A104480 Numbers m such that the period length P(m) of the Fibonacci sequence modulo m is a perfect square.

1, 7, 17, 21, 25, 34, 68, 97, 119, 127, 133, 136, 152, 175, 189, 238, 266, 275, 323, 337, 343, 357, 378, 381, 391, 399, 425, 437, 475, 476, 505, 525, 532, 544, 577, 608, 621, 625, 646, 647, 679, 707, 714, 749, 755, 756, 782, 798, 850, 864, 874, 889, 950, 952

Offset

1,2

Comments

144 appears to be the most common perfect square.

Links

Robert Israel, Table of n, a(n) for n = 1..5173

Example

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.

Maple

pisano:= proc(n) option remember; local f, F, k;
  F:= ifactors(n)[2];
  if nops(F)<>1 then ilcm(seq(procname(k[1]^k[2]), k=F))
  else
    f:= [0, 1];
    for k from 1 do f:=[f[2], f[1]+f[2] mod n];
                  if f=[0, 1] then return k fi
    od;
  fi
end:
select(t -> issqr(pisano(t)), [$1..1000]); # Robert Israel, Jun 13 2025

Mathematica

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 *)

Crossrefs

Cf. A001175.

Sequence in context: A234095 A354168 A287182 * A053746 A327830 A144695

Adjacent sequences: A104477 A104478 A104479 * A104481 A104482 A104483

Keyword

nonn

Author

William C. Brown (wcbrow00(AT)centre.edu), Apr 18 2005

Status

approved