A216494 G.f. satisfies A(x) = 1 + x*A(x)^3/(1 - x^4*A(x)^10).

1, 1, 3, 12, 55, 274, 1444, 7923, 44803, 259326, 1529034, 9151733, 55459124, 339595673, 2097962269, 13060078469, 81842038111, 515867610612, 3268440469234, 20803681980270, 132963257157430, 852981624781996, 5490522454007139, 35450567948693263, 229537971398979212, 1490074420399924169, 9696064337840077823

Offset: 0

Author

Paul D. Hanna, Sep 07 2012

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 274*x^5 + 1444*x^6 + 7923*x^7 +...
such that
A(x) = 1 + x*A(x)^3 + x^5*A(x)^13 + x^9*A(x)^23 + x^13*A(x)^33 + x^17*A(x)^43 +...
Given (1) A(x) = 1 + x*A(x)^3 / (1 - x^4*A(x)^10),
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 / (1 - x^4*A(x)^20/A(x)^10),
which illustrates that (2) is consistent with (1).
		

Crossrefs

Cf. A216493, A001764.

Programs

PARI

{a(n)=local(A=1+x); for(i=1, n, A=1+x*A^3/(1 - x^4*A^10 +x*O(x^n))); polcoeff(A, n)}
for(n=0, 30, print1(a(n), ", "))

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/4)} binomial(n-3*k-1,k) * binomial(3*n-2*k+1,n-4*k)/(3*n-2*k+1). - Seiichi Manyama, Aug 28 2023