A090356 - OEIS

A090356

G.f. A(x) satisfies A(x)^5 = BINOMIAL(A(x)^4); that is, the binomial transform of the coefficients in A(x)^4 yields the coefficients in A(x)^5.

1, 1, 5, 45, 595, 10475, 231255, 6148495, 191276600, 6815243040, 273601200136, 12217471594856, 600580173151560, 32224787998758280, 1873909224391774760, 117388347849375956328, 7880739469498103077588, 564440024187816634143380

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OFFSET

0,3

COMMENTS

In general, if A^n = BINOMIAL(A^(n-1)), then for all integer m>0 there exists an integer sequence B such that B^d = BINOMIAL(A^m) where d=gcd(m+1,n). Also, coefficients of A(k*x)^n = k-th binomial transform of coefficients in A(k*x)^(n-1) for all k>0.

LINKS

Vaclav Kotesovec, Table of n, a(n) for n = 0..310

FORMULA

G.f. satisfies: A(x)^5 = A(x/(1-x))^4/(1-x).

a(n) ~ (n-1)! / (20 * (log(5/4))^(n+1)). - Vaclav Kotesovec, Nov 19 2014

O.g.f.: A(x) = exp( Sum_{n >= 1} b(n)*x^n/n ), where b(n) = Sum_{k = 1..n} k!*Stirling2(n,k)*4^(k-1) = A050353(n) = 1/4*A094417(n) for n >= 1. - Peter Bala, May 26 2015

EXAMPLE

G.f.: A(x) = 1 + x + 5*x^2 + 45*x^3 + 595*x^4 + 10475*x^5 + 231255*x^6 + ...

The coefficients in A(x)^4 are given by A090357 and begin

A(x)^4: [1, 4, 26, 244, 3131, 52600, 1111940, ..., A090357(n), ...].

The binomial transform of A090357 yields the coefficients of A(x)^5:

A(x)^5: [1, 5, 35, 335, 4280, 70976, 1479800, ...]

as shown by

1 = 1*1,

5 = 1*1 + 1*4,

35 = 1*1 + 2*4 + 1*26,

335 = 1*1 + 3*4 + 3*26 + 1*244,

4280 = 1*1 + 4*4 + 6*26 + 4*244 + 1*3131, ...

MATHEMATICA

nmax = 17; sol = {a[0] -> 1};

Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^5 - A[x/(1 - 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 *)

With[{m = 40}, CoefficientList[Series[Exp[Sum[Sum[4^(j-1)*j!* StirlingS2[k, j], {j, k}]*x^k/k, {k, m+1}]], {x, 0, m}], x]] (* G. C. Greubel, Jun 09 2023 *)

PROG

(PARI) {a(n)=local(A); if(n<1, 0, A=1+x+x*O(x^n); for(k=1, n, B=subst(A^4, x, x/(1-x))/(1-x)+x*O(x^n); A=A-A^5+B); polcoeff(A, n, x))}

(Magma)

m:=40;

f:= func< n, x | Exp((&+[(&+[4^(j-1)*Factorial(j)* StirlingSecond(k, j)*x^k/k: j in [1..k]]): k in [1..n+2]])) >;

R<x>:=PowerSeriesRing(Rationals(), m+1); // A090356

Coefficients(R!( f(m, x) )); // G. C. Greubel, Jun 09 2023

(SageMath)

m=40

def f(n, x): return exp(sum(sum(4^(j-1)*factorial(j)* stirling_number2(k, j)*x^k/k for j in range(1, k+1)) for k in range(1, n+2)))

def A090356_list(prec):

P.<x> = PowerSeriesRing(QQ, prec)