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Self-convolution square of A001246, which is the squares of Catalan numbers.
2

%I #5 Mar 10 2018 06:14:35

%S 1,2,9,58,458,4120,40569,426842,4723890,54402904,646992474,7900772120,

%T 98642862232,1254984808672,16227116787737,212790354730842,

%U 2824992774357362,37915366854924952,513837166842215970

%N Self-convolution square of A001246, which is the squares of Catalan numbers.

%F 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.

%F 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.

%F From _Vaclav Kotesovec_, Mar 10 2018: (Start)

%F 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).

%F a(n) ~ (4/Pi - 1) * 2^(4*n + 3) / (Pi*n^3). (End)

%e G.f.: A(x) = 1 + 2*x + 9*x^2 + 58*x^3 + 458*x^4 + 4120*x^5 +...

%e 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 +...

%e A(x) satisfies: A(x/G(x)^2) = G(x)^2 where G(x) = g.f. of A006664:

%e G(x) = 1 + x + 2*x^2 + 8*x^3 + 46*x^4 + 322*x^5 + 2546*x^6 +...+ A006664(n)*x^n +...

%e 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 +...

%t Table[Sum[CatalanNumber[k]^2 * CatalanNumber[n-k]^2, {k, 0, n}], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 10 2018 *)

%o (PARI) {a(n)=local(C_2=vector(n+1, m, (binomial(2*m-2, m-1)/m)^2)); polcoeff(Ser(C_2)^2, n)}

%Y Cf. A168357, A168344, A001246, A006664, A000108.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Nov 23 2009