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%I #9 Mar 14 2015 16:06:07
%S 1,5,6,10,12,20,24,25,30,50,60,100,120,125,150,250,300,500,600,625,
%T 750,1250,1500,2500,3000,3125,3750,6250,7500,12500,15000,15625,18750,
%U 31250,37500,62500,75000,78125,93750,156250,187500,312500,375000,390625
%N Numbers n such that the period length P(n) of the Fibonacci sequence mod n is a multiple of n.
%C Numbers n such that A001175(n) (mod n) == 0.
%F Numbers of the form 2^a*5^b*6^c, b={0, 1, 2, ...}, c={0, 1} and a={0, 1, 2} but only equal 1 or 2 if b or c > 0. - _Robert G. Wilson v_
%F Equivalently, [1,6,10,12,20,24]*5^m, m>=0.
%e 6 is in the list because the first 24 Fibonacci numbers (A000045) are:
%e 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657;
%e mod(Fibonacci(k=0..inf.), 6) is:
%e 0,1,1,2,3,5,2,1,3,4,1,5,0,5,5,4,3,1,4,5,3,2,5,1 repeated;
%e this has period 24 which is a multiple of 6; therefore 6 is a member.
%t f[n_] := Block[{a = a0 = {1, 0}, k = 0}, While[k++; s = Mod[Plus @@ a, n]; a = RotateLeft[a]; a[[2]] = s; a != a0]; k]; lst = {1}; Do[ If[ Mod[ f[n], n] == 0, AppendTo[lst, n]; Print[n]], {n, 2, 469000}] (* _Robert G. Wilson v_, May 31 2005 *)
%Y Cf. A001175, A000045.
%K nonn
%O 1,2
%A James Higham-Kessler (James_Higham-Kessler(AT)Brown.edu), Apr 27 2005
%E Edited by _Robert G. Wilson v_, who also found a(30)-a(45), May 31 2005
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