A007887 a(n) = Fibonacci(n) mod 9.
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
Offset: 0
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).
Programs
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Magma
[Fibonacci(n) mod 9: n in [0..100]]; // Vincenzo Librandi, Feb 04 2014
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Mathematica
Mod[Fibonacci[Range[0,80]],9] (* Harvey P. Dale, Mar 23 2012 *)
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PARI
a(n)=fibonacci(n)%9 \\ Charles R Greathouse IV, Oct 07 2015
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