%I #20 Jun 06 2021 09:02:03
%S 1,72,4824,316736,20614104,1335305664,86248451520,5560325134848,
%T 357992555533272,23026456586057408,1479999826835627328,
%U 95071036081670530560,6104320340924619384256,391801560518407856592384
%N a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.
%C The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), A167713 (B=16).
%C The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)^3/B^k gives A079727 for B=1, A167867 (B=2), A167868 (B=3), A167869 (B=4), A167870 (B=16), A167871 (B=64).
%C p^2 divides all a(n) from n = (p-1)/2 to n = p-1 for prime p of the form p = 4k+3, p = {3,7,11,19,23,31,43,47,59,...} = A002145.
%C p^2 divides all a(n) from n = (2p-1 - (p-1)/2) to n = 2p-1 for prime p of the form p = 4k+3.
%C p^2 divides all a(n) from n = (3p-1 - (p-1)/2) to n = 3p-1 for prime p of the form p = 4k+3.
%C p^2 divides all a(n) from n = (p^2-1)/2 to n = p^2-1 for prime p of the form p = 4k+3.
%D W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd ed., Wiley, 1968, p. 361.
%H Vincenzo Librandi, <a href="/A167871/b167871.txt">Table of n, a(n) for n = 0..200</a>
%F a(n) = 64^n * Sum_{k=0..n} binomial(2*k,k)^3 / 64^k.
%F Recurrence: n^3*a(n) = 8*(4*n-1)*(4*n^2 - 2*n + 1)*a(n-1) - 512*(2*n-1)^3 *a(n-2). - _Vaclav Kotesovec_, Aug 14 2013
%F a(n) ~ 64^n*(Pi/GAMMA(3/4)^4 - 2/(Pi^(3/2)*sqrt(n))). - _Vaclav Kotesovec_, Aug 14 2013
%t Table[64^n Sum[Binomial[2k,k]^3/64^k,{k,0,n}],{n,0,20}] (* _Vincenzo Librandi_, Mar 26 2012 *)
%Y Cf. A079727, A167867, A167868, A167869, A167870, A167872.
%Y Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167859, A002145.
%K nonn
%O 0,2
%A _Alexander Adamchuk_, Nov 14 2009
%E More terms from _Sean A. Irvine_, Apr 25 2010