OFFSET
0,3
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
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
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, Sep 17 2013
STATUS
approved