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A323608
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The position function the fractalization of which yields A323607.
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1
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1, 1, 1, 2, 2, 3, 3, 5, 4, 6, 5, 8, 6, 9, 7, 12, 8, 12, 9, 15, 10, 15, 11, 19, 12, 18, 13, 22, 14, 21, 15, 27, 16, 24, 17, 29, 18, 27, 19, 34, 20, 30, 21, 36, 22, 33, 23, 42, 24, 36, 25, 43, 26, 39, 27, 49, 28, 42, 29, 50, 30, 45, 31, 58, 32, 48, 33, 57, 34, 51, 35, 64, 36, 54, 37, 64, 38, 57, 39, 73
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OFFSET
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1,4
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COMMENTS
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For a definition of the fractalization process, see comments in A194959. The sequence A323607, triangular array where row n is the list of the numbers from 1 to n sorted in Sharkovsky order, is clearly the result of a fractalization. Let {a(n)} (this sequence) be its position function.
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LINKS
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FORMULA
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Empirical observations: (Start)
For all odd numbers x >= 3,
a(x) = (1/2)*x - 1/2,
a(2x) = (3/4)*(2x) - 3/2,
a(4x) = (7/8)*(4x) - 5/2,
a(8x) = (15/16)*(8x) - 7/2,
etc.
For all c, a(2^c) = A000325(c) = 2^c-c.
Summarized by:
(End)
It appears that for all k > 0,
a(4k + 0) = 3k - 2 + a(k),
a(4k + 1) = 2k,
a(4k + 2) = 3k,
a(4k + 3) = 2k + 1.
(End)
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EXAMPLE
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- row 5 is: 3 5 4 2 1
- row 6 is: 3 5 6 4 2 1
Row 6 is row 5 in which 6 has been inserted in position 3, so a(6) = 3.
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MATHEMATICA
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lt[x_, y_] := Module[
{c, d, xx, yy, u, v},
{c, d} = IntegerExponent[#, 2] & /@ {x, y};
xx = x/2^c;
yy = y/2^d;
u = If[xx == 1, \[Infinity], c];
v = If[yy == 1, \[Infinity], d];
If[u != v, u < v, If[u == \[Infinity], c > d, xx < yy]]]
row[n_] := Sort[Range[n], lt]
a[n_] := First[FirstPosition[row[n], n]]
Table[a[n], {n, 1, 80}]
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CROSSREFS
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Cf. A194959 (introducing fractalization).
Cf. A323607 (fractalization of this sequence).
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KEYWORD
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AUTHOR
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STATUS
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approved
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