A219197 Self-convolution equals A199033.
1, 2, 9, 46, 253, 1452, 8570, 51594, 315225, 1948010, 12147881, 76316508, 482392198, 3064987460, 19560379470, 125309993974, 805458510441, 5192500350906, 33561539356277, 217429403317006, 1411572472199649, 9181398851046632, 59821825063376124, 390382132833183204
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + 2*x + 9*x^2 + 46*x^3 + 253*x^4 + 1452*x^5 +... where A(x)^2 = 1 + 4*x + 22*x^2 + 128*x^3 + 771*x^4 + 4744*x^5 +...+ A199033(n)*x^n +... Also, the g.f. A(x) satisfies: A(x) = G(x) * F(x*G(x)^2) where F(x) = 1 + x + 3*x^2 + 7*x^3 + 19*x^4 + 51*x^5 + 141*x^6 +...+ A002426(n)*x^n +... G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...+ A001764(n)*x^n +...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
Programs
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Mathematica
A002426[n_] := Sum[Binomial[n, 2*k]*Binomial[2*k, k], {k, 0, Floor[n/2]}]; Table[Sum[A002426[k]*Binomial[3*n - k + 1, n - k]*(2*k + 1)/(3*n - k + 1), {k, 0, n}], {n, 0, 50} ] (* G. C. Greubel, Mar 06 2017 *)
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PARI
{a(n)=local(A2=sum(m=0, n, sum(k=0, m, binomial(m+k+1, m-k)*binomial(2*m-k+1, k))*x^m+x*O(x^n))); polcoeff(A2^(1/2), n)} for(n=0, 30, print1(a(n), ", ")) -
PARI
{a(n)=local(G=1); for(i=0, n, G=1+x*G^3+O(x^(n+1))); polcoeff(G/sqrt(1-2*x*G^2-3*x^2*G^4), n)} for(n=0, 30, print1(a(n), ", ")) -
PARI
{A002426(n)=sum(k=0, n\2, binomial(n, 2*k)*binomial(2*k, k))} {a(n)=if(n==0, 1, sum(k=0, n, A002426(k)*binomial(3*n-k+1, n-k)*(2*k+1)/(3*n-k+1)))} for(n=0, 30, print1(a(n), ", ")) -
PARI
{A097893(n)=sum(m=0, n, sum(k=0, m\2, binomial(m, 2*k)*binomial(2*k, k)))} {a(n)=if(n==0, 1, sum(k=0, n, A097893(k)*binomial(3*n-k, n-k)*2*k/(3*n-k)))} for(n=0, 30, print1(a(n), ", "))
Formula
Sum_{k=0..n} a(n-k)*a(k) = Sum_{k=0..n} C(n+k+1,n-k)*C(2*n-k+1,k) = A199033(n).
G.f.: A(x) = G(x) / sqrt(1 - 2*x*G(x)^2 - 3*x^2*G(x)^4), where G(x) = 1 + x*G(x)^3 = g.f. of A001764.
G.f.: A(x) = Sum_{n>=0} A002426(n) * x^n * G(x)^(2*n+1), where A002426 are the central trinomial coefficients and G(x) = 1 + x*G(x)^3 = g.f. of A001764.
a(n) = Sum_{k=0..n} A002426(k) * C(3*n-k+1,n-k) * (2*k+1)/(3*n-k+1) for n>0, where A002426 are the central trinomial coefficients.
From Vaclav Kotesovec, Oct 05 2020: (Start)
Recurrence: 32*(n-1)*n*(2*n + 1)*(49*n^2 - 210*n + 222)*a(n) = 4*(n-1)*(10388*n^4 - 55104*n^3 + 96925*n^2 - 64446*n + 15006)*a(n-1) - 6*(22050*n^5 - 173439*n^4 + 536588*n^3 - 814340*n^2 + 604331*n - 174702)*a(n-2) - 81*(n-2)*(3*n - 7)*(3*n - 5)*(49*n^2 - 112*n + 61)*a(n-3).
a(n) ~ 3^(3*n + 7/4) / (Gamma(1/4) * n^(3/4) * 2^(2*n + 5/2)). (End)
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