|
|
A013709
|
|
a(n) = 4^(2*n+1).
|
|
16
|
|
|
4, 64, 1024, 16384, 262144, 4194304, 67108864, 1073741824, 17179869184, 274877906944, 4398046511104, 70368744177664, 1125899906842624, 18014398509481984, 288230376151711744, 4611686018427387904, 73786976294838206464, 1180591620717411303424, 18889465931478580854784
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
The partial sum of A000888(n) = Catalan(n)^2*(n + 1) resp. A267844(n) = Catalan(n)^2*(4n + 3) resp. A267987(n) = Catalan(n)^2*(4n + 4) divided by A013709(n) (this) a(n) = 2^(4n+2) absolutely converge to 1/Pi resp. 1 resp. 4/Pi. Thus this series is 1/Pi resp. 1 resp. 4/Pi. - Ralf Steiner, Jan 23 2016
|
|
LINKS
|
|
|
FORMULA
|
a(n) = 16*a(n-1), n > 0; a(0) = 4. G.f.: 4/(1 - 16*x). [Philippe Deléham, Nov 23 2008]
E.g.f.: 4*exp(16*x).
|
|
MAPLE
|
|
|
MATHEMATICA
|
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy,changed
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|