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Search: id:A030132
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| A030132 |
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Digital root of Fibonacci(n). |
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+0
10
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| 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9, 1, 1, 2
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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Every other (a(0),a(1)) pair of nonzero digits enters a cycle of length 24, except for (3,3) which enters a cycle of length 8, and (9,9) which is periodic of length 1. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Dec 29 2005
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REFERENCES
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S. Marivani and others, Problem 10974, Amer. Math. Monthly, 111 (No. 7, 2004), 628.
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LINKS
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Colm Mulcahy, Gibonacci Bracelets.
Marc Renault, The Fibonacci sequence modulo m
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FORMULA
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a(n+1) = sum of digits of (a(n) + a(n-1)).
Periodic with period 24 given by {1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 9, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 9}
a(n+1) = sum of digits of (a(n) + a(n-1)). a(n+1) = A007953(a(n) + a(n-1)). - Jonathan Vos Post (jvospost2(AT)yahoo.com), Dec 29 2005
a(n) + a(n+1) = A010077(n+4); a(A017641(n)) = 9. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jul 04 2007
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CROSSREFS
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Cf. A000045 (Fibonacci numbers), A010888 (digital roots), A004090, A007953.
Cf. A030133.
Sequence in context: A098906 A007887 A105472 this_sequence A130833 A004090 A104205
Adjacent sequences: A030129 A030130 A030131 this_sequence A030133 A030134 A030135
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KEYWORD
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nonn,base,nice
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AUTHOR
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youngelder(AT)webtv.net (Ana)
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EXTENSIONS
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Entry revised by njas, Aug 29 2004
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