Bernoulli numbers and arithmetic progressions

[T. D. Noe]

>I think Hans is on the right track.  I'm fairly sure that some
>computational errors obscure the truth.  It appears that the values of n in
>A090495 that yield the same irregular prime quotient _do_ form arithmetic
>progressions.  It appears that the irregularity index (see A073276 ,
>A073277, A060975, A061576, and just-submitted A091887) of an irregular
>prime is important.
>
>When the irregularity index of a prime p is 1 (as it is for p = 37, 59,
>67,...), then there is one progression n = n0 + k*p*(p-1)/2 that gives all
>the values of n.

Revised here:

When the irregularity index of a prime p is 2 (as it is for p = 157, 353,
379,...), then there are two progressions differing only in their starting
values:

   n = n1 + k*p*(p-1)/2   and   n = n2 + k*p*(p-1)/2

that give all the values of n.  For p = 157, n1 = 4789 and n2 = 8557.

Tony