|
|
A127699
|
|
Length of period of the sequence (1^1^1^..., 2^2^2^..., 3^3^3^..., 4^4^4^..., ...) modulo n.
|
|
2
|
|
|
1, 2, 6, 4, 20, 6, 42, 8, 18, 20, 220, 12, 156, 42, 60, 16, 272, 18, 342, 20, 42, 220, 5060, 24, 100, 156, 54, 84, 2436, 60, 1860, 32, 660, 272, 420, 36, 1332, 342, 156, 40, 1640, 42, 1806, 220, 180, 5060, 237820, 48, 294, 100, 816, 156, 8268, 54, 220, 168
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
For any positive integers a and m the sequence a, a^a, a^a^a, a^a^a^a,... becomes eventually constant modulo m. So the remainder of a^a^a^... modulo n is well-defined.
Shapiro and Shapiro treat this problem. - T. D. Noe, Jan 30 2009
|
|
LINKS
|
|
|
FORMULA
|
a(n) = lcm(n, a(lambda(n))), where lambda is Carmichael's reduced totient function. - T. D. Noe, Jan 30 2009
|
|
EXAMPLE
|
a(10)=20 because the last digit of 1^1^1^.. is 1; the sequence 2,2^2,2^2^2,.. ends with 2,4,6,6,...; the sequence 3,3^3,3^3^3,... with 3,7,7,...; 4,4^4,4^4^4,... with 4,6,6,...; and so on. We get as last digits 1,6,7,6,5,6,3,6,9,0, 1,6,3,6,5,6,7,6,9,0 and then the pattern repeats.
|
|
MAPLE
|
a:= proc(n) option remember; `if`(n=1, 1,
ilcm(n, a(numtheory[lambda](n))))
end:
|
|
MATHEMATICA
|
nn=100; a=Table[0, {nn}]; a[[1]]=1; Do[a[[n]]=LCM[n, a[[CarmichaelLambda[n]]]], {n, 2, nn}]; a (* T. D. Noe, Jan 30 2009 *)
|
|
PROG
|
(Python)
from functools import lru_cache
from math import lcm
from sympy import reduced_totient
@lru_cache(maxsize=None)
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
Extension and correction from T. D. Noe, Jan 30 2009
Incorrect formula removed by T. D. Noe, Feb 02 2009
|
|
STATUS
|
approved
|
|
|
|