p, 1) + w(p, 2)y 1 x(x +1) (mod p2). p ≡ ≡ − 2 If x 1/2 (mod p) (i.e., y 1/4 (mod p)), then Wp(y) 2p (mod p ) by Lemma≡ − 2.4. Thus (1.10) holds≡ for −x 0, 1, 1/2 (mod p). ≡ Now assume that x 0, 1, 1/2≡ (mod− p),− i.e., y 0, 1/4 (mod p). Then 6≡ − − 6≡ − 4y +1 (2x + 1)2 = =1 p p and u (2y +1,y2) u (x2 +(x + 1)2,x2(x + 1)2) p−1 ≡ p−1 ((x + 1)2)p−1 (x2)p−1 = − (x + 1)2 x2 − (x + 1)p−1 + xp−1 = (x + 1)p−1 xp−1 2x +1 − 2 (x + 1)p−1 xp−1 (mod p2). ≡2x +1 − Combining this with Lemma 2.4, we obtain (2x + 1)2 W (y) 2p + 1 xp−1(x + 1)p−1 p ≡ 2x(x + 1) − (2x + 1)2 2 (2x(x +1)+1) (x + 1)p−1 xp−1 − 4x(x + 1) 2x +1 − (2x + 1)2 2p + 1 xp−1 +1 (x + 1)p−1 ≡ 2x(x + 1) − − 2x +1 (2x2 +2x + 1) (x + 1)p−1 xp−1 − 2x(x + 1) − 2x +1 =2p + x2(xp−1 1) (x + 1)2 (x + 1)p−1 1 (mod p2). x(x + 1) − − − Therefore (1.10) holds in light of (2.12). So far we have completed the proof of Theorem 1.1. Proof of Corollary 1.2. It is known (cf. [G, (3.133)]) that 1 ( 1)k/2 k /2k if 2 k, D = P (0) = − k/2 | k −2 k ( 0 otherwise. PROPERTIES OF DELANNOY NUMBERS AND SCHRODER¨ NUMBERS 13 By (2.4) and [S11a, Lemma 4.3], k/2 k 1 1 ( 1) Ck/2/2 if 2 k, sk+1 =2Sk+1 = − | −2 −2 0 otherwise. Therefore n−1 1 1 D s k −2 k+1 −2 Xk=0 2 ⌊(n−1)/2⌋ 2j 2 ( 1)k/2 k = − C = j . 2k k/2 k/2 (j + 1)16j 06k