6=m^p-n^q?

[Jud McCranie]

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 |
+---------------------------------------------------------+

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://list.seqfan.eu/pipermail/seqfan/attachments/20021009/09cac983/attachment-0001.htm>