A228489 Period of trace(n*tau), where tau = (1+sqrt(5))/2 = golden ratio.

1, 1, 1, 1, 1, 4, 1, 1, 2, 4, 1, 1, 1, 6, 3, 1, 2, 1, 1, 10, 1, 8, 6, 1, 9, 4, 10, 8, 1, 18, 1, 6, 6, 1, 13, 1, 4, 1, 3, 10, 4, 4, 12, 6, 17, 1, 1, 1, 18, 21, 8, 10, 6, 8, 1, 8, 12, 4, 16, 16, 3, 6, 1, 14, 9, 14, 20, 1, 6, 36, 18, 1, 4, 13, 26, 1, 12, 20, 20

Offset: 0

Author

Clark Kimberling, Aug 23 2013

Keywords

• nonn

Comments

It is assumed that trace(n*tau) is purely periodic, as conjectured at A228487 where trace is defined.

Examples

			a(6) = 4 because the trace(6*tau) = 1001 (repeated) has period 4.
   n  trace(n*tau)
   -  ------------
   1  000000000...
   2  111111111...
   3  000000000...
   4  000000000...
   5  000000000...
   6  1001 (repeated)
   7  000000000...
   8  111111111...
   9  10 (repeated)
  10  0100 (repeated)
  14  110011 (repeated)
  17  10 (repeated)
  20  0101010010 (repeated)
  30  101000111111001010 (repeated)
  31  000000000...
  35  1100101010011 (repeated)
		

Crossrefs

Cf. A228487, A228488.

Programs

Mathematica

$MaxExtraPrecision = Infinity; period[seq_] := (If[Last[#1] == {} || Length[#1] == Length[seq] - 1, 0, Length[#1]] &)[NestWhileList[Rest, Rest[seq], #1 != Take[seq, Length[#1]] &, 1]]; periodicityReport[seq_] := ({Take[seq, Length[seq] - Length[#1]], period[#1], Take[#1, period[#1]]} &)[Take[seq, -Length[NestWhile[Rest[#1] &, seq, period[#1] == 0 &, 1, Length[seq]]]]]
(*output format {initial segment,period length,period}*)
t[{x_, y_, }] := t[{x, y}]; t[{x, y_}] := Prepend[If[# > y - #, {y - #, 1}, {#, 0}], y] &[Mod[x, y]]; userIn2[{x_, y_}] := Most[NestWhileList[t, {x, y}, (#[[2]] > 0) &]];
z = 160; pr = Table[p = Convergents[n*GoldenRatio, z];  pairs = Table[{Numerator[#], Denominator[#]} &[p[[k]]], {k, 1, z}]; periodicityReport[Most[Last[Map[Map[#[[3]] &, Rest[userIn2[#]]] &, pairs]]]], {n, 200}]
m = Map[#[[2]] &, pr]   (* Peter J. C. Moses, Aug 22 2013 *)