A234839 - OEIS

%I #48 Mar 03 2024 08:35:06

%S 1,-1,-1,8,-17,-1,116,-344,239,1709,-7001,9316,22276,-138412,268568,

%T 189008,-2608913,6809417,-1814851,-45852416,159116983,-155628353,

%U -720492928,3481793888,-5558713852,-9029921876,71541001076,-158672882224,-45300345128,1370202238072

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

%C 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

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

%H Seiichi Manyama, <a href="/A234839/b234839.txt">Table of n, a(n) for n = 0..1000</a>

%H Marc Chamberland and Karl Dilcher, <a href="http://dx.doi.org/10.1016/j.jnt.2009.05.010">A Binomial Sum Related to Wolstenholme's Theorem</a>, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672.

%H Robert Osburn, Brundaban Sahu, and Armin Straub, <a href="https://arxiv.org/abs/1312.2195">Supercongruences for sporadic sequences</a>, arXiv:1312.2195 [math.NT], 2014.

%F 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).

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

%F 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

%F From _Peter Bala_, Apr 02 2020: (Start)

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

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

%F From _Peter Bala_, Feb 23 2022: (Start)

%F 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.

%F 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.

%F 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]

%F 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.

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

%F a(n) =  Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(2*n+k, k)*2^(n-k). - _Peter Bala_, Feb 14 2024

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

%t Table[Hypergeometric2F1[-2*n, -n, 1, -1],{n,0,20}]

%o (PARI) a(n) = sum(k=0, n, (-1)^k*binomial(n,k)*binomial(2*n,k)); \\ _Michel Marcus_, Jan 13 2016

%Y Cf. A005809, A252355, A348410.

%K sign,easy

%O 0,4

%A _Vaclav Kotesovec_, Dec 31 2013