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A120974
G.f. satisfies: A(x/A(x)^4) = 1 + x; thus A(x) = 1 + series_reversion(x/A(x)^4).
6
1, 1, 4, 38, 532, 9329, 190312, 4340296, 108043128, 2890318936, 82209697588, 2467155342740, 77676395612884, 2554497746708964, 87449858261161216, 3107829518797739032, 114399270654847628768, 4353537522757357068296
OFFSET
0,3
LINKS
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.
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 *)
PROG
(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]}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jul 20 2006
STATUS
approved