%I #27 Aug 28 2023 10:51:23

%S 1,1,2,6,21,79,312,1277,5369,23049,100612,445214,1992606,9004260,

%T 41025315,188259072,869305315,4036286518,18832973733,88259024068,

%U 415252542641,1960718710035,9288106921038,44129146527731

%N Expansion of solution of functional equation.

%H Seiichi Manyama, <a href="/A112806/b112806.txt">Table of n, a(n) for n = 0..1000</a>

%H Gi-Sang Cheon, S.-T. Jin, L. W. Shapiro, <a href="http://dx.doi.org/10.1016/j.laa.2015.03.015">A combinatorial equivalence relation for formal power series</a>, Linear Algebra and its Applications, Available online 30 March 2015.

%F Given g.f. A(x), then series reversion of B(x)=x*A(x^3) is -B(-x).

%F Given g.f. A(x), then y=x*A(x^3) satisfies y=x+(xy)^2/(1-(xy)^3).

%F G.f. satisfies: A(x) = 1 + x*A(x)^2/(1 - x^2*A(x)^3). - _Paul D. Hanna_, Jun 06 2012

%F G.f. satisfies: A(x) = 1/A(-x*A(x)^3); note that the Catalan function C(x) = 1 + x*C(x)^2 (A000108) also satisfies this condition. - _Paul D. Hanna_, Jun 06 2012

%F a(n) = Sum_{i=0..n/2}((binomial(n+2*i+1,i)*Sum_{k=0..n-2*i}(binomial(k,n-k-2*i)*(-1)^(n-k)*binomial(n+k+2*i,k)))/(n+2*i+1)). - _Vladimir Kruchinin_, Mar 07 2016

%F a(n) = Sum_{k=0..floor(n/2)} binomial(n-k-1,k) * binomial(2*n-k+1,n-2*k)/(2*n-k+1). - _Seiichi Manyama_, Aug 28 2023

%o (PARI) {a(n)=local(A); if(n<0, 0, A=x+O(x^4); for(k=1,n, A=x+subst(x^2/(1-x^3),x,x*A)); polcoeff(A,3*n+1))}

%o (PARI) {a(n)=local(A=1+x+x*O(x^n));for(i=1,n,A=1+x*A^2/(1-x^2*A^3));polcoeff(A,n)} \\ _Paul D. Hanna_, Jun 06 2012

%o (Maxima)

%o a(n):=sum((binomial(n+2*i+1,i)*sum(binomial(k,n-k-2*i)*(-1)^(n-k)*binomial(n+k+2*i,k),k,0,n-2*i))/(n+2*i+1),i,0,n/2); /* _Vladimir Kruchinin_, Mar 07 2016 */

%Y Cf. A004148, A216490.

%K nonn

%O 0,3

%A _Michael Somos_, Sep 20 2005