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Rectangular array, by antidiagonals: row n gives the positions of n in the Fibonacci-products fractal sequence, A272900.
3

%I #8 May 30 2018 13:26:46

%S 1,2,4,3,6,8,5,9,11,15,7,12,14,19,23,10,16,18,24,28,34,13,20,22,29,33,

%T 40,46,17,25,27,35,39,47,53,61,21,30,32,41,45,54,60,69,77,26,36,38,48,

%U 52,62,68,78,86,96,31,42,44,55,59,70,76,87,95,106,116

%N Rectangular array, by antidiagonals: row n gives the positions of n in the Fibonacci-products fractal sequence, A272900.

%C This array is an interspersion. Every positive integer occurs exactly once, and each row is interspersed by each other row, except for initial terms.

%C Row 1: A033638 (quarter-squares plus 1)

%C Row 2: A002620 (quarter-squares)

%C Column 1: A267682 (conjectured)

%H Clark Kimberling, <a href="http://www.fq.math.ca/Papers1/42-1/quartkimberling01_2004.pdf">Orderings of products of Fibonacci numbers</a>, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

%e Northwest corner:

%e 1 2 3 4 6 9 12 15

%e 5 7 10 13 17 21 26 31

%e 8 11 14 18 2 27 32 38

%e 16 20 25 30 36 42 49 56

%e 23 28 33 39 45 52 59 67

%e 35 41 48 55 63 71 80 89

%e 46 53 60 68 76 85 94 104

%t z = 500; f[n_] := Fibonacci[n + 1]; u1 = Table[f[n], {n, 1, z}];

%t u2 = Sort[Flatten[Table[f[i]*f[j], {i, 1, z}, {j, i, z}]]];

%t uf = Table[Select[Range[80], MemberQ[u1, u2[[i]]/f[#]] &][[1]], {i, 1, z}]

%t r[n_, k_] := Flatten[Position[uf, n]][[k]]

%t TableForm[Table[r[n, k], {n, 1, 12}, {k, 1, 12}]] (* A272904 array *)

%t t = Table[r[n - k + 1, k], {n, 12}, {k, n, 1, -1}] // Flatten (* A272904 sequence *)

%Y Cf. A000045, A272900, A033638, A002620, A267682, A272908 (Lucas-products interspersion).

%K nonn,tabl,easy

%O 1,2

%A _Clark Kimberling_, May 10 2016