Help: A038537 and possible new seq

[Jack Brennen]

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.