0,2

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).

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

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

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.

Vincenzo Librandi, Table of n, a(n) for n = 0..300

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

a(n) ~ 2^(4*n+2)/(3*Pi*n). - Vaclav Kotesovec, Oct 20 2012

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

A167859 := proc(n)

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

4^n*% ;

end proc:

seq(A167859(n), n=0..20) ; # R. J. Mathar, Sep 21 2016

Table[4^n*Sum[Binomial[2*k, k]^2/4^k, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 20 2012 *)

(PARI) a(n) = 4^n*sum(k=0, n, binomial(2*k, k)^2/4^k) \\ Charles R Greathouse IV, Sep 21 2016

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

Sequence in context: A279266 A054915 A073555 * A243246 A113357 A030992

Adjacent sequences: A167856 A167857 A167858 * A167860 A167861 A167862

nonn,easy

Alexander Adamchuk, Nov 13 2009

More terms from Sean A. Irvine, Apr 14 2010

Further terms from Jon E. Schoenfield, May 09 2010

approved