A231404 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.

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, 60, 63, 64, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 128, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165

Offset: 1

Author

Thomas M. Bridge, Nov 08 2013

Keywords

• nonn

Comments

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.

Examples

			For n=0,...,4 we have u(n)= 0,1,2,0,-8. Clearly n=1,2,3,4 are in the sequence.
		

Links

• C. Smyth, The terms in Lucas sequences divisible by their indices, Journal of Integer Sequences, Vol.13 (2010), Article 10.2.4.

Crossrefs

Cf. A088138 (Lucas sequence).

Equal to union of A008585 (multiples of 3) and A000079 (powers of 2).

Programs

Mathematica

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 *)