%I #30 Sep 08 2022 08:44:35

%S 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,

%T 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,

%U 6,2,8,1,0,1,1,2,3,5,8,4,3

%N a(n) = Fibonacci(n) mod 9.

%H Vincenzo Librandi, <a href="/A007887/b007887.txt">Table of n, a(n) for n = 0..1000</a>

%H G. Wulczy, <a href="http://www.fq.math.ca/Scanned/13-1/advanced13-1.pdf">Unity with Fibonacci, Problem H-247 and solution</a>, Fib. Quarter. p. 89_90, Vol 15, 1, Feb. 1977. Mentions this sequence.

%H <a href="/index/Rec#order_24">Index entries for linear recurrences with constant coefficients</a>, 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).

%F 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

%F The numbers have period 24 since F_{n+24} = F_n + 9*(5152 F_{n+1} + 3184 F_n). - Olivier Wittenberg, Sep 28 2004

%t Mod[Fibonacci[Range[0,80]],9] (* _Harvey P. Dale_, Mar 23 2012 *)

%o (Magma) [Fibonacci(n) mod 9: n in [0..100]]; // _Vincenzo Librandi_, Feb 04 2014

%o (PARI) a(n)=fibonacci(n)%9 \\ _Charles R Greathouse IV_, Oct 07 2015

%K nonn,easy

%O 0,4

%A _N. J. A. Sloane_.