%I #27 Sep 14 2017 14:31:20
%S 1,3,2,8,11,7,4,11,28,3,9,12,23,19,9,16,11,7,28,5,12,23,9,48,40,35,19,
%T 4,59,12,19,15,16,39,9,36,6,27,28,19,19,43,11,59,23,15,9,55,148,35,38,
%U 52,35,6,21,31,16,26,57,28,12,21,43,68,51,67,14,19,119,32,7,72,112,99,5,33
%N Fibonacci winding number: the number of 'mod n' operations in one cycle of the Fibonacci sequence modulo n.
%C If pi(n) is the n-th Pisano number (A001175) then a(n) is usually about pi(n)/2 - and in any case a(n) > pi(n)/4.
%H T. D. Noe, <a href="/A088551/b088551.txt">Table of n, a(n) for n=2..10000</a>
%H R. C. Johnson, <a href="http://www.dur.ac.uk/bob.johnson/fibonacci/">Fibonacci Numbers and Resources</a>.
%H M. Merca, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Merca/merca3.html">Inequalities and Identities Involving Sums of Integer Functions</a>, J. Int. Seq. 14 (2011) # 11.9.1.
%F n*a(n) = sum{k=1..A001175(n)} fibonacci(k) mod n. [_Mircea Merca_, Jan 03 2011]
%e a(8)=4 because one cycle of the Fibonacci numbers modulo 8 is 0, 1, 1, 2, 3, 5; 0, 5, 5; 2, 7; 1; - including 4 'mod 8' operations, each marked with a semi-colon.
%t (* pp = Pisano period = A001175 *) pp[1] = 1;
%t pp[n_] := For[k = 1, True, k++, If[Mod[Fibonacci[k], n] == 0 && Mod[Fibonacci[k + 1], n] == 1, Return[k]]];
%t a[n_] := Sum[Mod[Fibonacci[k], n], {k, 1, pp[n]}]/n;
%t Table[a[n], {n, 2, 77}] (* _Jean-François Alcover_, Sep 05 2017 *)
%Y Cf. A001175, A015134, A214300.
%K easy,nice,nonn
%O 2,2
%A R C Johnson (bob.johnson(AT)dur.ac.uk), Nov 19 2003
%E More terms from _T. D. Noe_
%E Edited by _Ray Chandler_, Oct 26 2006