OFFSET
0,2
COMMENTS
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.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..300
FORMULA
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
MAPLE
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
MATHEMATICA
Table[4^n*Sum[Binomial[2*k, k]^2/4^k, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 20 2012 *)
PROG
(PARI) a(n) = 4^n*sum(k=0, n, binomial(2*k, k)^2/4^k) \\ Charles R Greathouse IV, Sep 21 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Alexander Adamchuk, Nov 13 2009
EXTENSIONS
More terms from Sean A. Irvine, Apr 14 2010
Further terms from Jon E. Schoenfield, May 09 2010
STATUS
approved