

A231404


Integers n dividing the Lucas sequence u(n), where u(i) = 2*u(i1)  4*u(i2) with initial conditions u(0)=0, u(1)=1.


0



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
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OFFSET

1,2


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.


LINKS

Table of n, a(n) for n=1..63.
C. Smyth, The terms in Lucas sequences divisible by their indices, Journal of Integer Sequences, Vol.13 (2010), Article 10.2.4.


EXAMPLE

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


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 *)


CROSSREFS

Cf. A088138 (Lucas sequence).
Equal to union of A008585 (multiples of 3) and A000079 (powers of 2).
Sequence in context: A033501 A336504 A331827 * A316860 A097273 A006446
Adjacent sequences: A231401 A231402 A231403 * A231405 A231406 A231407


KEYWORD

nonn


AUTHOR

Thomas M. Bridge, Nov 08 2013


STATUS

approved



