cp's OEIS Frontend

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)!).

%I A246460 #18 Sep 02 2014 03:18:32
%S A246460 1,7,77,1211,23009,489035,11203765,270937315,6825612185,177559028087,
%T A246460 4739821161173,129244697791951,3587524535220001,101099089948850323,
%U A246460 2886373390151379397,83343790441133767475,2430567530705659113545,71508611747063572974095,2120357936904537499679125,63315310358625743871987019
%N 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)!).
%C A246460 Conjecture: a(n) is always an integer.
%C A246460 The author proved this in the latest version of arXiv:1408.5381. - _Zhi-Wei Sun_, Sep 01 2014
%H A246460 Zhi-Wei Sun, <a href="/A246460/b246460.txt">Table of n, a(n) for n = 1..100</a>
%H A246460 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1408.5381">Two new kinds of numbers and related divisibility results</a>, arXiv:1408.5381, 2014.
%F A246460 Recurrence (obtained via the Zeilberger algorithm):
%F A246460 -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.
%F A246460 a(n) ~ 2^(1/4) * (17+12*sqrt(2))^n / (8 * Pi^(3/2) * n^(5/2)). - _Vaclav Kotesovec_, Aug 27 2014
%e A246460 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.
%p A246460 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
%t A246460 a[n_]:=Sum[(2k+1)*Binomial[n-1,k]^2*Binomial[n+k,k]^2,{k,0,n-1}]/n^2
%t A246460 Table[a[n],{n,1,20}]
%Y A246460 Cf. A245769, A246065, A246138, A246459.
%K A246460 nonn
%O A246460 1,2
%A A246460 _Zhi-Wei Sun_, Aug 26 2014
%E A246460 Typo in cross-reference corrected by _Vaclav Kotesovec_, Aug 27 2014