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A272900 Fibonacci-products fractal sequence.

%I A272900 #9 Jan 05 2025 19:51:40
%S A272900 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,
%T A272900 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,
%U A272900 8,6,4,2,1,3,5,7,9,8,6,4,2,1,3,5,7,9
%N A272900 Fibonacci-products fractal sequence.
%C A272900 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.
%H A272900 Clark Kimberling, <a href="/A272900/b272900.txt">Table of n, a(n) for n = 1..1000</a>
%H A272900 Clark Kimberling, <a href="https://web.archive.org/web/2024*/https://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.
%t A272900 z = 200; f[n_] := Fibonacci[n + 1]; u1 = Table[f[n], {n, 1, z}];
%t A272900 u2 = Sort[Flatten[Table[f[i]*f[j], {i, 1, z}, {j, i, z}]]];
%t A272900 Table[Select[Range[30], MemberQ[u1, u2[[i]]/f[#]] &][[1]], {i, 1, z}]
%Y A272900 Cf. A272904 (the associated interspersion), A000045, A049997, A272907 (Lucas-products fractal sequence).
%K A272900 nonn,easy
%O A272900 1,4
%A A272900 _Clark Kimberling_, May 10 2016