A168358 Self-convolution square of A001246, which is the squares of Catalan numbers.
1, 2, 9, 58, 458, 4120, 40569, 426842, 4723890, 54402904, 646992474, 7900772120, 98642862232, 1254984808672, 16227116787737, 212790354730842, 2824992774357362, 37915366854924952, 513837166842215970
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + 2*x + 9*x^2 + 58*x^3 + 458*x^4 + 4120*x^5 +... A(x)^(1/2) = 1 + x + 4*x^2 + 25*x^3 + 196*x^4 + 1764*x^5 + 17424*x^6 +...+ A001246(n)*x^n +... A(x) satisfies: A(x/G(x)^2) = G(x)^2 where G(x) = g.f. of A006664: G(x) = 1 + x + 2*x^2 + 8*x^3 + 46*x^4 + 322*x^5 + 2546*x^6 +...+ A006664(n)*x^n +... G(x)^2 = 1 + 2*x + 5*x^2 + 20*x^3 + 112*x^4 + 768*x^5 + 5984*x^6 +...+ A168357(n)*x^n +...
Programs
-
Mathematica
Table[Sum[CatalanNumber[k]^2 * CatalanNumber[n-k]^2, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 10 2018 *)
-
PARI
{a(n)=local(C_2=vector(n+1, m, (binomial(2*m-2, m-1)/m)^2)); polcoeff(Ser(C_2)^2, n)}
Formula
G.f.: A(x) = (1/x)*Series_Reversion(x/G(x)^2) where G(x) = g.f. of A006664, which is the number of irreducible systems of meanders.
G.f.: A(x) = G(x*A(x))^2 where A(x/G(x)^2) = G(x)^2 where G(x) = g.f. of A006664.
From Vaclav Kotesovec, Mar 10 2018: (Start)
Recurrence: (n+1)^2*(n+2)^3*(4*n^2 - 5*n - 3)*a(n) = 4*(n+1)^2*(48*n^5 - 12*n^4 - 136*n^3 + 15*n^2 + 49*n - 30)*a(n-1) - 32*(96*n^7 - 312*n^6 + 104*n^5 + 580*n^4 - 630*n^3 + 80*n^2 + 91*n - 12)*a(n-2) + 1024*(n-2)^3*(2*n - 3)^2*(4*n^2 + 3*n - 4)*a(n-3).
a(n) ~ (4/Pi - 1) * 2^(4*n + 3) / (Pi*n^3). (End)