OFFSET
0,3
COMMENTS
Hankel transform is 4^binomial(n+1,2) = A053763(n+1). Case k=2 of T(n,k) = (1/Pi)*2*k^2*(2*k)^n*Integral_{x=-1..1} x^n*sqrt(1-x^2)/(1+k^2-2*k*x) dx. T(n,k) has Hankel transform (k^2)^binomial(n+1,2). k=1 corresponds to C(n,floor(n/2)).
Series reversion of x*(1+x)/(1+2*x+5*x^2).
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
FORMULA
G.f.: (sqrt(1-16*x^2) + 2*x - 1)/(2*x*(1-5*x)) = c(4*x^2)/(1-x*c(4*x^2)), c(x) the g.f. of A000108.
a(n) = (1/(n+1))*Sum_{k=0..n+1} Sum_{j=0..k} C(n,k)*C(k,j)*C(2*n-2*k+j, n-2*k+j)*(-1)^(n-2*k+j)*2^j*5^(k-j).
a(n) = (1/Pi)*8*4^n*Integral_{x=-1..1} x^n*sqrt(1-x^2)/(5-4*x) dx.
a(n) = Sum_{k=0..floor(n/2)} A009766(n-k,k)*2^2k. - Philippe Deléham, Aug 18 2006
a(n) = Sum_{k=0..n} 4^(n-k)*A120730(n,k). - Philippe Deléham, Oct 16 2008
Conjecture: (n+1)*a(n) = 5*(n+1)*a(n-1) + 16*(n-2)*a(n-2) - 80*(n-2)*a(n-3). - R. J. Mathar, Nov 26 2012
a(n) ~ (9+(-1)^n) * 2^(2*n+5/2) / (9 * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Feb 13 2014
MATHEMATICA
CoefficientList[Series[(Sqrt[1-16*x^2]+2*x-1)/(2*x*(1-5*x)), {x, 0, 40}], x] (* Vaclav Kotesovec, Feb 13 2014 *)
PROG
(Magma) R<x>:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( (Sqrt(1-16*x^2)+2*x-1)/(2*x*(1-5*x)) )); // G. C. Greubel, Nov 07 2022
(SageMath)
def A120730(n, k): return 0 if (n>2*k) else binomial(n, k)*(2*k-n+1)/(k+1)
[A121724(n) for n in range(51)] # G. C. Greubel, Nov 07 2022
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Aug 17 2006, Feb 28 2007
EXTENSIONS
More terms from Vincenzo Librandi, Feb 15 2014
STATUS
approved