Search: id:A127699 Results 1-1 of 1 results found. %I A127699 %S A127699 1,2,6,4,20,6,42,8,18,20,220,12,156,42,60,16,272,18,342,20,42,220,5060, %T A127699 24,100,156,54,84,2436,60,1860,32,660,272,420,36,1332,342,156,40,1640, %U A127699 42,1806,220,180,5060,237820,48,294,100,816,156,8268,54,220,168 %N A127699 Length of period of the sequence (1^1^1^..., 2^2^2^..., 3^3^3^..., 4^4^4^..., ...) modulo n. %C A127699 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 A127699 Shapiro and Shapiro treat this problem. [From T. D. Noe (noe(AT)sspectra.com), Jan 30 2009] %H A127699 T. D. Noe, Table of n, a(n) for n=1..1000 %H A127699 Daniel B. Shapiro and S. David Shapiro, Iterated Exponents in Number Theory, Integers 7 (2007), #A23. [From T. D. Noe (noe(AT)sspectra.com), Jan 30 2009] %F A127699 a(n) = lcm(n, a(lambda(n))), where lambda is Carmichael's reduced totient function. [From T. D. Noe (noe(AT)sspectra.com), Jan 30 2009] %e A127699 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. %t A127699 nn=100; a=Table[0,{nn}]; a[[1]]=1; Do[a[[n]]=LCM[n,a[[CarmichaelLambda[n]]]], {n,2,nn}]; a [From T. D. Noe (noe(AT)sspectra.com), Jan 30 2009] %Y A127699 Cf. A000010. %Y A127699 Sequence in context: A100695 A100140 A009262 this_sequence A124838 A088659 A052100 %Y A127699 Adjacent sequences: A127696 A127697 A127698 this_sequence A127700 A127701 A127702 %K A127699 easy,nonn %O A127699 1,2 %A A127699 Jan Fricke (jfricke(AT)math.uni-jena.de), Apr 11 2007 %E A127699 Extension and correction from T. D. Noe (noe(AT)sspectra.com), Jan 30 2009 %E A127699 Removed incorrect formula -- T. D. Noe (noe(AT)sspectra.com), Feb 02 2009 Search completed in 0.001 seconds