|
|
A154623
|
|
Sequence with g.f. 1+(x/(1-5*x))*c(x/(1-5*x)), c(x) the g.f. of A000108.
|
|
4
|
|
|
1, 1, 6, 37, 235, 1539, 10392, 72267, 516474, 3783115, 28317562, 215969271, 1673702191, 13148444197, 104494340880, 838670818365, 6788255966595, 55346471893395, 454123503938490, 3746885525588175, 31066887028255065
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Hankel transform is F(4n+1) (A033889).
a(n+1) is the 5th binomial transform of A000108.
|
|
LINKS
|
|
|
FORMULA
|
G.f.: (1/2)*(3-sqrt((1-9*x)/(1-5*x))).
a(n)=(4/5)*0^n+sum{k=0..n-1, C(n-1,k)*A000108(k)*5^(n-k-1)}.
Conjecture: n*a(n) +2*(8-7n)*a(n-1) +45*(n-2)*a(n-2) = 0. - R. J. Mathar, Dec 14 2011
a(n) = (-1)^(n-1)*(GegenbauerC(n-1,-n+1,7/2) + GegenbauerC(n-2,-n+1,7/2)) for n>=1. - Peter Luschny, May 13 2016
|
|
MATHEMATICA
|
CoefficientList[Series[1/2*(3-Sqrt[(1-9*x)/(1-5*x)]), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|