Help: A038537 and possible new seq

[David Wilson]

You can even do better than this.  Since a wonderful prime is prime,
you can strike out 3 (mod 6) and 3 (mod 9) as well.  I assume you
did this in your original calculation, since no multiples of 3 show up
in your list of 24 numbers (mod 2520).
  
----- Original Message ----- 
From: Jack Brennen 
To: seqfan at ext.jussieu.fr 
Sent: Monday, June 23, 2003 12:44 PM
Subject: Re: Help: A038537 and possible new seq


I also found the following note to myself, which was an invaluable
aid in searching for these "wonderful" numbers:

The number must end with 1, 3, 7, or 9 in each base from 2 to 10; thus,
the number must be congruent to:

  1 (mod 2)
  1 (mod 3)
  1 or 3 (mod 4)
  1 or 3 (mod 5)
  1 or 3 (mod 6)
  1 or 3 (mod 7)
  1 or 3 or 7 (mod 8)
  1 or 3 or 7 (mod 9)
  1 or 3 or 7 or 9 (mod 10)

Combining all of these relations, we find that there are 24 possible
residues modulo 2520:

       1   43   73  241  451  631  673  703
     871  883 1051 1081 1123 1303 1513 1681
    1711 1753 1891 1963 2131 2143 2311 2383

Note that if a number falls into one of these residue classes, it will
not be divisible by 2 or 5 in any base B.
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