## Re: Square factors of b^p-1 [SG double Wieferich sequence]

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• ... Congrats on establishing that substantial sequence, David. It must be worthy of submission to Neil s OEIS. Your upper limit on q is about 5 times higher
Message 1 of 81 , Oct 4 4:40 AM
>
> Prime q=2*p+1 with primes b<c<p such that q^2|b^p-1 and q^2|c^p-1
>
> 555383, 1767407, 2103107, 7400567, 12836987, 14668163, 15404867,
> 16238303, 19572647, 25003799, 26978663, 27370727, 35182919,
> 36180527, 38553023, 39714083, 52503587, 53061143, 53735699,
> 55072427, 63302159, 70728839, 77199743, 77401679, 86334299,
> 97298759, 97375319, 103830599, 106208783, 106710287, 108711599,
> 112590683, 120441239, 124581719, 126236879, 128538659, 129881603,
> 133833983, 141132143, 141194387, 145553399, 151565087
>
> Comments: (p,q) is a Sophie Germain prime pair; (b,q) and (c,q)
> are Wieferich prime pairs; each of (b,c) is a square modulo q^2.
> The sequence is now complete up to the 42nd term, q=151565087.
> Mike Oakes set a puzzle on a more general case with primes such
> that b<p1<q, c<p2<q, b<c, q^2|b^p1-1 and q^2|c^p2-1. His sequence
> is complete only up q=27370727, containing only two new known
> primes, q=2452757 and q=22796069, with p1=p2=(q-1)/4, found in
> a simple analysis of http://www.cecm.sfu.ca/~mjm/WieferichBarker

Congrats on establishing that substantial sequence, David.
It must be worthy of submission to Neil's OEIS.

Your upper limit on q is about 5 times higher that I have reached for the "more general case" - which my code would take about 5^2*4=100 days on one 3.6GHz core to complete.

Mike
• Hi , all , Not even draft yet , let alone published : http://upforthecount.com/math/nnp.html still needs a lot of work . But , I have updated the top of the
Message 81 of 81 , Nov 29, 2011
Hi , all ,

Not even draft yet , let alone published :
http://upforthecount.com/math/nnp.html
still needs a lot of work .
But , I have updated the top of the main table :
http://upforthecount.com/math/nnp2.txt
with complete factorizations now extending through
np ( 90 ) .

When is np ( n ) prime ?
So far , only for 1 , 2 and 3 .
When n mod 6 = 4 , np ( n ) has at least 3 algebraic
factors , 3 and n^2 + n + 1 ( twice ) .
But what about the residue ( primitive part ? ) ?
When is this prime ?
n t n^n + (n+1)^(n+1)
4 4 3 * 7 * 7 * 23
16 6 3 * 7 * 7 * 13 * 13 * 34041259347101651
Any others ?

Cheers ,

Walter
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