A127927 G.f. A(x) satisfies: [x^(2n)] A(x)/Catalan(x)^n = A001764(n) = C(3n,n)/(2n+1) and [x^(2n+1)] A(x)/Catalan(x)^n = A001764(n+1) for n>=0, where Catalan(x) is the g.f. of A000108.
1, 1, 3, 9, 31, 108, 391, 1431, 5319, 19926, 75252, 285750, 1090491, 4177774, 16060401, 61916977, 239307063, 926929746, 3597296770, 13984508500, 54448030092, 212282062488, 828673761978, 3238495227846, 12669206034339
Offset: 0
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
Programs
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Magma
[1] cat [Binomial(2*n,n) - (-1)^(n-1)*(&+[Binomial(3*k, k)*Binomial(k-n - 1, n-2*k-1)/(2*k+1): k in [0..Floor((n-1)/2)]]): n in [1..50]]; // G. C. Greubel, Apr 30 2018
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Mathematica
a[n_] := Binomial[2*n, n] - (-1)^(n-1)*Sum[ Binomial[3*k, k]*Binomial[k - n-1, n-1-2*k]/(2*k+1), {k, 0, Floor[(n-1)/2]}]; Table[a[n], {n, 0, 50}] (* G. C. Greubel, Apr 30 2018 *) -
PARI
{a(n)=binomial(2*n,n)+(-1)^n*sum(i=0,(n-1)\2, binomial(3*i,i) *binomial(i-n-1,n-1-2*i)/(2*i+1))}
Formula
a(n) = C(2*n,n) - (-1)^(n-1)*Sum_{i=0..[(n-1)/2]} C(3*i,i)*C(i-n-1,n-1-2*i)/(2*i+1).
From Vaclav Kotesovec, May 01 2018: (Start)
Recurrence: 2*(n-1)*n*(2*n + 1)*(5*n - 6)*a(n) = (n-1)^2*(115*n^2 - 138*n + 56)*a(n-1) + 4*(n-2)*(n+1)*(2*n - 3)*(5*n - 11)*a(n-2) - 36*(n-2)*(2*n - 5)*(2*n - 3)*(5*n - 1)*a(n-3).
a(n) ~ 4^n / (phi^2 * sqrt(Pi*n)), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. (End)
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