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%I #29 Apr 18 2022 22:40:27
%S 1,1,2,10,80,1000,17600,418000,12848000,496672000,23576960000,
%T 1348404640000,91442700800000,7255463564800000,665885747225600000,
%U 69994087116448000000,8354181454767104000000,1123646013779238400000000,169165728883243642880000000
%N a(n+1) = sum{j = 0,...n}[C(2n,2j)a(j)a(n-j)] with a(0) = 1.
%C Define e.g.f. A(x) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!. Then A(x) = P(z + x/sqrt(6)) where P is the Weierstrass P-function with g_2 = 0, g_3 = 4 and z = 2.10327313773966...*i satisfies P(z) = 1, P'(z) = 0. - _Michael Somos_, Apr 18 2022
%H Vaclav Kotesovec, <a href="/A063902/b063902.txt">Table of n, a(n) for n = 0..270</a>
%H Markus Kuba, Alois Panholzer, <a href="http://arxiv.org/abs/1411.4587">Combinatorial families of multilabelled increasing trees and hook-length formulas</a>, arXiv:1411.4587 [math.CO], 2014.
%F E.g.f. satisfies: A(x) = exp( Integral 1/A(x) * Integral A(x)^2 dx dx ), where A(x) = Sum_{n>=0} a(n)*x^(1*n)/(2*n)! and the constant of integration is zero. - _Paul D. Hanna_, Jun 02 2015
%F From _Vaclav Kotesovec_, Jun 14 2015: (Start)
%F a(n) ~ c * d^n * n!^2 * sqrt(n), where d = A258895 = 32*Pi / (Gamma(1/6) * Gamma(1/3))^2 = 2^(17/3) * Pi^2 / (3*Gamma(1/3)^6) = 0.452104299183420528841..., c = 1.53043521765866544548745... = 2^(20/3) * Pi^(3/2) / Gamma(1/3)^6 = 192*sqrt(Pi) / (Gamma(1/3)*Gamma(1/6))^2.
%F a(n) ~ 3 * 2^(5*n+7) * Pi^(n+3/2) * n^(2*n+3/2) / (exp(2*n) * Gamma(1/6)^(2*n+2) * Gamma(1/3)^(2*n+2)).
%F a(n) ~ 2^((17*n+23)/3) * Pi^(2*n+5/2) * n^(2*n+3/2) / (3^n * exp(2*n) * Gamma(1/3)^(6*n+6)).
%F (End)
%F Define e.g.f. A(x) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)!. Then 3*A'(x)^2 = 2*A(x)^3 - 2. - _Michael Somos_, Jan 07 2022
%e a(3) = 1*a(0)a(2) + 6*a(1)a(1) + 1*a(2)a(0) = 2+6+2 = 10.
%e E.g.f.: A(x) = 1 + x^2/2! + 2*x^4/4! + 10*x^6/6! + 80*x^8/8! + ...
%e G.f. = 1 + x + 2*x^2 + 10*x^3 + 80*x^4 + 1000*x^8 + 17600*x^9 + ...
%t Clear[a]; a[n_]:=a[n]=If[n<2,1,Sum[a[j]*a[n-1-j]*Binomial[2*n-2,2*j],{j,0,n-1}]]; Table[a[n], {n,0,30}] (* _Vaclav Kotesovec_, Jun 14 2015 *)
%t a[ n_] := If[n < 0, 0, Module[{x, y}, Nest[Dt[#] /. {Dt[x] -> y, Dt[y] -> x^2} &, x, 2*n] /. {x -> 1, y -> 0}]]; (* _Michael Somos_, Apr 18 2022 *)
%o (PARI) /* E.g.f. A(x) = exp( Integral 1/A(x) * Integral A(x)^2 dx dx ) */
%o {a(n) = local(A=1+x); for(i=1,n, A = exp( intformal( 1/A * intformal( A^2 + x*O(x^n)) ) ) ); n!*polcoeff(A,n)}
%o for(n=0,20,print1(a(2*n),", ")) \\ _Paul D. Hanna_, Jun 02 2015
%o (PARI) /* By definition: */
%o {a(n) = if(n==0,1,sum(k=0,n-1, binomial(2*n-2,2*k)*a(k)*a(n-k-1)))}
%o for(n=0,20,print1(a(n),", ")) \\ _Paul D. Hanna_, Jun 02 2015
%o (PARI) {a(n) = if(n<0, 0, my(x='x, y = 1 + O(x^2)); for(i=1, n, y = 1 + intformal(intformal(y^2))); polcoeff(y, 2*n)*(2*n)!)}; /* _Michael Somos_, Apr 18 2022 */
%Y a(n+1) = sum[C(n, j)a(j)a(n-j)] would give factorials A000142, a(n+1) = sum[C(2n, 2j)a(j)a(n-j)]/a(n) would give Catalan numbers A000108, a(n+1) = sum[C(n, j)a(j)a(n-j)]/a(n) would give central binomials A001405.
%Y Cf. A007558, A258895.
%K nonn
%O 0,3
%A _Henry Bottomley_, Aug 30 2001
%E More terms from _Vaclav Kotesovec_, Jun 14 2015
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