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A103354 a(n) = floor(x), where x is the solution to x = 2^(n-x).

%I A103354 #19 Aug 20 2017 17:19:47
%S A103354 1,1,2,2,3,4,4,5,6,7,8,8,9,10,11,12,13,14,15,16,16,17,18,19,20,21,22,
%T A103354 23,24,25,26,27,28,29,30,31,32,32,33,34,35,36,37,38,39,40,41,42,43,44,
%U A103354 45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,64,65,66,67
%N A103354 a(n) = floor(x), where x is the solution to x = 2^(n-x).
%C A103354 Partial sums seem to be A006697(n-1) = A094913(n-1) + 1. - _M. F. Hasler_, Dec 14 2007 [Confirmed by Allouche and Shallit, 2016. - _N. J. A. Sloane_, Mar 24 2017]
%H A103354 Robert Israel, <a href="/A103354/b103354.txt">Table of n, a(n) for n = 1..10000</a>
%H A103354 J.-P. Allouche, J. Shallit, <a href="http://arxiv.org/abs/1605.02361">On the subword complexity of the fixed point of a -> aab, b -> b, and generalizations</a>, arXiv preprint arXiv:1605.02361 [math.CO], 2016.
%F A103354 a(n) is approximately n/(1+log_2(n)/n).
%F A103354 a(n) = floor(LambertW(log(2)*2^n)/log(2)) = floor(n - log_2(n) + log_2(n)/(n log(2)) + O((log(n)/n)^2)) = floor(n - log_2(n) + 1.5*log_2(n)/n) at least for all n < 10^7. - _M. F. Hasler_, Dec 14 2007
%p A103354 A[1]:= 1;
%p A103354 for n from 2 to 100 do
%p A103354   for x from A[n-1]  while x <= 2^(n-x) do od;
%p A103354   A[n]:= x-1;
%p A103354 od:
%p A103354 seq(A[i],i=1..100); # _Robert Israel_, Dec 04 2016
%t A103354 a[n_] := Floor[ FullSimplify[ ProductLog[ 2^n*Log[2]]/Log[2]]]; Table[a[n], {n, 1, 74}] (* _Jean-François Alcover_, Dec 13 2011, after _M. F. Hasler_ *)
%o A103354 (PARI) A103354(n)=floor(solve(X=1,log(n)*2,X-2^(n-X))+1e-9) \\ _M. F. Hasler_, Dec 14 2007
%o A103354 (PARI) A103354(n)=floor(n-log(n)/log(2)*(1-1.5/n)) \\ _M. F. Hasler_, Dec 14 2007
%Y A103354 Cf. A006697 (partial sums), A094913.
%K A103354 nonn
%O A103354 1,3
%A A103354 _N. J. A. Sloane_, Mar 21 2005
%E A103354 Edited by _M. F. Hasler_, Dec 14 2007