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Period length of trace(sqrt(n)).
4

%I #7 Dec 04 2016 19:46:32

%S 0,1,1,0,1,1,1,1,0,1,1,1,3,1,1,0,1,1,4,1,2,4,1,1,0,1,1,1,4,1,6,1,1,1,

%T 1,0,1,1,1,1,1,1,6,1,4,8,1,1,0,1,1,4,4,4,1,1,2,4,4,1,7,1,1,0,1,1,8,1,

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

%N Period length of trace(sqrt(n)).

%C It is assumed that trace(sqrt(n)) is purely periodic, as conjectured at A228487 where trace is defined.

%C If n is a square, then trace(sqrt(n)) is the empty word, denoted by E. Examples:

%C n ........... trace(sqrt(n))

%C 1 ........... E

%C 2 ........... 000000000...

%C 3 ........... 111111111...

%C 4 ........... E

%C 5 ........... 000000000...

%C 6 ........... 000000000...

%C 7 ........... 111111111...

%C 8 ........... 111111111...

%C 9 ........... E

%C 10 .......... 000000000...

%C 11 .......... 000000000...

%C 12 .......... 000000000...

%C 13 .......... 110(repeated)

%C 19 .......... 0110(repeated)

%C 22 .......... 1001(repeated)

%C 31 .......... 100010(repeated)

%C 46 .......... 11000101(repeated)

%e a(13) = 3 because the trace(sqrt(13)) = 110(repeated) has period length 3.

%t $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]]]]]

%t (*output format: {initial segment, period length, period}*)

%t 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) &]];

%t z = 160; pr = Table[If[IntegerQ[Sqrt[n]], {0, 0}, p = Convergents[Sqrt[n], z]; pairs = Table[{Numerator[#], Denominator[#]} &[p[[k]]], {k, 1, z}]; periodicityReport[ Most[Last[Map[Map[#[[3]] &, Rest[userIn2[#]]] &, pairs]]]]], {n, 120}]

%t m = Map[#[[2]] &, pr] (* _Peter J. C. Moses_, Aug 22 2013 *)

%Y Cf. A228487, A228489.

%K nonn

%O 0,13

%A _Clark Kimberling_, Aug 23 2013