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A082991
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a(1) = 1 and for n > 1, a(n) = 2 * length of the cycle reached for the map x -> A062401(x), starting at n [where A062401(n) = phi(sigma(n))], or -1 if no finite cycle is ever reached.
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1
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1, 2, 2, 4, 2, 4, 4, 2, 2, 4, 4, 2, 4, 2, 2, 6, 4, 6, 2, 2, 6, 2, 2, 6, 6, 2, 6, 6, 2, 6, 6, 4, 6, 6, 6, 4, 6, 6, 6, 6, 2, 4, 2, 6, 6, 6, 6, 4, 4, 4, 6, 4, 6, 4, 6, 4, 4, 6, 6, 4, 6, 4, 4, 4, 6, 4, 4, 4, 4, 4, 6, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6, 4, 4, 6, 4, 4, 6, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
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OFFSET
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1,2
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COMMENTS
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From the original definition: Define a sequence u_n as follows: u_n(1) = n, thereafter u_n(2k) = sigma(u_n(2k-1)), u_n(2k+1) = phi(u_n(2k)); then a(n) is the length of the ultimate period of u_n(k) (which is conjectured to become ultimately periodic for any n>=1).
Conjecture: despite results for small terms, all even number are obtained as values. (For example, 12 occurs since a(12102) = 12).
Because for all n > 1, A000010(n) < n and A062401(n) > 1, such sequences u_n cannot end in odd cycle when n > 1. From this follows that for n > 1, a(n) = 2 * length of the cycle reached for the map x->A062401(x), starting at n, or -1 if no finite cycle is ever reached.
See entry A095955 for further notes about the occurrence of cycles.
(End)
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REFERENCES
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J. Berstel et al., Combinatorics on Words: Christoffel Words and Repetitions in Words, Amer. Math. Soc., 2008. See p. 83.
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LINKS
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FORMULA
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EXAMPLE
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If n=6, u(1)=6, u(2)=sigma(6)=12, u(3)=phi(12)=4, u(4)=sigma(4)=7 u(5)=phi(7)=6, hence u(k) becomes periodic with period (6,12,4,7) of length 4 and a(6)=4.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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