A234839 a(n) = Sum_{k = 0..n} (-1)^k * binomial(n,k) * binomial(2*n,k).

1, -1, -1, 8, -17, -1, 116, -344, 239, 1709, -7001, 9316, 22276, -138412, 268568, 189008, -2608913, 6809417, -1814851, -45852416, 159116983, -155628353, -720492928, 3481793888, -5558713852, -9029921876, 71541001076, -158672882224, -45300345128, 1370202238072

Offset

0,4

Comments

For each n > 0, a(p-n) == 2^(2 - 3*n)*A252355(n) (mod p), for all primes p >= 2*n+1 [Chamberland, et al., Thm. 2.3]. - L. Edson Jeffery, Dec 17 2014

References

R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Links

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Marc Chamberland and Karl Dilcher, A Binomial Sum Related to Wolstenholme's Theorem, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672.

Robert Osburn, Brundaban Sahu, and Armin Straub, Supercongruences for sporadic sequences, arXiv:1312.2195 [math.NT], 2014.

Formula

Recurrence: 2*n*(2*n-1)*(7*n-10)*a(n) = -(91*n^3 - 221*n^2 + 160*n - 36)*a(n-1) - 16*(n-1)*(2*n-3)*(7*n-3)*a(n-2).

Lim sup n->infinity |a(n)|^(1/n) = 2*sqrt(2).

exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 - x + 3*x^3 - 7*x^4 + 4*x^5 + 24*x^6 - 85*x^7 + 99*x^8 + 215*x^9 - 1196*x^10 + ... appears to have integer coefficients. - Peter Bala, Jan 04 2016

From Peter Bala, Apr 02 2020: (Start)

a(n) = Sum_{k = 0..floor(n/2)} (-1)^(n+k)*binomial(n,k)*binomial(n,2*k).

a(n) = hypergeom([-n, -n/2, 1/2 - n/2], [1, 1/2], 1). (End)

From Peter Bala, Feb 23 2022: (Start)

a(n) = [x^n] ((1 - x)*(1 - x^2))^n. This implies that exp( Sum_{n >= 1} a(n)*x^n/n ) has integer coefficients as suggested above.

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.

Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k. [added Apr 12 2022: this was proved in 2014 by Osburn et al.; see Example 3.3]

The o.g.f. A(x) is the diagonal of the bivariate rational function 1/(1 - t*(1-x)*(1-x^2) ) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.

Let F(x) = 1/x*Series_Reversion( x/((1-x)*(1-x^2)) ). Then A(x) = 1 + x*d/dx(Log(F(x))). (End)

a(n) = Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(2*n+k, k)*2^(n-k). - Peter Bala, Feb 14 2024

Mathematica

Table[Sum[(-1)^k*Binomial[n, k]*Binomial[2*n, k], {k, 0, n}], {n, 0, 20}]
Table[Hypergeometric2F1[-2*n, -n, 1, -1], {n, 0, 20}]

Prog

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

Crossrefs

Cf. A005809, A252355, A348410.

Sequence in context: A024280 A115434 A024107 * A066554 A302976 A244537

Adjacent sequences: A234836 A234837 A234838 * A234840 A234841 A234842

Keyword

sign,easy

Author

Vaclav Kotesovec, Dec 31 2013

Status

approved