A272907 Lucas-products fractal sequence.

1, 1, 1, 1, 2, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 4, 2, 3, 1, 4, 2, 3, 5, 1, 4, 2, 3, 5, 1, 4, 2, 3, 5, 1, 6, 4, 2, 3, 5, 1, 6, 4, 2, 3, 5, 7, 1, 6, 4, 2, 3, 5, 7, 1, 6, 4, 2, 3, 5, 7, 1, 8, 6, 4, 2, 3, 5, 7, 1, 8, 6, 4, 2, 3, 5, 7, 9, 1, 8, 6, 4, 2, 3, 5, 7, 9

Offset: 1

Author

Clark Kimberling, May 10 2016

Keywords

• nonn
• easy

Comments

Let L = A000032, the Lucas numbers. Let s be the sequence of all products L(i)L(j), for 1 <= i < = j, arranged in increasing order; viz., (1,3,4,7,9,11,12,16,18,21,...) = (L(1)L(1), L(1)L(2), L(1)L(3), L(1)L(4), L(2)L(2), L(1)L(5), L(2)L(3), L(3)L(3), L(1)L(6), L(2)L(4),...). The sequence of first factors is represented by indices A272907 = (1,1,1,1,2,1,2,3,1,2,...), which is a fractal sequence; i.e., the removal of the first occurrence of each term in A272907 leaves A272907, so that the sequence contains itself infinitely many times.

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A272908 (associated interspersion), A000032, A272900 (Fibonacci-products fractal sequence).

Programs

Mathematica

z = 200; f[n_] := LucasL[n]; u1 = Table[f[n], {n, 1, z}];
u2 = Sort[Flatten[Table[f[i]*f[j], {i, 1, z}, {j, i, z}]]];
Table[Select[Range[30], MemberQ[u1, u2[[i]]/f[#]] &][[1]], {i, 1, z}]