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A228488
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Period length of trace(sqrt(n)).
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4
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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, 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, 6, 4, 6, 1, 5, 3, 1, 8, 4, 1, 1, 1, 0, 1, 1, 1, 4, 6
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OFFSET
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0,13
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COMMENTS
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It is assumed that trace(sqrt(n)) is purely periodic, as conjectured at A228487 where trace is defined.
If n is a square, then trace(sqrt(n)) is the empty word, denoted by E. Examples:
n ........... trace(sqrt(n))
1 ........... E
2 ........... 000000000...
3 ........... 111111111...
4 ........... E
5 ........... 000000000...
6 ........... 000000000...
7 ........... 111111111...
8 ........... 111111111...
9 ........... E
10 .......... 000000000...
11 .......... 000000000...
12 .......... 000000000...
13 .......... 110(repeated)
19 .......... 0110(repeated)
22 .......... 1001(repeated)
31 .......... 100010(repeated)
46 .......... 11000101(repeated)
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LINKS
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EXAMPLE
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a(13) = 3 because the trace(sqrt(13)) = 110(repeated) has period length 3.
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MATHEMATICA
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$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[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}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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