A272900 Fibonacci-products fractal sequence.

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

Offset

1,4

Comments

Let F = A000045, the Fibonacci numbers. Let s be the sequence of all products F(i)F(j), for 2 <= i < = j, arranged in increasing order; viz., (1,2,3,4,5,6,8,9,10,13,15,...) = (F(2)F(2), F(2)F(3), F(2)F(4), F(3)F(3), F(2)F(5), ... ), as at A049997. The sequence of first factors is (F(2), F(2), F(2), F(3), F(2),...), represented by indices (2,2,2,3,2,...). Subtracting 1 from each term leaves A272900, which is a fractal sequence; i.e., the removal of the first occurrence of each term in A272900 leaves A272900, so that the sequence contains itself infinitely many times.

Links

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

Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.

Mathematica

z = 200; f[n_] := Fibonacci[n + 1]; 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}]

Crossrefs

Cf. A272904 (the associated interspersion), A000045, A049997, A272907 (Lucas-products fractal sequence).

Sequence in context: A288738 A214651 A196059 * A023116 A084822 A292224

Adjacent sequences: A272897 A272898 A272899 * A272901 A272902 A272903

Keyword

nonn,easy

Author

Clark Kimberling, May 10 2016

Status

approved