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%I #11 Jun 20 2021 02:46:38
%S 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,
%T 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,
%U 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
%N Let f(n) be the inverse of the function g(x) = x^x. Then a(n) = floor(f(n)).
%C a(n) is the value of x that solves the equation x^x = n, truncated to an integer.
%F a(n) = floor(g^-1(n)) where g(x) = x^x.
%e a(32)=3 because the solution to the equation x^x = 32 is x = 3.080448349..., and floor(3.080448349...) = 3.
%t f[n_] := Floor[ N[ Log[n]/ProductLog[Log[ n]]]]; Join[{1}, Table[ f[n], {n, 2, 105}]] (* _Robert G. Wilson v_, Oct 21 2005 *)
%Y Cf. A000312, A095703.
%K easy,nonn
%O 1,4
%A Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Sep 16 2003
%E Edited by _Jon E. Schoenfield_, Sep 09 2017
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