A127699 - OEIS

%I #21 Jun 19 2023 09:12:29

%S 1,2,6,4,20,6,42,8,18,20,220,12,156,42,60,16,272,18,342,20,42,220,

%T 5060,24,100,156,54,84,2436,60,1860,32,660,272,420,36,1332,342,156,40,

%U 1640,42,1806,220,180,5060,237820,48,294,100,816,156,8268,54,220,168

%N Length of period of the sequence (1^1^1^..., 2^2^2^..., 3^3^3^..., 4^4^4^..., ...) modulo n.

%C 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.

%C Shapiro and Shapiro treat this problem. - _T. D. Noe_, Jan 30 2009

%H Alois P. Heinz, <a href="/A127699/b127699.txt">Table of n, a(n) for n = 1..20000</a> (first 1000 terms from T. D. Noe)

%H Daniel B. Shapiro and S. David Shapiro, <a href="http://www.emis.de/journals/INTEGERS/papers/h23/h23.Abstract.html">Iterated Exponents in Number Theory,</a> Integers 7 (2007), #A23.

%F a(n) = lcm(n, a(lambda(n))), where lambda is Carmichael's reduced totient function. - _T. D. Noe_, Jan 30 2009

%e 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.

%p a:= proc(n) option remember; `if`(n=1, 1,

%p       ilcm(n, a(numtheory[lambda](n))))

%p     end:

%p seq(a(n), n=1..56);  # _Alois P. Heinz_, Jan 03 2023

%t 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 *)

%o (Python)

%o from functools import lru_cache

%o from math import lcm

%o from sympy import reduced_totient

%o @lru_cache(maxsize=None)

%o def A127699(n): return 1 if n == 1 else lcm(n, A127699(reduced_totient(n))) # _Chai Wah Wu_, Jan 03 2023

%Y Cf. A002322.

%K easy,nonn

%O 1,2

%A _Jan Fricke_, Apr 11 2007

%E Extension and correction from _T. D. Noe_, Jan 30 2009

%E Incorrect formula removed by _T. D. Noe_, Feb 02 2009