A088222 -id:A088222 - cp's OEIS frontend

1, 1, 3, 15, 108, 1032, 12388, 179572, 3052986, 59555338, 1310677726, 32114051862, 866766965308, 25547102523604, 816335926158372, 28107705687291892, 1037367351120788551, 40852168787823027351, 1709792654612819858341

Offset: 0

Paul D. Hanna, Nov 26 2003

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.

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

Vaclav Kotesovec, Table of n, a(n) for n = 0..390

Cf. A004123, A050351, A084784, A090352, A090353, A090356, A090358, A088222.

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:=PowerSeriesRing(Rationals(), m+1);  // A090351
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 08 2023

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-François Alcover, Nov 02 2019 *)
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 *)

{a(n) = my(A); if(n<0,0,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); polcoef(A,n,x))}
for(n=0,25,print1(a(n),", "))

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)))
def A090351_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A090351_list(m-9) # G. C. Greubel, Jun 08 2023

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
G.f. satisfies [x^n] 1/A(x)^(2*n-2) = [x^n] 1/A(x)^(3*n-3) = -(n-1)*A088222(n) for n >= 0. - Paul D. Hanna, Apr 28 2025
G.f.: Product_{k>=1} 1/(1 - k*x)^((1/6) * (2/3)^k). - Seiichi Manyama, May 26 2025

cp's OEIS Frontend

A088794 Coefficient of x^n in A(x)^(2n) is A004123(n); self-convolution is A088222.

Views

Author

Keywords

Crossrefs

A090351 G.f. satisfies A^3 = BINOMIAL(A^2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula