A090358 G.f. satisfies A^6 = BINOMIAL(A^5).

1, 1, 6, 66, 1071, 23151, 627236, 20452976, 779947641, 34050858041, 1674497370602, 91575747294582, 5512402585832847, 362148111801511407, 25783279860096503952, 1977349647140061768364, 162508269041154881377519

Offset: 0

Author

Paul D. Hanna, Nov 26 2003

Keywords

• nonn
• easy

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.

In general, if g.f. satisfies A(x)^(k+1) = A(x/(1-x))^k / (1-x), k>0, then a(n) ~ (n-1)! / (k*(k+1) * (log((k+1)/k))^(n+1)). - Vaclav Kotesovec, Nov 19 2014

Examples

			A^6 = BINOMIAL(A090362), since A090362=A^5. Also,
BINOMIAL(A) = A090359^2 since 2=gcd(1+1,6),
BINOMIAL(A^2) = A090360^3 since 3=gcd(2+1,6) and
BINOMIAL(A^3) = A090361^2 since 2=gcd(3+1,6).
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..280

Crossrefs

Cf. A084784, A090351, A090353, A090356, A090359, A090360, A090361, A090362, A094418.

Programs

Magma

m:=40;
f:= func< n,x | Exp((&+[(&+[5^(j-1)*Factorial(j)*StirlingSecond(k,j) *x^k/k: j in [1..k]]): k in [1..n+2]])) >;
R:=PowerSeriesRing(Rationals(), m+1); // A090358
Coefficients(R!( f(m,x) )); // G. C. Greubel, Jun 08 2023

Mathematica

nmax = 16; sol = {a[0] -> 1};
Do[A[x_] = Sum[a[k] x^k, {k, 0, n}] /. sol; eq = CoefficientList[A[x]^6 - A[x/(1 - x)]^5/(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[5^(j-1)*j!* StirlingS2[k,j], {j,k}]*x^k/k, {k,m+1}]], {x,0,m}], x]] (* G. C. Greubel, Jun 08 2023 *)

PARI

{a(n)=local(A); if(n<1,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A^5,x,x/(1-x))/(1-x)+x*O(x^n); A=A-A^6+B);polcoeff(A,n,x))}

SageMath

m=50
def f(n, x): return exp(sum(sum( 5^(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 A090358_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( f(m,x) ).list()
A090358_list(m-5) # G. C. Greubel, Jun 08 2023

Formula

G.f. satisfies: A(x)^6 = A(x/(1-x))^5/(1-x).

a(n) ~ (n-1)! / (30 * (log(6/5))^(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)*5^(k-1) = 1/5*A094418(n) for n >= 1. - Peter Bala, May 26 2015

G.f.: Product_{k>=1} 1/(1 - k*x)^((1/30) * (5/6)^k). - Seiichi Manyama, May 26 2025