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A090356 Satisfies A^5 = BINOMIAL(A^4). 6

%I

%S 1,1,5,45,595,10475,231255,6148495,191276600,6815243040,273601200136,

%T 12217471594856,600580173151560,32224787998758280,1873909224391774760,

%U 117388347849375956328,7880739469498103077588,564440024187816634143380

%N Satisfies A^5 = BINOMIAL(A^4).

%C 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.

%H Vaclav Kotesovec, <a href="/A090356/b090356.txt">Table of n, a(n) for n = 0..310</a>

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

%F a(n) ~ (n-1)! / (20 * (log(5/4))^(n+1)). - _Vaclav Kotesovec_, Nov 19 2014

%F 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

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

%o (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))}

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

%K nonn,easy

%O 0,3

%A _Paul D. Hanna_, Nov 26 2003

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Last modified March 23 04:46 EDT 2019. Contains 321422 sequences. (Running on oeis4.)