A361777 Expansion of e.g.f. A(x) satisfying A(x) = exp( x * A(x)^x ).
1, 1, 1, 7, 25, 241, 1561, 19951, 188497, 3032065, 37720081, 734331511, 11341504681, 259658249137, 4792613587945, 126280535523871, 2712093428032801, 80881163134899841, 1981706113050012577, 66009436508505875815, 1817280748378601067961, 66887742743997848317681
Offset: 0
Keywords
Examples
E.g.f.: A(x) = 1 + x + x^2/2! + 7*x^3/3! + 25*x^4/4! + 241*x^5/5! + 1561*x^6/6! + 19951*x^7/7! + 188497*x^8/8! + 3032065*x^9/9! + 37720081*x^10/10! + ... where A(x) = exp( x * A(x)^x ) and A(x) = exp(x + 2*x^3/2! + 3^2*x^5/3! + 4^3*x^7/4! + 5^4*x^9/5! + 6^5*x^11/6! + 7^6*x^13/7! + 8^7*x^15/8! + ... + n^(n-1)*x^(2*n-1)/n! + ...). Also, A(x) = 1 + x + (1+2*x)*x^2/2! + (1+3*x)^2*x^3/3! + (1+4*x)^3*x^4/4! + (1+5*x)^4*x^5/5! + (1+6*x)^5*x^6/6! + ... + (1 + n*x)^(n-1)*x^n/n! + ... SPECIFIC VALUES. A( exp(-1/2) ) = exp( exp(1/2) ) = 5.2003257647899611368422256424148878...
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..416
Crossrefs
Cf. A000272.
Programs
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Mathematica
nmax = 20; CoefficientList[Series[E^(-LambertW[-x^2]/x), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Apr 17 2023 *) -
PARI
{a(n) = my(A=1); for(i=1,n, A = exp( x*A^x +x*O(x^n) ) ); n!*polcoeff(A,n)} for(n=0,25,print1(a(n),", ")) -
PARI
{a(n) = if(n==0,1, sum(k=0,n\2, (n-2*k)*(n-k)^(k-1) * n!/((n-2*k)!*k!) ) )} for(n=0,25,print1(a(n),", ")) -
PARI
my(N=30, x='x+O('x^N)); Vec(serlaplace(exp(x*exp(-lambertw(-x^2))))) \\ Seiichi Manyama, Apr 25 2023
Formula
E.g.f. A(x) = Sum_{n>=0} a(n)*x^n/n! may be defined by the following.
(1) A(x) = exp( x * A(x)^x ).
(2) A(x) = ( LambertW(-x^2)/(-x^2) )^(1/x).
(3) A(x) = Sum_{n>=0} (1 + n*x)^(n-1) * x^n/n!.
(4) A(x) = exp( -LambertW(-x^2) / x ).
(5) A(x) = exp( Sum_{n>=1} n^(n-1) * x^(2*n-1) / n! ).
(6) A(x) = 1 + Sum_{n>=1} x^n/n! * Sum_{k>=0} n*(n+k)^(k-1) * x^(2*k) / k!.
(7) a(n) = Sum_{k=0..floor(n/2)} (n-2*k)*(n-k)^(k-1) * n!/((n-2*k)!*k!) for n > 0, with a(0) = 1.
(8) A'(x) = A(x)^(1+x) * (1 + x^2*A(x)^x) / (1 - x^2*A(x)^x).
a(n) ~ sqrt(2) * (exp(2*exp(1/2)) - (-1)^n) * n^(n-1) / exp((n-1)/2 + exp(1/2)). - Vaclav Kotesovec, Apr 17 2023
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