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A090351
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G.f. satisfies A^3 = BINOMIAL(A^2).
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7
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1, 1, 3, 15, 108, 1032, 12388, 179572, 3052986, 59555338, 1310677726, 32114051862, 866766965308, 25547102523604, 816335926158372, 28107705687291892, 1037367351120788551, 40852168787823027351, 1709792654612819858341
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OFFSET
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0,3
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COMMENTS
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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.
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LINKS
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FORMULA
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G.f. satisfies: A(x)^3 = A(x/(1-x))^2/(1-x).
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
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EXAMPLE
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MATHEMATICA
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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;
With[{m=40}, CoefficientList[Series[Exp[Sum[Sum[2^(j-1)*j!* StirlingS2[k, j], {j, k}]*x^k/k, {k, m+1}]], {x, 0, m}], x]] (* G. C. Greubel, Jun 08 2023 *)
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PROG
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(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))}
(Magma)
m:=40;
f:= func< n, x | Exp((&+[(&+[2^(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); // A090351
(SageMath)
m=50
def f(n, x): return exp(sum(sum(2^(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)))
P.<x> = PowerSeriesRing(QQ, prec)
return P( f(m, x) ).list()
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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