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A194029
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Natural fractal sequence of the Fibonacci sequence (1,2,3,5,8,...).
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38
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1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28
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OFFSET
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1,4
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COMMENTS
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Suppose that c(1), c(2), c(3),... is a strictly increasing sequence of positive integers with c(1)=1, and that the sequence c(k+1)-c(k) is strictly increasing. The natural fractal sequence f of c is here introduced by the following rule:
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If c(k)<=n<c(k+1), then f(n)=1+n-c(k).
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The natural interspersion of c is here introduced as the array given by T(n,k)=(position of k-th n in f). Note that c=(row 1 of T).
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As an example, let c=A000217=(1,3,6,10,15,...), the triangular numbers, so that
f=(1,2,1,2,3,1,2,3,4,1,2,3,4,5,1,2,3,4,5,6,1,...), and a northwest corner of T is
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1 3 6 10 15
2 4 7 11 16
5 8 12 17 23
9 13 18 24 31
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Since every number in the set N of positive integers occurs exactly once in this (and every) interspersion, a listing of the terms of T by antidiagonals comprises a permutation, p, of N; letting q denote the inverse of p, we thus have for each c a fractal sequence, an interspersion T, and two permutations of N:
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c..........f..........T, as p....q......
A000045....A194029....A194030....A194031
A000290....A071797....A194032....A194033
A000217....A002260....A066182....A066181
A028387....A074294....A194034....A194035
A028872....A071797....A194036....A194037
A034856....A002260....A194038....A194040
It appears that this is also a triangle read by rows in which row n lists the first A000045(n) positive integers, n >= 1 (see example). - Omar E. Pol, May 28 2012
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REFERENCES
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Clark Kimberling, "Fractal sequences and interspersions," Ars Combinatoria 45 (1997) 157-168.
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LINKS
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Table of n, a(n) for n=1..82.
Clark Kimberling, Numeration systems and fractal sequences, Acta Arithmetica 73 (1995) 103-117.
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EXAMPLE
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The sequence (1,2,3,5,8,13,...) is used to place 1's in positions numbered 1,2,3,5,8,13,... Then gaps are filled in with consecutive counting numbers:
1,1,1,2,1,2,3,1,2,3,4,5,1,...
From Omar E. Pol, May 28 2012: (Start)
Written as an irregular triangle the sequence begins:
1;
1;
1,2;
1,2,3;
1,2,3,4,5;
1,2,3,4,5,6,7,8;
1,2,3,4,5,6,7,8,9,10,11,12,13;
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21;
The row lengths are A000045(n).
(End)
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MATHEMATICA
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z = 40;
c[k_] := Fibonacci[k + 1];
c = Table[c[k], {k, 1, z}] (* A000045 *)
f[n_] := If[MemberQ[c, n], 1, 1 + f[n - 1]]
f = Table[f[n], {n, 1, 800}] (* A194029 *)
r[n_] := Flatten[Position[f, n]]
t[n_, k_] := r[n][[k]]
TableForm[Table[t[n, k], {n, 1, 8}, {k, 1, 7}]]
p = Flatten[Table[t[k, n - k + 1], {n, 1, 13}, {k, 1, n}]] (* A194030 *)
q[n_] := Position[p, n]; Flatten[Table[q[n], {n, 1, 80}]] (* A194031 *)
Flatten[Range[Fibonacci[Range[66]]]] (* Birkas Gyorgy, Jun 30 2012 *)
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CROSSREFS
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Cf. A000045, A194030, A194031.
Sequence in context: A194844 A138528 A037125 * A194055 A162192 A329795
Adjacent sequences: A194026 A194027 A194028 * A194030 A194031 A194032
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KEYWORD
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nonn,tabf
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AUTHOR
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Clark Kimberling, Aug 12 2011
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STATUS
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approved
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