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Search: id:A106290
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| A106290 |
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Number of different orbit lengths of the 5-step recursion mod n. |
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2
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| 1, 3, 4, 4, 2, 9, 2, 6, 7, 6, 2, 11, 2, 6, 8, 8, 2, 9, 3, 8, 8, 6, 4, 12, 3, 6, 10, 8, 3, 18, 2, 10, 8, 6, 4, 11, 2, 6, 8, 12, 2, 18, 4, 8, 14, 9, 4, 16, 3, 9, 8, 8, 2, 12, 4, 12, 10, 6, 3, 22
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OFFSET
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1,2
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COMMENT
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Consider the 5-step recursion x(k)=x(k-1)+x(k-2)+x(k-3)+x(k-4)+x(k-5) mod n. For any of the n^5 initial conditions x(1), x(2), x(3), x(4) and x(5) in Zn, the recursion has a finite period. Each of these n^5 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths for each n. For n=8, there are 6 different lengths: 1, 2, 3, 6, 12, and 24. The maximum possible length of an orbit is A106303(n), the period of the Fibonacci 5-step sequence mod n.
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LINKS
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Eric Weisstein's World of Mathematics, Fibonacci n-Step
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CROSSREFS
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Cf. A106287 (orbits of 5-step sequences), A106309 (primes that yield a simple orbit structure in 5-step recursions).
Sequence in context: A120650 A103121 A081573 this_sequence A073498 A105736 A090283
Adjacent sequences: A106287 A106288 A106289 this_sequence A106291 A106292 A106293
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), May 02 2005
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