

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
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OFFSET

0,4


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
G. Wulczy, Unity with Fibonacci, Problem H247 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

N. J. A. Sloane.


STATUS

approved



