login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A098665
a(n) = Sum_{k = 0..n} binomial(n,k) * binomial(n+1,k+1) * 4^k.
4
1, 6, 43, 332, 2661, 21810, 181455, 1526040, 12939145, 110413406, 947052723, 8157680228, 70518067309, 611426078346, 5315138311383, 46308989294640, 404274406256145, 3535479068797110, 30966952059306555, 271616893912241532, 2385412594943633781, 20973327081776664546
OFFSET
0,2
COMMENTS
Fifth binomial transform of A098664.
LINKS
FORMULA
G.f.: ((1+3*x)-sqrt(1-10*x+9*x^2))/(8*x*sqrt(1-10*x+9*x^2)).
E.g.f.: exp(5x)*(BesselI(0, 4x)+BesselI(1, 4x)/2).
Recurrence: (n+1)*(2*n-1)*a(n) = 4*(5*n^2-2)*a(n-1) - 9*(n-1)*(2*n+1)*a(n-2). - Vaclav Kotesovec, Oct 15 2012
a(n) ~ 9^(n+1)/(4*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 15 2012
From Peter Bala, Jan 07 2022: (Start)
The following formulas assume an offset of 1:
a(n) = (1/4) * Sum_{k = 0..n} binomial(n,k)*A119259(k).
a(n) = (1/4) * Sum_{k = 0..n} binomial(n,k)*binomial(2*n-k-1,n-k)*3^k.
a(n) = (1/4) * [x^n] ((1 + 3*x)/(1 - x))^n.
The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p >= 3 and positive integers n and k. (End)
a(n) = 4^n*hypergeom([-n, -n - 1], [1], 1/4). - Detlef Meya, May 21 2024
MATHEMATICA
Table[SeriesCoefficient[((1+3*x)-Sqrt[1-10*x+9*x^2])/(8*x*Sqrt[1-10*x+9*x^2]), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 15 2012 *)
a[n_] := 4^n*HypergeometricPFQ[{-n, -n - 1}, {1}, 1/4]; Flatten[Table[a[n], {n, 0, 21}]] (* Detlef Meya, May 21 2024 *)
PROG
(PARI) my(x='x+O('x^66)); Vec(((1+3*x)-sqrt(1-10*x+9*x^2))/(8*x*sqrt(1-10*x+9*x^2))) \\ Joerg Arndt, May 12 2013
CROSSREFS
Sequence in context: A091128 A349302 A025594 * A153397 A005786 A071541
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Sep 20 2004
STATUS
approved