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 A007887 a(n) = Fibonacci(n) mod 9. 7
 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 0, 1, 1, 2, 3, 5, 8, 4, 3, 7, 1, 8, 0, 8, 8, 7, 6, 4, 1, 5, 6, 2, 8, 1, 0, 1, 1, 2, 3, 5, 8, 4, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 G. Wulczy, Unity with Fibonacci, Problem H-247 and solution, Fib. Quarter. p. 89_90, Vol 15, 1, Feb. 1977. Mentions this sequence. Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1). FORMULA Period 24 = A001175(9). Proof: F_{n+24} = F_n + 9*(5152 F_{n+1} + 3184 F_n). - Olivier Wittenberg, following a conjecture by Ralf Stephan, Sep 28 2004 The numbers have period 24 since F_{n+24} = F_n + 9*(5152 F_{n+1} + 3184 F_n). - Olivier Wittenberg, Sep 28 2004 MATHEMATICA Mod[Fibonacci[Range[0, 80]], 9] (* Harvey P. Dale, Mar 23 2012 *) PROG (MAGMA) [Fibonacci(n) mod 9: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014 (PARI) a(n)=fibonacci(n)%9 \\ Charles R Greathouse IV, Oct 07 2015 CROSSREFS Sequence in context: A246558 A307638 A098906 * A105472 A030132 A004090 Adjacent sequences:  A007884 A007885 A007886 * A007888 A007889 A007890 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified May 9 00:09 EDT 2021. Contains 343685 sequences. (Running on oeis4.)