J. Number Theory 131(2011), no. 12, 2387–2397.
arXiv:1009.2486v4 [math.NT] 22 Aug 2011
ON DELANNOY NUMBERS ¨ AND SCHRODER NUMBERS
Zhi-Wei Sun Department of Mathematics, Nanjing University Nanjing 210093, People’s Republic of China
[email protected] http://math.nju.edu.cn/∼zwsun Abstract. The nth Delannoy number and the nth Schr¨ oder number given by Dn =
n X n n + k
k=0
and Sn =
k
k
n X n n + k 1 k k k+1 k=0
respectively arise naturally from enumerative combinatorics. Let p be an odd prime. We mainly show that p−1 X
k=1
Dk ≡2 k2
−1 p
Ep−3 (mod p)
and p−1 X
k=1
2 Sk m2 − 6m + 1 m − 6m + 1 ≡ 1 − (mod p), mk 2m p
where (−) is the Legendre symbol, E0 , E1 , E2 , . . . are Euler numbers, and m is any integer not divisible by p. We also conjecture that p−1 X
k=1
Dk2 k2
≡ −2qp (2)2 (mod p)
where qp (2) denotes the Fermat quotient (2p−1 − 1)/p.
2010 Mathematics Subject Classification. Primary 11A07, 11B75; Secondary 05A15, 11B39, 11B68, 11E25. Keywords. Congruences, central Delannoy numbers, Euler numbers, Schr¨ oder numbers. Supported by the National Natural Science Foundation (grant 10871087) of China.
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ZHI-WEI SUN
1. Introduction For n ∈ N = {0, 1, 2, . . . }, the (central) Delannoy number Dn denotes the number of lattice paths from the point (0, 0) to (n, n) with steps (1, 0), (0, 1) and (1, 1), while the Schr¨oder number Sn represents the number of such paths that never rise above the line y = x. It is known that X n n X 2k n+k n n+k = Dn = k 2k k k k=0
k=0
and
n X n n+k
n X 1 n+k Sn = Ck , = k k 2k k+1 k=0 k=0 2k 2k where Ck stands for the Catalan number 2k /(k + 1) = − k k k+1 . For information on Dn and Sn , the reader may consult [CHV], [S], and p. 178 and p. 185 of [St]. Despite their combinatorial backgrounds, surprisingly Delannoy numbers and Schr¨ oder numbers have some nice number-theoretic properties. As usual, for an odd prime p we let ( p· ) denote the Legendre symbol. Recall that Euler numbers E0 , E1 , E2 , . . . are integers defined by E0 = 1 and the recursion: n X n En−k = 0 for n = 1, 2, 3, . . . . k k=0 2|k
Our first theorem is concerned with Delannoy numbers and their generalization. Theorem 1.1. Let p be an odd prime. Then p−1 X −1 Dk Ep−3 (mod p) ≡2 k2 p
(1.1)
k=1
and
p−1 X Dk k=1
k
≡ −qp (2) (mod p),
(1.2)
where qp (2) denotes the Fermat quotient (2p−1 − 1)/p. If we set n X n n+k k x (n ∈ N), Dn (x) = k k k=0
then for any p-adic integer x we have √ √ p−1 X Dk (x) (−1 + −x)p + (−1 − −x)p + 2 ≡ (mod p). k p k=1
(1.3)
¨ ON DELANNOY NUMBERS AND SCHRODER NUMBERS
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Corollary 1.1. Let p be an odd prime. We have p−1 X Dk (3)
k
k=1 p−1
X Dk (−4) k
k=1
p−1 X Dk (−9)
k
k=1
and also
≡ − 2qp (2) (mod p) provided p 6= 3,
(1.4)
3 − 3p ≡ (mod p), p
(1.5)
≡ − 6qp (2) (mod p),
(1.6)
p−1 X Dk (−2)
k
k=1
4 ≡ − Pp−( p2 ) (mod p), p
(1.7)
where the Pell sequence {Pn }n>0 is given by P0 = 0, P1 = 1, and Pn+1 = 2Pn + Pn−1 (n = 1, 2, 3, . . . ). If p 6= 5, then p−1 X Dk (−5)
k
k=1
5 ≡ −2qp (2) − Fp−( p5 ) (mod p), p
(1.8)
where the Fibonacci sequence {Fn }n>0 is defined by F0 = 0, F1 = 1, and Fn+1 = Fn + Fn−1 (n = 1, 2, 3, . . . ). Now we propose two conjectures which seem challenging in the author’s opinion. Conjecture 1.1. Let p > 3 be a prime. We have p−1 X D2 k
k=1
k2
p−1 X Dk
k=1 p−1 X
k=1
k
≡ − 2qp (2)2 (mod p),
(1.9)
≡ − qp (2) + p qp (2)2 (mod p2 ),
Dk Sk ≡ − 2p
p−1 X (−1)k + 3 k=1
k
(1.10)
(mod p4 ),
and (p−1)/2
X
k=1
Dk Sk ≡
4x2 (mod p) 0 (mod p)
if p ≡ 1 (mod 4) & p = x2 + y 2 (2 ∤ x, 2 | y), if p ≡ 3 (mod 4).
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ZHI-WEI SUN
Also,
Pp−1
2 n=1 sn /n
≡ −6 (mod p), where n X 2k n+k = Dn − Sn . sn := k+1 2k k=0
Remark 1.1. Let p be an odd prime. Though there are many congruences for qp (2) mod p, (1.9) is curious since its left-hand side is a sum Pp−1 of squares. It is known that k=1 1/k ≡ −p2 Bp−3 /3 (mod p3 ) if p > 3, wherePB0 , B1 , B2 , . . . are Bernoulli numbers. In addition, we can prove Pp−1 p−1 −1 2 3 that k=0 Dk ≡ ( p ) − p Ep−3 (mod p ) and k=0 Dk2 ≡ ( p2 ) (mod p).
Conjecture 1.2. Let p > 3 be a prime. Then 3 p−1 3 p−1 p−1 X X 1 −2 X 1 k 3 k k (−1) Dk (2) ≡ (−1) Dk − ≡ (−1) Dk 4 p 8 k=0 k=0 k=0 ( −1 ( p )(4x2 − 2p) (mod p2 ) if p ≡ 1 (mod 3) & p = x2 + 3y 2 (x, y ∈ Z), ≡ 0 (mod p2 ) if p ≡ 2 (mod 3).
Also, p−1 3 1 −1 X k (−1) Dk p 2 k=0 2 2 2 2 4x − 2p (mod p ) if p ≡ 1, 7 (mod 24) and p = x + 6y (x, y ∈ Z), ≡ 8x2 − 2p (mod p2 ) if p ≡ 5, 11 (mod 24) and p = 2x2 + 3y 2 (x, y ∈ Z), 0 (mod p2 ) if ( −6 ) = −1. p
And p−1 X
k
3
(−1) Dk (−4) ≡
k=0 2 2 4x − 2p (mod p ) ≡ 12x2 − 2p (mod p2 ) 0 (mod p2 )
−5 p
X p−1 k=0
3 1 (−1) Dk − 16 k
if p ≡ 1, 4 (mod 15) and p = x2 + 15y 2 (x, y ∈ Z), if p ≡ 2, 8 (mod 15) and p = 3x2 + 5y 2 (x, y ∈ Z),
if ( −15 p ) = −1.
Remark 1.2. Note that (−1)n Dn (x) = Dn (−x − 1) for any n ∈ N, since k n X n −n − 1 X k j x Dn (−x − 1) = j k k j=0 k=0 n n X −n − 1 n − j X n j x = n−k k j j=0 k=0 n X n j −j − 1 = (−1)n Dn (x). x = n j j=0 Concerning Schr¨ oder numbers we establish the following result.
¨ ON DELANNOY NUMBERS AND SCHRODER NUMBERS
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Theorem 1.2. Let p be an odd prime and let m be an integer not divisible by p. Then 2 p−1 X m − 6m + 1 m2 − 6m + 1 Sk 1− (mod p). ≡ mk 2m p
(1.11)
k=1
Example 1.1. Theorem 1.2 in the case m = 6 gives that p−1 X Sk
k=1
6k
≡ 0 (mod p)
for any prime p > 3.
(1.12)
For technical reasons, we will prove Theorem 1.2 in the next section and show Theorem 1.1 and Corollary 1.1 in Section 3. 2. Proof of Theorem 1.2 Lemma 2.1. Let p be an odd prime and let m be any integer not divisible by p. Then p−1 X Ck m(m − 4) m−4 1− (mod p). ≡ mk 2 p
(2.1)
k=1
Proof. This follows from [Su10, Theorem 1.1] in which the author even Pp−1 k 2 determined k=1 Ck /m mod p . However, we will give here a simple proof of (2.1). For each k = 1, . . . , p − 1, we clearly have
(p − 1)/2 k
2k −1/2 k (mod p). = ≡ (−4)k k
Note also that
Cp−1
p−1 1 Y p+k ≡ −1 (mod p). = 2p − 1 k k=1
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ZHI-WEI SUN
Therefore p−1 X Ck ≡ mk k=1
X
00 is given by L0 = 2, L1 = 1, and Ln+1 = Ln + Ln−1 (n = 1, 2, 3, . . . ). It is well known that Ln =
√ !n 1+ 5 + 2
√ !n 1− 5 for all n ∈ N. 2
Putting x = −5 in (1.3) we get p−1 X Dk (−5) k=1
k
≡
2 − 2p L p 2p (1 − Lp ) + 2 − 2p = p p
2 ≡ − (Lp − 1) − 2qp (2) (mod p). p
It is known that 2(Lp − 1) ≡ 5Fp−( p5 ) (mod p2 ) provided p 6= 5 (see the proof of [ST, Corollary 1.3]). So (1.8) holds if p 6= 5. We are done.
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References [CHV] J. S. Caughman, C. R. Haithcock and J. J. P. Veerman, A note on lattice chains and Delannoy numbers, Discrete Math. 308 (2008), 2623–2628. [GKP] R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, New York, 1994. [L] E. Lehmer, On congruences involving Bernoulli numbers and the quotients of Fermat and Wilson, Ann. of Math. 39 (1938), 350–360. [PWZ] M. Petkovˇsek, H. S. Wilf and D. Zeilberger, A = B, A K Peters, Wellesley, 1996. [S] N. J. A. Sloane, Sequences A001850, A006318 in OEIS (On-Line Encyclopedia of Integer Sequences), http://oeis.org/. [St] R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999. [Su10] Z. W. Sun, Binomial coefficients, Catalan numbers and Lucas quotients, Sci. China Math. 53 (2010), 2473–2488. http://arxiv.org/abs/0909.5648. [Su11] Z. W. Sun, On congruences related to central binomial coefficients, J. Number Theory 131 (2011), 2219–2238. [ST] Z. W. Sun and R. Tauraso, New congruences for central binomial coefficients, Adv. in Appl. Math. 45 (2010), 125–148.
