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%I #14 Mar 15 2020 02:43:47
%S 0,2,1,4,4,2,6,8,7,4,3,8,12,6,7,16,16,14,18,4,19,8,22,8,33,12,7,40,11,
%T 26,23,32,8,16,6,32,5,18,37,24,40,38,42,8,7,22,10,32,61,84,38,12,35,
%U 32,46,40,32,28,24,44
%N a(n) is the smallest nonnegative k such that n divides 2^k + k.
%C a(2^m) = 2^m for m > 0. If p is an odd prime then by Fermat, a(p) <= p-1. 25 is the smallest n with a(n) > n.
%H Harvey P. Dale, <a href="/A135366/b135366.txt">Table of n, a(n) for n = 1..1000</a>
%H International Mathematical Olympiad, <a href="http://www.imo-official.org/problems/IMO2006SL.pdf">Problem N7</a>, IMO-2006, p. 63.
%e a(9)=7 since 2^7 + 7 = 9*15 and 2^k + k is not divisible by 9 for 0 <= k < 7.
%t sk[n_]:=Module[{k=0},While[!Divisible[2^k+k,n],k++];k]; Array[sk,60] (* _Harvey P. Dale_, Jun 01 2013 *)
%o (PARI) a(n) = for(m=0, oo, if(Mod(2, n)^m==-m, return(m))); \\ _Jinyuan Wang_, Mar 15 2020
%Y Cf. A135359, A247248.
%K nonn
%O 1,2
%A _John L. Drost_, Feb 16 2008
%E Corrected by _Harvey P. Dale_, Jun 01 2013
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