A229237 E.g.f. A(x) satisfies: A(x)^A(x) = 1/(1 - x*A(x)^4).

1, 1, 8, 135, 3544, 126980, 5778606, 319234454, 20755549256, 1552791269232, 131408062049040, 12411898074678432, 1294418444771718168, 147733436055601473168, 18315901821846419101416, 2451257290708213030681080, 352217918432527724627871936, 54082428426583359310449351168

Offset: 0

Author

Vaclav Kotesovec, Sep 17 2013

Keywords

• nonn

Comments

Generally, for A(x)^A(x) = 1/(1-x*A(x)^p) is limit n->infinity a(n)^(1/n)/n = exp(p*(1-r)/(r-p))*(p-r+exp(r/(p-r))), where r is the root of the equation exp(r/(p-r)) = (r-p)/r*(r + LambertW(-1,-r*exp(-r))

Generally, if e.g.f. A(x) satisfies A(x)^A(x) = 1/(1-x*A(x)^p), then a(n) ~ s*sqrt((s^s-1)/(p*(s^s-1)*(p*s^s-1)-s)) * n^(n-1) * (s^(p+s)/(s^s-1))^n / exp(n), where s is the root of the equation (1+log(s))*s = (s^s-1)*p. Compared with my previous result, limit n->infinity a(n)^(1/n)/n = s^(p+s)/(s^s-1)/exp(1). - Vaclav Kotesovec, Dec 28 2013

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..326

Crossrefs

Cf. A141209, A216135, A216136, A349561.

Programs

Mathematica

Table[Sum[(4*n-k+1)^(k-1)*(-1)^(n-k)*StirlingS1[n,k],{k,0,n}],{n,0,20}]
p=4; E^(p*(1-r)/(r-p))*(p-r+E^(r/(p-r)))/.FindRoot[E^(r/(p-r))==(r-p)/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 *)

Formula

Limit n->infinity a(n)^(1/n)/n = exp(4*(1-r)/(r-4))*(4-r+exp(r/(4-r))) = 3.635561077783029..., where r = 0.8373821681637... is the root of the equation exp(r/(4-r)) = (r-4)/r*(r + LambertW(-1,-r*exp(-r))

a(n) ~ s*sqrt((s^s-1)/(4*(s^s-1)*(4*s^s-1)-s)) * n^(n-1) * (s^(4+s)/(s^s-1))^n / exp(n), where s = 1.3031377498774256189193761312... is the root of the equation (1+log(s))*s = 4*(s^s-1). - Vaclav Kotesovec, Dec 28 2013