A141209 E.g.f. satisfies A(x)^A(x) = 1/(1 - x*A(x)).
1, 1, 2, 9, 64, 620, 7626, 113792, 1997192, 40316544, 920271840, 23438308872, 658947505272, 20270099889624, 677226678369528, 24420959694718680, 945370712175873216, 39103903755819561984, 1721215383181421110848, 80329148928437231089152
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + x + 2*x^2/2! + 9*x^3/3! + 64*x^4/4! + 620*x^5/5! +...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..385
Programs
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Mathematica
Table[Sum[(n-k+1)^(k-1)*Abs[StirlingS1[n,k]], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Sep 17 2013 *) E^((2*r-1)/(1-r))*(1+(1-r)*E^(r/(r-1)))/.FindRoot[E^(r/(1-r))==(r-1)/r*(r+LambertW[-1,-r*E^(-r)]), {r,1/2}, WorkingPrecision->50] (* program for numerical value of the limit n->infinity a(n)^(1/n)/n, Vaclav Kotesovec, Sep 17 2013 *) -
PARI
{a(n)=local(A=1+x);for(i=0,n,A=exp(-log(1-x*(A+O(x^n)))/A));n!*polcoeff(A,n)} -
PARI
{a(n,m=1)=sum(k=0,n,m*(n-k+m)^(k-1)*polcoeff(prod(j=1,n-1,1+j*x),n-k))} \\ Paul D. Hanna, Jul 08 2009 -
PARI
{a(n,m=1)=n!*sum(k=0,n,m*(n-k+m)^(k-1)*polcoeff((-log(1-x+x*O(x^n))/x)^k/k!,n-k))} \\ Paul D. Hanna, Jul 08 2009 -
PARI
{a(n,m=1)=sum(k=0,n,m*(n-k+m)^(k-1)*(-1)^(n-k)*stirling(n,k,1))} \\ Paul D. Hanna, Jul 08 2009
Formula
From Paul D. Hanna, Jul 08 2009: (Start)
(1) a(n) = Sum_{k=0..n} (n-k+1)^(k-1) *(-1)^(n-k) *Stirling1(n,k).
Let A(x)^m = Sum_{n>=0} a(n,m)*x^n/n!, then
(2) a(n,m) = Sum_{k=0..n} m*(n-k+m)^(k-1) *(-1)^(n-k) *Stirling1(n,k) ;
which is equivalent to the following:
(3) a(n,m) = Sum_{k=0..n} m*(n-k+m)^(k-1) * {[x^(n-k)] Product_{j=1..n-1} (1+j*x)};
(4) a(n,m) = n!*Sum_{k=0..n} m*(n-k+m)^(k-1) * {[x^(n-k)] (-log(1-x)/x)^k/k!}.
(End)
Limit_{n->oo} a(n)^(1/n)/n = exp((2*r-1)/(1-r))*(1+(1-r)*exp(r/(r-1))) = 0.97848198198076..., where r = 0.42324001455512542... is the root of the equation exp(r/(1-r)) = (r-1)/r*(r + LambertW(-1,-r*exp(-r))). - Vaclav Kotesovec, Sep 17 2013
a(n) ~ s*sqrt((s^s-1)/((s^s-1)^2-s)) * n^(n-1) * (s^(1+s)/(s^s-1))^n / exp(n), where s = 2.083029805648017585241865819... is the root of the equation (1+log(s))*s = (s^s-1). - Vaclav Kotesovec, Dec 28 2013
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