A246460 a(n) = (sum_{k=0}^{n-1} (2k+1)*C(n-1,k)^2*C(n+k,k)^2)/n^2, where C(n,k) denotes the binomial coefficient n!/(k!(n-k)!).

1, 7, 77, 1211, 23009, 489035, 11203765, 270937315, 6825612185, 177559028087, 4739821161173, 129244697791951, 3587524535220001, 101099089948850323, 2886373390151379397, 83343790441133767475, 2430567530705659113545, 71508611747063572974095, 2120357936904537499679125, 63315310358625743871987019

Offset: 1

Author

Zhi-Wei Sun, Aug 26 2014

Keywords

• nonn

Comments

Conjecture: a(n) is always an integer.

The author proved this in the latest version of arXiv:1408.5381. - Zhi-Wei Sun, Sep 01 2014

Examples

			a(2) = 7 since sum_{k=0,1} (2k+1)C(1,k)^2*C(2+k,k)^2 = 1 + 3*3^2 = 28 = 2^2*7.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..100
• Zhi-Wei Sun, Two new kinds of numbers and related divisibility results, arXiv:1408.5381, 2014.

Crossrefs

Cf. A245769, A246065, A246138, A246459.

Programs

Maple

A246460:=n->add((2*k+1)*binomial(n-1,k)^2*binomial(n+k,k)^2/n^2, k=0..n-1): seq(A246460(n), n=1..20); # Wesley Ivan Hurt, Aug 26 2014

Mathematica

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

Formula

Recurrence (obtained via the Zeilberger algorithm):

-n^3*(2*n + 5)*(3*n^2 + 12*n + 11)*a(n) + (2*n + 5)*(105*n^5 + 675*n^4 + 1579*n^3 + 1663*n^2 + 768*n+126)*a(n+1) - (2*n + 1)*(105*n^5 + 900*n^4 + 2929*n^3 + 4448*n^2 + 3048*n + 684)*a(n+2) + (n + 3)^3*(2*n + 1)*(3*n^2 + 6*n + 2)*a(n+3) = 0.

a(n) ~ 2^(1/4) * (17+12*sqrt(2))^n / (8 * Pi^(3/2) * n^(5/2)). - Vaclav Kotesovec, Aug 27 2014

Extensions

Typo in cross-reference corrected by Vaclav Kotesovec, Aug 27 2014