A098665 a(n) = Sum_{k = 0..n} binomial(n,k) * binomial(n+1,k+1) * 4^k.
1, 6, 43, 332, 2661, 21810, 181455, 1526040, 12939145, 110413406, 947052723, 8157680228, 70518067309, 611426078346, 5315138311383, 46308989294640, 404274406256145, 3535479068797110, 30966952059306555, 271616893912241532, 2385412594943633781, 20973327081776664546
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
Programs
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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 *)
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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
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
Comments