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Numbers n dividing the Lucas sequence u(n), defined by u(i) = 2*u(i-1) - 3*u(i-2) with initial conditions u(0)=0, u(1)=1.
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%I #13 Nov 20 2013 12:24:09

%S 1,2,4,8,16,32,56,64,112,128,224,256,272,392,448,512,544,728,784,896,

%T 992,1024,1088,1456,1568,1792,1904,1984,2048,2176,2408,2744,2912,3136,

%U 3584,3808,3968,4096,4352,4624,4816,5096,5488,5824,6176,6272,6944,7168,7616

%N Numbers n dividing the Lucas sequence u(n), defined by u(i) = 2*u(i-1) - 3*u(i-2) with initial conditions u(0)=0, u(1)=1.

%C Except for 1 and 2, all other terms are divisible by 4. This sequence contains every nonnegative power of 2.

%H C. Smyth, <a href="http://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,...,5 we have u(n)=0,1,2,1,-4,-11. Clearly n=1,2,4 divide their respective u(n).

%t nn = 10000; s = LinearRecurrence[{2, -3}, {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. A000079 (powers of 2 (subsequence)).

%Y Cf. A088137 (Lucas sequence).

%K nonn

%O 1,2

%A _Thomas M. Bridge_, Nov 08 2013