|
|
A105953
|
|
Numbers n such that the period length P(n) of the Fibonacci sequence mod n is a multiple of n.
|
|
0
|
|
|
1, 5, 6, 10, 12, 20, 24, 25, 30, 50, 60, 100, 120, 125, 150, 250, 300, 500, 600, 625, 750, 1250, 1500, 2500, 3000, 3125, 3750, 6250, 7500, 12500, 15000, 15625, 18750, 31250, 37500, 62500, 75000, 78125, 93750, 156250, 187500, 312500, 375000, 390625
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
Numbers n such that A001175(n) (mod n) == 0.
|
|
LINKS
|
|
|
FORMULA
|
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
Equivalently, [1,6,10,12,20,24]*5^m, m>=0.
|
|
EXAMPLE
|
6 is in the list because the first 24 Fibonacci numbers (A000045) are:
0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657;
mod(Fibonacci(k=0..inf.), 6) is:
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;
this has period 24 which is a multiple of 6; therefore 6 is a member.
|
|
MATHEMATICA
|
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 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
James Higham-Kessler (James_Higham-Kessler(AT)Brown.edu), Apr 27 2005
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|