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A090356 Satisfies A^5 = BINOMIAL(A^4). 6
1, 1, 5, 45, 595, 10475, 231255, 6148495, 191276600, 6815243040, 273601200136, 12217471594856, 600580173151560, 32224787998758280, 1873909224391774760, 117388347849375956328, 7880739469498103077588, 564440024187816634143380 (list; graph; refs; listen; history; text; internal format)
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

A^5 = BINOMIAL(A090357), since A090357=A^4.

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))}

CROSSREFS

Cf. A084784, A090353, A090357, A090358; A050353, A094417.

Sequence in context: A051539 A007696 A090136 * A201365 A112940 A294332

Adjacent sequences:  A090353 A090354 A090355 * A090357 A090358 A090359

KEYWORD

nonn,easy

AUTHOR

Paul D. Hanna, Nov 26 2003

STATUS

approved

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Last modified February 22 01:45 EST 2019. Contains 320381 sequences. (Running on oeis4.)