A244973 a(n) = Sum_{k=0..n} (-1)^k*C(n, k)^2*C(2*k, k), where C(n, k) denotes the binomial coefficient n!/(k!*(n-k)!).

1, -1, -1, 17, -65, 49, 881, -5489, 12223, 42785, -479951, 1746271, 440881, -39651457, 212039855, -326783183, -2817155137, 23175692033, -68726927071, -128775914225, 2285692892785, -10156877725985, 6169206210815, 196882990135745, -1274770281690575

Offset: 0

Author

Zhi-Wei Sun, Jul 08 2014

Keywords

• sign
• easy

Comments

Zhi-Wei Sun introduced this sequence in arXiv:1407.0967. For any prime p > 5, he proved that Sum_{k=1..p-1} a(k)/k^2 == 0 (mod p) and Sum_{k=1..p-1} a(k)/k == 0 (mod p^2). This is quite similar to Wolstenholme's congruences Sum_{k=1..p-1} 1/k^2 == 0 (mod p) and Sum_{k=1..p-1} 1/k == 0 (mod p^2) for any prime p > 3.

Conjecture: For any prime p > 5 and positive integer n, the number (a(p*n) - a(n))/(p*n)^3 is always a p-adic integer.

The author proved a weaker version of this in arXiv:1610.03384. - Zhi-Wei Sun, Nov 12 2016

Examples

			a(3) = 17 since C(3,0)^2*C(2*0,0) - C(3,1)^2*C(2,1) + C(3,2)^2*C(4,2) - C(3,3)^2*C(6,3) = 1 - 18 + 54 - 20 = 17.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..200
• V. J. Guo, G.-S. Mao and H. Pan, Proof of a conjecture involving Sun polynomials, J. Difference Equ. Appl., 22(2016), no. 8, 1184-1197; also arXiv:1511.04005 [math.NT], 2015.
• Zhi-Wei Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, arXiv:1407.0967 [math.NT], 2014-2016.
• Zhi-Wei Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, Ramanujan J. 40 (2016), no. 3, 511-533.
• Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1016.03384 [math.NT], 2016.

Crossrefs

Cf. A000172, A002893, A275027, A277640.

Programs

Maple

a := n -> hypergeom([1/2, -n, -n], [1, 1], -4):
seq(simplify(a(n)), n = 0..24);  # Peter Luschny, Mar 16 2025

Mathematica

s[n_]:=Sum[Binomial[n,k]^2*Binomial[2k,k](-1)^k,{k,0,n}]
Table[s[n],{n,0,20}]

PARI

a(n) = sum(k=0, n, (-1)^k*binomial(n,k)^2*binomial(2*k,k)); \\ Michel Marcus, Nov 13 2016

Formula

Recurrence (obtained via the Zeilberger algorithm):

(n+3)^2*(4n+5)*a(n+3) + (20n^3+125n^2+254n+165)*a(n+2) + (76n^3+399n^2+678n+375)*a(n+1) - 25*(n+1)^2*(4n+9)*a(n) = 0.

Lim_sup_{n->oo} |a(n)|^(1/n) = 5. - Vaclav Kotesovec, Jul 13 2014

a(n) = Sum_{k = 0..n} (-1)^(n-k)*C(n,2*k)^2*C(2*k,k) = Sum_{k = 0..n} (-1)^(n-k)*C(n,k)*C(n,2*k)*C(n-k,k). - Zhi-Wei Sun, Nov 12 2016

Conjecture: a(n) = Sum_{k = 0..n} binomial(n, k)*b(k), where b(n) = Sum_{k = 0..n} (-1)^k*binomial(n, k)^2*binomial(2*k, n). [Added Mar 16 2025: this conjecture can be verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package for Maple]. - Peter Bala, Jul 19 2024