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Search: id:A127699
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| A127699 |
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Length of period of the sequence (1^1^1^..., 2^2^2^..., 3^3^3^..., 4^4^4^..., ...) modulo n. |
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+0
1
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| 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; listen)
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OFFSET
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1,2
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COMMENT
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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. [From T. D. Noe (noe(AT)sspectra.com), Jan 30 2009]
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..1000
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]
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FORMULA
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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]
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EXAMPLE
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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.
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MATHEMATICA
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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]
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CROSSREFS
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Cf. A000010.
Sequence in context: A100695 A100140 A009262 this_sequence A124838 A088659 A052100
Adjacent sequences: A127696 A127697 A127698 this_sequence A127700 A127701 A127702
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KEYWORD
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easy,nonn
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AUTHOR
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Jan Fricke (jfricke(AT)math.uni-jena.de), Apr 11 2007
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EXTENSIONS
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Extension and correction from T. D. Noe (noe(AT)sspectra.com), Jan 30 2009
Removed incorrect formula -- T. D. Noe (noe(AT)sspectra.com), Feb 02 2009
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