[seqfan] Re: Computing more terms of draft sequence A343745

[Tom Duff]

The Wells Johnson paper is in Math Comp (Jan 1975, pp 113-120). JSTOR
has it. In the introduction he says

    A prime p is called regular if it does not divide the numerator of
any of the Bernoulli numbers B_2, B_4, ... , B_{p-1}.

and later

   If p is an irregular prime and p divides the numerator of the
Bernoulli number B_{2k} for 0<2k<p-1, we shall refer to (p,2k) as an
irregular pair.

Still later we have

   It has been known for a long time that consecutive irregular pairs
(those of the form (p,2k) and (p, 2k+2)) occur for p=491 and 587.

So it looks like you only have to check that prime p divides B_{2k}
for some 2k<p-1 to determine that p is an irregular prime, and then if
p divides both B_{2k} and B_{2k+2} (again for some 2k<p-1, or more
likely 2k<p-3 -- it's not 100% clear) we have consecutive irregular
prime pairs.

It looks like the name of the sequence should be

Consecutive irregular *prime* pairs, i.e., primes p where an integer k
exists such that p divides the numerators of the Bernoulli numbers
B_{2k} and B_{2k+2}, *where 2k+2<p-1*.
(changes marked with *...*)

On Fri, Apr 30, 2021 at 4:42 PM D. S. McNeil <dsm054 at gmail.com> wrote:
>
> I think the 'order in which they appear' definition is going to be more
> practical than the pure "this is the set of primes that satisfy the
> condition" case, because otherwise it might take a good bit of effort to
> prove that different primes aren't in the set.
>
> The numbers generated by a quick scan I did match rgwv's, so I think we're
> interpreting the definition in the same way.  OTOH something must be wrong
> somewhere, because 37 shows up at such a small k that it should've been
> caught.
>
> Is there an additional condition on p which rules out numbers like 37 and
> 103?
>
>
> Doug
>
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