login
A167713
a(n) = 16^n * Sum_{k=0..n} binomial(2*k, k) / 16^k.
8
1, 18, 294, 4724, 75654, 1210716, 19372380, 309961512, 4959397062, 79350401612, 1269606610548, 20313706474200, 325019306291356, 5200308911062296, 83204942617113336, 1331279082028930896, 21300465313063974726
OFFSET
0,2
COMMENTS
p^2 divides a((p-3)/2) for prime p of the form p = 6k + 1 (A002476).
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), a(n) = A167713 (B=16).
LINKS
Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
Eric Weisstein's World of Mathematics, Binomial Sums.
FORMULA
a(n) = 16^n * Sum_{k=0..n} ((2k)!/(k!)^2) / 16^k.
a(n) = 16^n * Sum_{k=0..n} binomial(2k,k) / 16^k.
G.f.: 1/((1-16*x)*sqrt(1-4*x)). - R. J. Mathar, Nov 13 2009
Recurrence: n*a(n) = 2*(10*n-1)*a(n-1) - 32*(2*n-1)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(4*n+1)/sqrt(3). - Vaclav Kotesovec, Oct 20 2012
MAPLE
A167713 := proc(n) coeftayl( 1/(1-16*x)/sqrt(1-4*x), x=0, n) ; end proc: seq(A167713(n), n=0..40) ; # R. J. Mathar, Nov 13 2009
MATHEMATICA
16^n * Sum[ (2k)!/(k!)^2 / 16^k, {k, 0, 50} ].
CoefficientList[Series[1 / ((1 - 16 x) Sqrt[1 - 4 x]), {x, 0, 20}], x] (* Vincenzo Librandi, May 27 2013 *)
KEYWORD
nonn
AUTHOR
Alexander Adamchuk, Nov 10 2009
EXTENSIONS
Extended by R. J. Mathar, Nov 13 2009
STATUS
approved