6=m^p-n^q?

[ZAKIRS]

ok but i'd like also to draw all's attn to: 
 Alf van der Poorten, Remarks on the
<http://www.research.att.com/~njas/sequences/a023057.txt> sequence of
'perfect' powers
is it smth illuminating here? zak
           

-----Original Message-----
From: Jud McCranie [mailto:judmccr at bellsouth.net]
Sent: Thursday, October 10, 2002 12:05 AM
To: ZAKIRS
Cc: seqfan at ext.jussieu.fr
Subject: RE: 6=m^p-n^q?


At 10:53 PM 10/9/2002 +0200, ZAKIRS wrote:


i don't know but have a look at A001597, A023055, A023057. zak



A23057 is for _adjacent_ perfect powers.  There are solutions to the
equation that aren't in A23057, for instance 29 = 15^2-14^2, but these are
not adjacent perfect powers because 6^3 is between 14^2 and 15^2.



+---------------------------------------------------------+
|       Jud McCranie                                      |
|                                                         |
| Programming Achieved with Structure, Clarity, And Logic |
+---------------------------------------------------------+


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