A049998

[franktaw at netscape.net]

Here's a sketch of a proof.
 
If you look at the matrix of products of Fibonacci numbers:
 
 1  2  3  5  8 13 . . .
 2  4  6 10 16 26
 3  6  9 15 24 39
 5 10 15 25 40 65
 8 16 24 40 64 104
13 26 39 65 104 169
. . .
 
each antidiagonal, such as 21 26 24 25 24 26 21
has it's minimum value at the start, with increasing values (with a Fibonacci difference) every second value until you get to the middle, then an increase of one to an adjacent value, and increases again every second value (with a Fibonacci difference) until you get back next to the outside.  (The other side, of course, is just the mirror image.)  It should be simple enough to establish these relationships.  And then the jump from one antidiagonal to the next is F_{n+2} - 2*F_n = F_{n+1} - F_{n} = F_{n-1}, so those differences are also Fibonacci numbers.
 
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
 
 
-----Original Message-----
From: David Wilson <davidwwilson at comcast.net>
To: Sequence Fans <seqfan at ext.jussieu.fr>
Sent: Wed, 14 Dec 2005 13:07:28 -0500
Subject: A049998


Is A049998 composed entirely of Fibonacci numbers? 
 
-------------------------------- 
- David Wilson 
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