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Search: id:A079002
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| A079002 |
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Numbers n such that the Fibonacci residues F(k) mod n form the complete set (0,1,2,....,n-1). |
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+0
2
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| 1, 2, 3, 4, 5, 6, 7, 9, 10, 14, 15, 20, 25, 27, 30, 35, 45, 50, 70, 75, 81, 100, 125, 135, 150, 175, 225, 243, 250, 350, 375, 405, 500, 625, 675, 729, 750, 875, 1125, 1215, 1250, 1750, 1875, 2025, 2187, 2500, 3125, 3375, 3645, 3750, 4375, 5625, 6075, 6250, 6561
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OFFSET
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1,2
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REFERENCES
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R. L. Graham, D. E. Knuth and O. Patashnick, "Concrete Mathematics", second edition, Addison Wesley, ex. 6.85, p. 318, p. 562
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FORMULA
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Consists of the integers of the form: 5^k, 2*5^k, 4*5^k, 3^j*5^k, 6*5^k, 7*5^k and 14*5^k [see Concrete Mathematics]
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EXAMPLE
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Fibonacci numbers (A000045) are 0,1,1,2,3,5,8,.. and mod 5 these are 0,1,1,2,3,0,3,3,4,... i.e. all possible remainders mod 5 occur in the Fib series mod 5, so 5 is in the series. This is not true for n=8 so 8 is not in the series.
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CROSSREFS
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Cf. A066853, A001175.
Sequence in context: A087950 A060527 A152493 this_sequence A119984 A059879 A049537
Adjacent sequences: A078999 A079000 A079001 this_sequence A079003 A079004 A079005
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KEYWORD
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nonn
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AUTHOR
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Benoit Cloitre (benoit7848c(AT)orange.fr), Feb 01 2003
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EXTENSIONS
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Corrected by Ron Knott (ron(AT)ronknott.com), Jan 05 2005
Entry revised by N. J. A. Sloane (njas(AT)research.att.com), Nov 28 2006, following a suggestion from Martin Fuller (martin_n_fuller(AT)btinternet.com)
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