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A090351 Satisfies A^3 = BINOMIAL(A^2). 7
1, 1, 3, 15, 108, 1032, 12388, 179572, 3052986, 59555338, 1310677726, 32114051862, 866766965308, 25547102523604, 816335926158372, 28107705687291892, 1037367351120788551, 40852168787823027351, 1709792654612819858341 (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..390

FORMULA

G.f. satisfies: A(x)^3 = A(x/(1-x))^2/(1-x).

a(n) ~ (n-1)! / (6 * (log(3/2))^(n+1)). - Vaclav Kotesovec, Nov 18 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)*2^(k-1) = A050351(n) = 1/2*A004123(n+1) for n >= 1. - Peter Bala, May 26 2015

EXAMPLE

A^3 = BINOMIAL(A090352), since A090352=A^2.

MATHEMATICA

nmax = 18; sol = {a[0] -> 1};

Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^3 - A[x/(1 - x)]^2/(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-Fran├žois 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^2, x, x/(1-x))/(1-x)+x*O(x^n); A=A-A^3+B); polcoeff(A, n, x))}

CROSSREFS

Cf. A084784, A090352, A090353, A090356, A090358. A004123, A050351.

Sequence in context: A105618 A120732 A245835 * A136221 A153305 A110328

Adjacent sequences:  A090348 A090349 A090350 * A090352 A090353 A090354

KEYWORD

nonn,easy

AUTHOR

Paul D. Hanna, Nov 26 2003

STATUS

approved

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Last modified October 25 06:30 EDT 2021. Contains 348239 sequences. (Running on oeis4.)