A049999 - OEIS

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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, 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

(list; graph; refs; listen; history; text; internal format)

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

Table of n, a(n) for n=1..84.

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

Cf. A000045, A049997, A049998, A094563, A226857, A271354.

Sequence in context: A348608 A342805 A363901 * A126015 A247169 A144336