A323608 - OEIS

%I #24 Apr 02 2019 05:54:45

%S 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,

%T 14,21,15,27,16,24,17,29,18,27,19,34,20,30,21,36,22,33,23,42,24,36,25,

%U 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

%N The position function the fractalization of which yields A323607.

%C 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.

%F Empirical observations: (Start)

%F For all odd numbers x >= 3,

%F a(x) = (1/2)*x - 1/2,

%F a(2x) = (3/4)*(2x) - 3/2,

%F a(4x) = (7/8)*(4x) - 5/2,

%F a(8x) = (15/16)*(8x) - 7/2,

%F etc.

%F For all c, a(2^c) = A000325(c) = 2^c-c.

%F Summarized by:

%F a((2^c)*(2k+1)) = A126646(c)*k + A000295(c) + A000007(k) = (2^(c+1)-1)*k + (2^c-1-c) + [k==0].

%F (End)

%F From _Luc Rousseau_, Apr 01 2019: (Start)

%F It appears that for all k > 0,

%F a(4k + 0) = 3k - 2 + a(k),

%F a(4k + 1) = 2k,

%F a(4k + 2) = 3k,

%F a(4k + 3) = 2k + 1.

%F (End)

%e In A323607 in triangular form,

%e - row 5 is:  3  5  4  2  1

%e - row 6 is:  3  5  6  4  2  1

%e Row 6 is row 5 in which 6 has been inserted in position 3, so a(6) = 3.

%t lt[x_, y_] := Module[

%t   {c, d, xx, yy, u, v},

%t   {c, d} = IntegerExponent[#, 2] & /@ {x, y};

%t   xx = x/2^c;

%t   yy = y/2^d;

%t   u = If[xx == 1, \[Infinity], c];

%t   v = If[yy == 1, \[Infinity], d];

%t   If[u != v, u < v, If[u == \[Infinity], c > d, xx < yy]]]

%t row[n_] := Sort[Range[n], lt]

%t a[n_] := First[FirstPosition[row[n], n]]

%t Table[a[n], {n, 1, 80}]

%Y Cf. A194959 (introducing fractalization).

%Y Cf. A323607 (fractalization of this sequence).

%Y Cf. A126646, A000295, A000007.

%Y Cf. A000325.

%K nonn,look

%O 1,4

%A _Luc Rousseau_, Jan 19 2019