Re: Lucas super-pseudoprime puzzle

Expand Messages
• ... Indeed :-) Here my Chinese speed-up is huge. Please note that I counted *all* the (q,n) pairs with square-free non-Carmichael n, coprime to 6, and more
Message 1 of 46 , Nov 2, 2010
• 0 Attachment
"mikeoakes2" <mikeoakes2@...> wrote:

> if(fac>3
> then the run times collapse drastically

Indeed :-) Here my Chinese speed-up is huge.

Please note that I counted *all* the (q,n) pairs with
square-free non-Carmichael n, coprime to 6, and more
than 3 prime divisors. You stopped when you found
just one q, for a given n, not so?
So you may have missed the dramatic existence of
more than 70,000 (q,n) pairs here:

> n=7056721 [7, 1; 47, 1; 89, 1; 241, 1] is ok

We *already* knew that it was "ok", since q=1 comes from
> "Mike Oakes" Chebyshev NMBRTHRY

What I did not know (but very soon found)
is that for n = 7056721 there are

> precisely 75383 integers q
> such that n > q > 0 and V(p,q,n) = p mod n,
> for every integer p, namely those with
> q = 1 mod 47 and kronecker(q,241) > -1

I still haven't fully digested that very interesting kronecker.
Why does 241 care about the kronecker, while 7 and 89 do not?

David
• ... http://physics.open.ac.uk/~dbroadhu/cert/dbmo116.out gives my 116, in the format [n, factors, number of solutions] With n
Message 46 of 46 , Nov 9, 2010
• 0 Attachment
"mikeoakes2" <mikeoakes2@...> wrote:

> > My revised count up to 2*10^10 is 116.
> My (original) count up to 2*10^10 was 105.
> So it must have missed 11, i.e. a bigger proportion.