OFFSET
0,1
COMMENTS
This sequence seems to be cyclic with period 12, but the equivalent sequence for the Fibonacci numbers (A152727) is not.
Lucas numbers modulo 12 are cyclic with period 24 and no 0 in the cycle (unlike Fibonacci numbers): 2, 1, 3, 4, 7, 11, 6, 5, 11, 4, 3, 7, 10, 5, 3, 8, 11, 7, 6, 1, 7, 8, 3, 11. It follows that this sequence is cyclic with period 12: 3, 2, 2, 3, 2, 2, 4, 2, 2, 3, 2, 2. - Jens Kruse Andersen, Jul 15 2014
LINKS
Jens Kruse Andersen, Table of n, a(n) for n = 0..1000
FORMULA
For n >= 12, a(n) = a(n-12). - Jens Kruse Andersen, Jul 15 2014
EXAMPLE
a(6) = 4 because lucas(6) = 18, both 2 and 3 divide 18, but 4 does not.
PROG
(PARI) lucas(n) = if(n==0, 2, 2*fibonacci(n-1)+fibonacci(n));
vector(1000, n, m=lucas(n-1); d=2; while(m%d==0, d++); d)
CROSSREFS
KEYWORD
nonn
AUTHOR
Colin Barker, Jul 12 2014
STATUS
approved