A120976 G.f. A(x) satisfies A(x/A(x)^5) = 1 + x ; thus A(x) = 1 + series_reversion(x/A(x)^5).

1, 1, 5, 60, 1060, 23430, 602001, 17281760, 541258390, 18210836060, 651246905140, 24566101401035, 971933892729980, 40156993344526860, 1726753293393763625, 77065076699967844390, 3561820706538663354320, 170162336673835615653925, 8389644485709060522744640

Offset: 0

Author

Paul D. Hanna, Jul 20 2006

Keywords

• nonn

Crossrefs

Cf. A120977; variants: A120970, A120972, A120974, A030266, A067145, A107096.

Programs

PARI

{a(n)=local(A=[1,1]);for(i=2,n,A=concat(A,0); A[ #A]=-Vec(subst(Ser(A),x,x/Ser(A)^5))[ #A]);A[n+1]}

PARI

b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(5*n+k, j)/(5*n+k)*b(n-j, 5*j)));
a(n) = if(n==0, 1, b(n-1, 5)); \\ Seiichi Manyama, Jun 04 2025

Formula

G.f. satisfies: A(x) = 1 + x*B(x)^5 = 1 + (1 + x*C(x)^5 )^5 where B(x) and C(x) satisfy: C(x) = B(x)*B(A(x)-1), B(x) = A(A(x)-1), B(A(x)-1) = A(B(x)-1), B(x/A(x)^5) = A(x), B(x) = A(x*B(x)^5) and B(x) is g.f. of A120977.

From Seiichi Manyama, Jun 04 2025: (Start)

Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120977.

b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(5*n+k,j)/(5*n+k) * b(n-j,5*j).

a(n) = b(n-1,5) for n > 0. (End)