6=m^p-n^q?

[David Wilson]

You original sequence was

10, 14, 22, 29, 31, 34, 42, 50, 52, 54, 58, 60, 62, 66, 68, 70, 72, 
  76, 78, 82, 84, 86, 88, 90, 92, 94, 96, 98, 102, 108, 110, 111, 
  112, 114, 118, 120, 122, 124, 126, 130, 132, 134, 136, 140, 142, 
  146, 150, 153, 156, 158, 160, 162, 164, 174, 176, 177, 178, 182, 
  186, 188, 190, 192, 193, 194

Besides the exclusion of 6 and inclusion of 10, this sequence contains
several odd number and multiples of 4, all of which can be represented
as differences of squares (someone already mentioned this).  You can
understand how these errors throw doubt on the remaining values in
your sequence.

Also, there has been ample discussion to the effect that it takes some
seriously high-powered math to show that any particular element is in
this sequence.  In other words, the elements in the seqence you describe
are at this point conjectural.  This is not bad thing, there are many
conjectural sequences in the OEIS, for example, there are several
conjectural sequences associated with the Goldbach conjecture.
I myself submitted conjectural sequence  A023057, related to the current
discussion.

I know that Neil is down on adding new keywords to the database at
this point in history, but a "conj" keyword for conjectural sequences might
really be useful.

----- Original Message ----- 
  From: ZAKIRS 
  To: seqfan at ext.jussieu.fr 
  Sent: Tuesday, October 08, 2002 1:08 AM
  Subject: 6=m^p-n^q?


  Dear All, thank you very much for the instructive discn.

  as to 6  - this is my another typo: 6 should be in the list 
  (and 10 removed hence subject).

  OK, and what the number theory says now:
  can 6 be represented as difference of two full powers (not powerful numbers) or not? 
  Only imagine  these lots of fullpowers and why not differ by 6? zak

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