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
Alois P. Heinz, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)
Daniel B. Shapiro and S. David Shapiro, Iterated Exponents in Number Theory, Integers 7 (2007), #A23.
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:
seq(a(n), n=1..56); # Alois P. Heinz, Jan 03 2023
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)
def A127699(n): return 1 if n == 1 else lcm(n, A127699(reduced_totient(n))) # Chai Wah Wu, Jan 03 2023
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Jan Fricke, Apr 11 2007
EXTENSIONS
STATUS
approved