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A177405
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Form triangle of weighted Farey fractions; read numerators by rows.
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6
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0, 1, 0, 1, 2, 1, 0, 1, 2, 1, 4, 5, 2, 5, 4, 1, 0, 1, 2, 1, 4, 5, 2, 5, 4, 1, 2, 3, 4, 13, 14, 5, 4, 3, 2, 9, 12, 5, 14, 13, 4, 9, 6, 1, 0, 1, 2, 1, 4, 5, 2, 5, 4, 1, 2, 3, 4, 13, 14, 5, 4, 3, 2, 9, 12, 5, 14, 13, 4, 9, 6, 1, 4, 5, 2, 7, 8, 3, 10, 11
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refs;
listen;
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OFFSET
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0,5
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COMMENTS
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Start with the list of fractions 0/1, 1/1 and repeatedly insert the weighted mediants (2a+c)/(2b+d) and (a+2c)/(b+2d) between every pair of adjacent elements a/b and c/d of the list. The fractions are to be reduced before the insertion step.
James Propp asks: Does every fraction between 0 and 1 with odd denominator appear in the triangle?
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REFERENCES
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James Propp, Posting to the Math Fun Mailing List, Dec 10 2010.
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LINKS
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EXAMPLE
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Triangle begins:
0 1
- -
1 1
0 1 2 1
- - - -
1 3 3 1
0 1 2 1 4 5 2 5 4 1
- - - - - - - - - -
1 5 7 3 9 9 3 7 5 1
0 1 .2 1 .4 .5 2 .5 .4 1 2 3 4 13 14 5 4 3 2 .9 12 5 14 13 4 .9 6 1
- - -- - -- -- - -- -- - - - - -- -- - - - - -- -- - -- -- - -- - -
1 7 11 5 17 19 7 17 13 3 5 7 9 27 27 9 7 5 3 13 17 7 19 17 5 11 7 1
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MATHEMATICA
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Lengthen[L_] :=
Module[{i, M}, M = Table[0, {3 Length[L]}];
M[[1]] = Numerator[L[[1]]]/(2 + Denominator[L[[1]]]);
M[[2]] = 2*Numerator[L[[1]]]/(1 + 2 Denominator[L[[1]]]);
For[i = 1, i < Length[L], i++, M[[3 i]] = L[[i]];
M[[3 i + 1]] = (2 Numerator[L[[i]]] +
Numerator[L[[i + 1]]])/(2 Denominator[L[[i]]] +
Denominator[L[[i + 1]]]);
M[[3 i + 2]] = (Numerator[L[[i]]] +
2 Numerator[L[[i + 1]]])/(Denominator[L[[i]]] +
2 Denominator[L[[i + 1]]])]; M[[3 Length[L]]] = L[[Length[L]]];
Return[M]]
WF[n_] := WF[n] = If[n == 0, {1}, Lengthen[WF[n - 1]]]
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CROSSREFS
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KEYWORD
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nonn,frac,tabf,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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