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a(n) = b(n)-b(n-1), where b=A049997 are numbers of the form Fibonacci(i)*Fibonacci(j).
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%I #31 Jan 05 2025 19:51:36

%S 1,1,1,1,1,1,2,1,1,3,2,1,5,3,1,1,8,5,1,2,13,8,1,1,3,21,13,2,1,5,34,21,

%T 3,1,1,8,55,34,5,1,2,13,89,55,8,1,1,3,21,144,89,13,2,1,5,34,233,144,

%U 21,3,1,1,8,55,377,233,34,5,1,2,13,89,610,377,55,8,1,1,3,21,144,987,610,89

%N a(n) = b(n)-b(n-1), where b=A049997 are numbers of the form Fibonacci(i)*Fibonacci(j).

%C _David W. Wilson_ conjectured (Dec 14 2005) that this sequence consists only of Fibonacci numbers. Proofs were found by _Franklin T. Adams-Watters_ and _Don Reble_, Dec 14 2005. The following is Reble's proof:

%C Rearrange A049997, as suggested by _Bernardo Boncompagni_:

%C 1

%C 2

%C 3 4

%C 5 6

%C 8 9 10

%C 13 15 16

%C 21 24 25 26

%C 34 39 40 42

%C 55 63 64 65 68

%C 89 102 104 105 110

%C 144 165 168 169 170 178

%C 233 267 272 273 275 288

%C 377 432 440 441 442 445 466

%C Then we know that

%C F(a+1) * F(a-1) - F(a) * F(a) = (-1)^a

%C F(a+1) * F(b-1) - F(a-1) * F(b+1)

%C = + (-1)^b F(a-b), if a>b

%C = - (-1)^a F(b-a), if a<b

%C Use these to show that from F(x) to F(x+1), the representable numbers are

%C F(x) = F(x) * F(2)

%C < F(x-2) * F(4)

%C < F(x-4) * F(6)

%C < ...

%C < F(x-3) * F(5)

%C < F(x-1) * F(3)

%C < F(x+1) * F(1) = F(x+1)

%C (If x is even, the first identity is needed when the parity changes in the middle.)

%C Each Fibonacci-product is in one of those subsequences and the identities show that each difference is a Fibonacci number.

%H 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. (Includes a proof of the conjecture proved in Comments.)

%t t = Take[ Union@Flatten@Table[ Fibonacci[i]Fibonacci[j], {i, 0, 20}, {j, 0, i}], 85]; Drop[t, 1] - Drop[t, -1] (* _Robert G. Wilson v_, Dec 14 2005 *)

%Y A049997 gives numbers of the form F(i)*F(j), when these Fibonacci-products are arranged in order without duplicates.

%Y A049999(n) gives the smallest index k such that Fibonacci(k) equals a(n).

%K nonn

%O 1,7

%A _Clark Kimberling_

%E More terms from _Robert G. Wilson v_, Dec 14 2005

%E Name edited by _Michel Marcus_, Mar 11 2016