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%I #18 Nov 20 2013 12:25:21
%S 1,2,3,4,6,8,9,12,15,16,18,21,24,27,30,32,33,36,39,42,45,48,51,54,57,
%T 60,63,64,66,69,72,75,78,81,84,87,90,93,96,99,102,105,108,111,114,117,
%U 120,123,126,128,129,132,135,138,141,144,147,150,153,156,159,162,165
%N Integers n dividing the Lucas sequence u(n), where u(i) = 2*u(i-1) - 4*u(i-2) with initial conditions u(0)=0, u(1)=1.
%C The sequence consists of all nonnegative powers of 2, together with all positive multiples of 3. There are infinitely many pairs of consecutive integers in this sequence.
%H C. Smyth, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Smyth/smyth2.html">The terms in Lucas sequences divisible by their indices</a>, Journal of Integer Sequences, Vol.13 (2010), Article 10.2.4.
%e For n=0,...,4 we have u(n)= 0,1,2,0,-8. Clearly n=1,2,3,4 are in the sequence.
%t nn = 500; s = LinearRecurrence[{2, -4}, {1, 2}, nn]; t = {}; Do[If[Mod[s[[n]], n] == 0, AppendTo[t, n]], {n, nn}]; t (* _T. D. Noe_, Nov 08 2013 *)
%Y Cf. A088138 (Lucas sequence).
%Y Equal to union of A008585 (multiples of 3) and A000079 (powers of 2).
%K nonn
%O 1,2
%A _Thomas M. Bridge_, Nov 08 2013