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 A272900 Fibonacci-products fractal sequence. 3
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified January 24 19:12 EST 2021. Contains 340411 sequences. (Running on oeis4.)