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a(n) = 4^n * Sum_{k=0..n} binomial(2*k, k)^2 / 4^k.
7

%I #21 Jun 07 2021 04:43:34

%S 1,8,68,672,7588,93856,1229200,16695424,232418596,3293578784,

%T 47309094672,686870685312,10059942413584,148412250014336,

%U 2202990595617344,32873407393419776,492791264816231204

%N a(n) = 4^n * Sum_{k=0..n} binomial(2*k, k)^2 / 4^k.

%C Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p = {7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, ...} = A167860, apparently a subset of primes of the form 8n+7 (A007522).

%C 7^3 divides a(13) and 7^2 divides a(10)-a(13).

%C Every a(n) from a(kp-1 - (p-1)/2) to a(kp-1) is divisible by prime p from A167860.

%C Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p from A167860. For p=7 every a(n) from a((p^3-1)/2) to a(p^3-1) and from a((p^4-1)/2) to a(p^4-1)is divisible by p^2.

%H Vincenzo Librandi, <a href="/A167859/b167859.txt">Table of n, a(n) for n = 0..300</a>

%F Recurrence: n^2*a(n) = 4*(5*n^2 - 4*n + 1)*a(n-1) - 16*(2*n - 1)^2*a(n-2). - _Vaclav Kotesovec_, Oct 20 2012

%F a(n) ~ 2^(4*n+2)/(3*Pi*n). - _Vaclav Kotesovec_, Oct 20 2012

%F G.f.: 2*EllipticK(4*sqrt(x))/(Pi*(1-4*x)), where EllipticK is the complete elliptic integral of the first kind, using the Gradshteyn and Ryzhik convention, also used by Maple. In the convention of Abramowitz and Stegun, used by Mathematica, this would be written as 2*K(16*x)/(Pi*(1-4*x)). - _Robert Israel_, Sep 21 2016

%p A167859 := proc(n)

%p add( (binomial(2*k,k)/2^k)^2,k=0..n) ;

%p 4^n*% ;

%p end proc:

%p seq(A167859(n),n=0..20) ; # _R. J. Mathar_, Sep 21 2016

%t Table[4^n*Sum[Binomial[2*k,k]^2/4^k,{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Oct 20 2012 *)

%o (PARI) a(n) = 4^n*sum(k=0,n, binomial(2*k,k)^2/4^k) \\ _Charles R Greathouse IV_, Sep 21 2016

%Y Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167860, A007522.

%K nonn,easy

%O 0,2

%A _Alexander Adamchuk_, Nov 13 2009

%E More terms from _Sean A. Irvine_, Apr 14 2010

%E Further terms from _Jon E. Schoenfield_, May 09 2010