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A007887 a(n) = Fibonacci(n) mod 9. 6
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: A064737 A246558 A098906 * A105472 A030132 A004090

Adjacent sequences:  A007884 A007885 A007886 * A007888 A007889 A007890

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane.

STATUS

approved

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Last modified September 26 00:10 EDT 2017. Contains 292500 sequences.