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%I #4 Aug 08 2012 19:53:22
%S 1,7,17,21,25,34,68,97,119,127,133,136,152,175,189,238,266,275,323,
%T 337,343,357,378,381,391,399,425,437,475,476,505,525,532,544,577,608,
%U 621,625,646,647,679,707,714,749,755,756,782,798,850,864,874,889,950,952
%N Numbers n such that the period length P(n) of the Fibonacci sequence modulo n is a perfect square.
%C 144 appears to be the most common perfect square.
%e Let P(n) be the period length of the modulo n Fibonacci sequence (also called the Pisano period). Then {P(n)}=1,3,8,6,20,24,16,12,... and a(2)=7 because the second perfect square in {P(n)} occurs when n=7.
%t t = {1}; Do[a = {1, 0}; a0 = a; k = 0; While[k++; s = Mod[Plus @@ a, n]; a = RotateLeft[a]; a[[2]] = s; a != a0]; If[IntegerQ[Sqrt[k]], AppendTo[t, n]], {n, 2, 1000}]; t (* _T. D. Noe_, Aug 08 2012 *)
%Y Cf. A001175.
%K nonn
%O 1,2
%A William C. Brown (wcbrow00(AT)centre.edu), Apr 18 2005
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