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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. 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 *) 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: A007696 A090136 A357322 * A201365 A112940 A343710 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 November 30 08:36 EST 2022. Contains 358438 sequences. (Running on oeis4.)