A120974 G.f. A(x) satisfies A(x/A(x)^4) = 1 + x; thus A(x) = 1 + series_reversion(x/A(x)^4).

1, 1, 4, 38, 532, 9329, 190312, 4340296, 108043128, 2890318936, 82209697588, 2467155342740, 77676395612884, 2554497746708964, 87449858261161216, 3107829518797739032, 114399270654847628768, 4353537522757357068296, 171010040645759712226048

Offset: 0

Author

Paul D. Hanna, Jul 20 2006

Keywords

• nonn

Links

• Robert Israel, Table of n, a(n) for n = 0..250

Crossrefs

Cf. A120975; variants: A120970, A120972, A120976, A030266, A067145, A107096.

Programs

Maple

A:= x -> 1:
for m from 1 to 30 do
  Ap:= unapply(A(x)+c*x^m,x);
  S:= series(Ap(x/Ap(x)^4)-1-x, x, m+1);
  cs:= solve(convert(S,polynom),c);
  A:= subs(c=cs, eval(Ap));
od:
seq(coeff(A(x),x,m),m=0..30);# Robert Israel, Oct 25 2019

Mathematica

nmax = 17; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[ A[x/A[x]^4] - 1 - x + O[x]^(n + 1), x] == 0 /. sol; sol = sol ~Join~ Solve[eq][[1]], {n, 1, nmax}];
sol /. Rule -> Set;
a /@ Range[0, nmax] (* Jean-François Alcover, Nov 02 2019 *)

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)^4))[ #A]);A[n+1]}

PARI

b(n, k) = if(k==0, 0^n, k*sum(j=0, n, binomial(4*n+k, j)/(4*n+k)*b(n-j, 4*j)));
a(n) = if(n==0, 1, b(n-1, 4)); \\ Seiichi Manyama, Jun 04 2025

Formula

G.f. satisfies: A(x) = 1 + x*B(x)^4 = 1 + (1 + x*C(x)^4 )^4 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)^4) = A(x), B(x) = A(x*B(x)^4) and B(x) is g.f. of A120975.

From Seiichi Manyama, Jun 04 2025: (Start)

Let b(n,k) = [x^n] B(x)^k, where B(x) is the g.f. of A120975.

b(n,0) = 0^n; b(n,k) = k * Sum_{j=0..n} binomial(4*n+k,j)/(4*n+k) * b(n-j,4*j).

a(n) = b(n-1,4) for n > 0. (End)