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<DIV><FONT face=Arial size=2>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:</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>1. The product uniquely determines m and n,
and</FONT></DIV>
<DIV><FONT face=Arial size=2>2. All the products in a given antidiagonal (m+n)
are greater than all the products in the smaller antidiagonal</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>I went on to discover the following:</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>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.</FONT></DIV>
<DIV><FONT face=Arial size=2>MaxAntiDiag(x) is the largest value in that
AntiDiagonal.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>MinAntiDiag(x) - MaxAntiDiag(x-1) = F_(x-5), as
long as x is at least 7</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>For example, <FONT face=Arial
size=2>MinAntiDiag(13) - MaxAntiDiag(12) = F_8, which you can see, as
follows:</FONT></FONT></DIV>
<DIV><FONT face=Arial size=2>The largest product in AntiDiagonal 12 is F_9 * F_3
= 34 * 2 = 68</FONT></DIV>
<DIV><FONT face=Arial size=2>The smallest product in AntiDiagonal 13 is F_11 *
F_2 = 89 * 1 = 89</FONT></DIV>
<DIV><FONT face=Arial size=2>The difference (89-68) is 21, which is
F_8</FONT> </DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>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</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>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.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>--Graeme</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<BLOCKQUOTE
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><B>From:</B>
<A title=franktaw@netscape.net
href="mailto:franktaw@netscape.net">franktaw@netscape.net</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A title=seqfan@ext.jussieu.fr
href="mailto:seqfan@ext.jussieu.fr">seqfan@ext.jussieu.fr</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Wednesday, December 14, 2005 5:39
PM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: A049998</DIV>
<DIV><FONT face=Arial size=2></FONT><FONT face=Arial size=2></FONT><FONT
face=Arial size=2></FONT><BR></DIV>
<DIV style="FONT-SIZE: 10pt; FONT-FAMILY: 'Verdana'">
<DIV>
<DIV>. . .</DIV>
<DIV><FONT face=Arial></FONT> </DIV>
<DIV>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).</DIV>
<DIV><FONT face=Arial></FONT> </DIV>
<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL
60067<BR>847-776-7645</DIV></DIV></DIV></BLOCKQUOTE></BODY></HTML>