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A049999
a(n) = smallest index k such that Fibonacci(k) = d(n), where d = A049998 (sequence of first differences of ordered products of Fibonacci numbers, i.e., of A049997, with no duplicates).
1
1, 1, 1, 1, 1, 1, 3, 1, 1, 4, 3, 1, 5, 4, 1, 1, 6, 5, 1, 3, 7, 6, 1, 1, 4, 8, 7, 3, 1, 5, 9, 8, 4, 1, 1, 6, 10, 9, 5, 1, 3, 7, 11, 10, 6, 1, 1, 4, 8, 12, 11, 7, 3, 1, 5, 9, 13, 12, 8, 4, 1, 1, 6, 10, 14, 13, 9, 5, 1, 3, 7, 11, 15, 14, 10, 6, 1, 1, 4, 8, 12, 16, 15, 11
OFFSET
1,7
COMMENTS
"David W. Wilson conjectured (Dec 14 2005) that" sequence A049998 "consists only of Fibonacci numbers. Proofs were found by Franklin T. Adams-Watters and Don Reble, Dec 14 2005." - Petros Hadjicostas, Nov 08 2019 [This comment was copied from A049998, which includes Don Reble's proof of the conjecture.]
LINKS
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35. (Includes a proof of the conjecture proved in the Comments for sequence A049998.)
FORMULA
A000045(a(n)) = A049998(n) = A049997(n) - A049997(n-1) for n >= 1. - Petros Hadjicostas, Nov 08 2019
EXAMPLE
From Petros Hadjicostas, Nov 08 2019: (Start)
A049998(1) = 1 = Fibonacci(1) = Fibonacci(2), so a(1) = min(1,2) = 1.
A049998(7) = 2 = Fibonacci(3), so a(7) = 3.
A049998(10) = 3 = Fibonacci(4), so a(10) = 4.
A049998(13) = 5 = Fibonacci(5), so a(13) = 5.
A049998(17) = 8 = Fibonacci(6), so a(17) = 6. (End)
CROSSREFS
KEYWORD
nonn
EXTENSIONS
Name edited by and more terms from Petros Hadjicostas, Nov 08 2019
STATUS
approved