%I #16 Aug 28 2023 10:51:36

%S 1,1,3,12,55,274,1444,7923,44803,259325,1529008,9151327,55454164,

%T 339543312,2097460255,13055579858,81803671623,515552408141,

%U 3265924761595,20784056808550,132812937949820,851847261569025,5482066256568375,35388168141000935,229081418808206500,1486757986305948780,9672120691595571320

%N G.f. satisfies A(x) = 1 + x*A(x)^3 + x^5*A(x)^13.

%F G.f. satisfies: A(x) = 1/A(-x*A(x)^5); note that the g.f. of A001764, G(x) = 1 + x*G(x)^3, also satisfies this condition.

%F a(n) = Sum_{k=0..floor(n/5)} binomial(n-4*k,k) * binomial(3*n-2*k+1,n-4*k)/(3*n-2*k+1). - _Seiichi Manyama_, Aug 28 2023

%e G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 274*x^5 + 1444*x^6 + 7923*x^7 +...

%e Related expansions:

%e A(x)^3 = 1 + 3*x + 12*x^2 + 55*x^3 + 273*x^4 + 1431*x^5 + 7806*x^6 + 43893*x^7 +...

%e A(x)^13 = 1 + 13*x + 117*x^2 + 910*x^3 + 6578*x^4 + 45643*x^5 + 309127*x^6 +...

%e Given (1) A(x) = 1 + x*A(x)^3 + x^5*A(x)^13,

%e suppose (2) A(x) = 1/A(-x*A(x)^5),

%e then substituting x in (1) with -x*A(x)^5 yields:

%e 1/A(x) = 1 - x*A(x)^5/A(x)^3 - x^5*A(x)^25/A(x)^13,

%e which illustrates that (2) is consistent with (1).

%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=1+x*A^3+x^5*A^13 +x*O(x^n)); polcoeff(A, n)}

%o for(n=0, 30, print1(a(n), ", "))

%Y Cf. A216494, A216490, A001764.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Sep 07 2012