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Search: id:A106286
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| A106286 |
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Number of orbits of the 4-step recursion mod n. |
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+0
5
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| 1, 4, 6, 28, 3, 24, 10, 220, 91, 12, 130, 240, 343, 40, 168, 1756, 19, 364, 22, 132, 81, 2068, 26, 1968, 253, 1372, 2336, 448, 2557, 672, 16, 14044, 1143, 76, 108, 4612, 1411, 88, 3084, 1860, 11815, 324, 22, 32092, 13213, 104, 50, 15792, 2467, 4012, 168, 17812
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OFFSET
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1,2
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COMMENT
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Consider the 4-step recursion x(k)=x(k-1)+x(k-2)+x(k-3)+x(k-4) mod n. For any of the n^4 initial conditions x(1), x(2), x(3) and x(4) in Zn, the recursion has a finite period. Each of these n^4 vectors belongs to exactly one orbit. In general, there are only a few different orbit lengths (A106289) for each n. For instance, the 220 orbits mod 8 have lengths of 1, 5, 10 and 20.
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REFERENCES
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D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, Vol. 67, 1960, 525-532.
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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. A015134 (orbits of Fibonacci sequences), A106285 (orbits of 3-step sequences), A106287 (orbits of 5-step sequences), A106289 (number of different orbit lengths), A106308 (n producing a simple orbit structure).
Sequence in context: A068321 A012896 A013078 this_sequence A066293 A050881 A122113
Adjacent sequences: A106283 A106284 A106285 this_sequence A106287 A106288 A106289
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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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