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URL: https://oeis.org/A216493

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A216493
G.f. satisfies A(x) = 1 + x*A(x)^3 + x^5*A(x)^13.
1
1, 1, 3, 12, 55, 274, 1444, 7923, 44803, 259325, 1529008, 9151327, 55454164, 339543312, 2097460255, 13055579858, 81803671623, 515552408141, 3265924761595, 20784056808550, 132812937949820, 851847261569025, 5482066256568375, 35388168141000935, 229081418808206500, 1486757986305948780, 9672120691595571320
OFFSET
0,3
FORMULA
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.
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
EXAMPLE
G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 274*x^5 + 1444*x^6 + 7923*x^7 +...
Related expansions:
A(x)^3 = 1 + 3*x + 12*x^2 + 55*x^3 + 273*x^4 + 1431*x^5 + 7806*x^6 + 43893*x^7 +...
A(x)^13 = 1 + 13*x + 117*x^2 + 910*x^3 + 6578*x^4 + 45643*x^5 + 309127*x^6 +...
Given (1) A(x) = 1 + x*A(x)^3 + x^5*A(x)^13,
suppose (2) A(x) = 1/A(-x*A(x)^5),
then substituting x in (1) with -x*A(x)^5 yields:
1/A(x) = 1 - x*A(x)^5/A(x)^3 - x^5*A(x)^25/A(x)^13,
which illustrates that (2) is consistent with (1).
MATHEMATICA
terms=27; A[_]=1; Do[A[x_] = 1 + x*A[x]^3 + x^5*A[x]^13 + O[x]^terms//Normal, terms]; CoefficientList[Series[A[x], {x, 0, terms}], x] (* Stefano Spezia, Sep 08 2025 *)
PROG
(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)}
for(n=0, 30, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 07 2012
STATUS
approved