A086858 Let f(n) be the inverse of the function g(x) = x^x. Then a(n) = floor(f(n)).
1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
Offset: 1
Examples
a(32)=3 because the solution to the equation x^x = 32 is x = 3.080448349..., and floor(3.080448349...) = 3.
Programs
-
Mathematica
f[n_] := Floor[ N[ Log[n]/ProductLog[Log[ n]]]]; Join[{1}, Table[ f[n], {n, 2, 105}]] (* Robert G. Wilson v, Oct 21 2005 *) -
PARI
a(n)=exp(lambertw(log(x)))\1 \\ Charles R Greathouse IV, Nov 29 2024
Formula
a(n) = floor(g^-1(n)) where g(x) = x^x.
a(n) ~ log n/log log n. - Charles R Greathouse IV, Nov 29 2024
Extensions
Edited by Jon E. Schoenfield, Sep 09 2017
Comments