|
|
A115257
|
|
Partial sums of binomial(2n,n)^2.
|
|
4
|
|
|
1, 5, 41, 441, 5341, 68845, 922621, 12701245, 178338145, 2542242545, 36677022081, 534311328705, 7846771001041, 116019251361041, 1725360846921041, 25786805857871441, 387084441100423541, 5832802431123111941
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
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
|
|
|
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
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
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)
|
|
MAPLE
|
series( 2*EllipticK(4*x^(1/2))/(Pi*(1-x)) , x=0, 20); # Mark van Hoeij, Apr 06 2013
|
|
MATHEMATICA
|
Accumulate[(Binomial[2#, #])^2&/@Range[0, 20]] (* Harvey P. Dale, Mar 04 2011 *)
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|