A115257 Partial sums of binomial(2n,n)^2.

1, 5, 41, 441, 5341, 68845, 922621, 12701245, 178338145, 2542242545, 36677022081, 534311328705, 7846771001041, 116019251361041, 1725360846921041, 25786805857871441, 387084441100423541, 5832802431123111941

Offset: 0

Author

Paul Barry, Jan 18 2006

Keywords

• easy
• nonn

Comments

Central coefficients of number triangle A115255.

p divides all a(n) from a((p-1)/2) to a(p-1) for Gaussian primes p=7,23,31,79,167,431,479,983, ... of the form 4n+3, A002145(n) and for primes of the form 8n+7, A007522(n). - Alexander Adamchuk, Jul 05 2006

Conjecture: For any positive integer n, the polynomials Sum_{k=0}^n binomial(2k,k)^2*x^k and Sum_{k=0}^n binomial(2k,k)^2*x^k/(k+1) are irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• DLMF Digital Library of Mathematical Functions, Elliptic Integrals, NIST, 2016.

Crossrefs

Cf. A000984, A002145, A007522, A115255, A228002.

Programs

Maple

series( 2*EllipticK(4*x^(1/2))/(Pi*(1-x)) ,x=0,20); # Mark van Hoeij, Apr 06 2013

Mathematica

Table[Sum[((2k)!/(k!)^2)^2,{k,0,n}], {n,0,40}] (* Alexander Adamchuk, Jul 05 2006 *)
Accumulate[(Binomial[2#,#])^2&/@Range[0,20]]  (* Harvey P. Dale, Mar 04 2011 *)

Maxima

makelist(sum(binomial(2*k,k)^2,k,0,n),n,0,12); /* Emanuele Munarini, Oct 28 2016 */

PARI

a(n) = sum(k=0, n, binomial(2*k, k)^2); \\ Michel Marcus, Oct 30 2016

Formula

a(n) = Sum_{k=0..n} C(2k, k)^2. a(n) = A115255(2n, n).

a(n) = C(2n,n)^2 + C(2n-2,n-1)^2 + ... + C(2k,k)^2 + ... + C(2,1)^2 + C(0,0)^2, where C(2k,k) = (2k)!/(k!)^2 are the central binomial coefficients A000984(k). - Alexander Adamchuk, Jul 05 2006

a(n) = Sum_{k=0..n} ((2k)!/(k!)^2)^2. a(n) = Sum_{k=0..n} A000984[k]^2. - Alexander Adamchuk, Jul 05 2006

Recurrence: n^2*a(n) = (17*n^2-16*n+4)*a(n-1) - 4*(2*n-1)^2*a(n-2). - Vaclav Kotesovec, Oct 19 2012

a(n) ~ 16^(n+1)/(15*Pi*n). - Vaclav Kotesovec, Oct 19 2012

From Emanuele Munarini, Oct 28 2016: (Start)

Let K(x) be the complete elliptic integral of the first kind as defined in [DLMF, 19.2.4] for phi = Pi/2.

a(n) = (2/Pi)*K(16)-((16^(n+1)*Gamma(n+3/2)^2)/(Pi*Gamma(n+2)^2))*hypergeometric (1,n+3/2,n+3/2;n+2,n+2;16).

G.f.: A(t) = (2/Pi)*(K(16*t)/(1-t)).

Diff. eq. satisfied by the g.f. t*(1-17*t+16*t^2)*A''(t)+(1-35*t+64*t^2)*A'(t)-(5-36*t)*A(t)=0. (End)