A049998

[Graeme McRae]

This is fascinating.  Following Adams-Watters' lead, I investigated the Fibonacci products of F_m * F_n, where m and n are at least 2, and m >= n, and I verified two things:

1. The product uniquely determines m and n, and
2. All the products in a given antidiagonal (m+n) are greater than all the products in the smaller antidiagonal

I went on to discover the following:

I defined MinAntiDiag(x) as the smallest value of F_m * F_n such that m+n=x, in other words the smallest value of AntiDiagonal x in the Fibonacci Product Table.
MaxAntiDiag(x) is the largest value in that AntiDiagonal.

MinAntiDiag(x) - MaxAntiDiag(x-1) = F_(x-5), as long as x is at least 7

For example, MinAntiDiag(13) - MaxAntiDiag(12) = F_8, which you can see, as follows:
The largest product in AntiDiagonal 12 is F_9 * F_3 = 34 * 2 = 68
The smallest product in AntiDiagonal 13 is F_11 * F_2 = 89 * 1 = 89
The difference (89-68) is 21, which is F_8 

Within a diagonal, d'Ocagne's identity provides a list of fibonacci numbers (with alternating sign) such that each successive pair adds up to a small fibonacci number, giving an ordering of elements within the diagonal that starts near the edge F_(x-2) * F_2, and takes every second element until it "bounces" off the center of the product table, back near the edge F_(x-3) * F_3

It will take a little more work on my part to prove these identities, but with Adams-Watters' insight, I see it's not very difficult at all.

--Graeme

  ----- Original Message ----- 
  From: franktaw at netscape.net 
  To: seqfan at ext.jussieu.fr 
  Sent: Wednesday, December 14, 2005 5:39 PM
  Subject: Re: A049998


  . . .

  Incidently, this also shows that the only duplicates in the products of Fibonaccis are the trivial ones: F_i * F_j = F_j * F_i (using only the distinct positive Fibonaccis to make products).

  Franklin T. Adams-Watters
  16 W. Michigan Ave.
  Palatine, IL 60067
  847-776-7645
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