A209917 E.g.f. A(x) satisfies: A(x/(1-x))/(1-x) = (1/x^2) * d/dx x^3*A(x)/3.
1, 3, 21, 249, 4356, 103932, 3213216, 124146432, 5834291328, 326570493312, 21408981213888, 1621281984076224, 140205279698051328, 13711076231477352192, 1503606581609959001088, 183562416179374733411328, 24787906630769478567297024
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + 3*x + 21*x^2/2! + 249*x^3/3! + 4356*x^4/4! + 103932*x^5/5! + ... Related expansions: A(x/(1-x))/(1-x) = 1 + 4*x + 35*x^2/2! + 498*x^3/3! + 10164*x^4/4! + ... A(x) + x*A'(x)/3 = 1 + 4*x + 35*x^2/2! + 498*x^3/3! + 10164*x^4/4! + ... Also, a(n) appears in the expansion: B(x) = 1 + 3*x + 21*x^2/2!^2 + 249*x^3/3!^2 + 4356*x^4/4!^2 + 103932*x^5/5!^2 + ... such that log(B(x)) = 3*x + 3*x^2/(2*2!) + 3*x^3/(3*3!) + 3*x^4/(4*4!) + 3*x^5/(5*5!) + ...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..100
Programs
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Mathematica
a = ConstantArray[0,21]; a[[1]]=1; a[[2]]=3; Do[a[[n+1]] = (n-1)!*Sum[3*Binomial[n, k]*a[[k+1]]/k!,{k,0,n-1}],{n,2,20}]; a (* Vaclav Kotesovec, Feb 23 2014 *) Table[Sum[BellY[n, k, 3/Range[n]], {k, 0, n}] n!, {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *) -
PARI
{a(n)=local(A=1+x, B); for(i=1, n, B=subst(A, x, x/(1-x+x*O(x^n)))/(1-x); A=1+3*intformal((B-A)/x)); n!*polcoeff(A, n)} -
PARI
{a(n)=if(n<0, 0, if(n==0, 1, (n-1)!*sum(k=0, n-1, 3*binomial(n, k)*a(k)/k!)))} -
PARI
{a(n)=n!^2*polcoeff(exp(sum(m=1, n, 3*x^m/(m*m!))+x*O(x^n)), n)} for(n=0,30,print1(a(n),", ")) -
Sage
@CachedFunction def a(n): return 1 if (n==0) else factorial(n-1)*sum( 3*binomial(n, j)*a(j)/factorial(j) for j in (0..n-1) ) [a(n) for n in (0..20)] # G. C. Greubel, Jun 23 2021
Formula
E.g.f.: exp( Sum_{n>=1} 3*x^n/(n*n!) ) = Sum_{n>=0} a(n)*x^n/n!^2.
a(n) = (n-1)!* Sum_{k=0..n-1} 3*binomial(n,k)*a(k)/k! for n>0 with a(0)=1.