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A118106
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Period of the vector sequence d(n)^k mod n for k=1,2,3,..., where d(n) is the vector of divisors of n.
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2
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1, 1, 1, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 3, 4, 1, 1, 6, 1, 4, 6, 10, 1, 2, 1, 12, 1, 6, 1, 4, 1, 1, 10, 8, 12, 6, 1, 18, 3, 4, 1, 6, 1, 10, 12, 11, 1, 4, 1, 20, 16, 12, 1, 18, 5, 6, 18, 28, 1, 4, 1, 5, 6, 1, 4, 10, 1, 8, 22, 12, 1, 6, 1, 36, 20, 18, 30, 12, 1, 4, 1, 20, 1, 6, 16, 14, 28, 10, 1, 12, 12
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,6
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COMMENTS
| This sequence is related to the period of sigma_k(n) mod n. Note that a(n)=1 iff n is a power of a prime.
The record period lengths of p-1 occur at n=2p, where p is a prime with primitive root 2 (A001122). - T. D. Noe, Oct 25 2007
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..1000
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EXAMPLE
| a(35)=12 because d(35)=(1,5,7,35) and (1,5,7,35)^k (mod 35) is the sequence of vectors (1,5,7,0), (1,25,14,0), (1,20,28,0), (1,30,21,0), (1,10,7,0), (1,15,14,0), (1,5,28,0), (1,25,21,0), (1,20,7,0), (1,30,14,0), (1,10,28,0), (1,15,21,0), (1,5,7,0),..., which has a period of 12.
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MATHEMATICA
| Table[d=Divisors[n]; k=0; found=False; While[i=0; While[i<k-1 && !found, i++; found=(dk[i]==dk[k])]; !found, k++; dk[k]=PowerMod[d, k, n]]; k-i, {n, 100}]
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CROSSREFS
| Cf. A118107 (period of the vector sequence d(n)^2^k mod n).
Sequence in context: A157113 A139320 A174204 * A143201 A158298 A112331
Adjacent sequences: A118103 A118104 A118105 * A118107 A118108 A118109
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KEYWORD
| nonn
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AUTHOR
| T. D. Noe (noe(AT)sspectra.com), Apr 13 2006
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