Re: Lucas super-pseudoprimes for Q <> 1
- --- In email@example.com, "djbroadhurst" <d.broadhurst@...> wrote:
>Thanks for that, David.
> Excellent: post-hoc speed-ups are always welcome.
> Yet first credit should always go to the first
> discovery. Happily this also goes to you:
> > I have just finished a 1 GHz-yr investigation of this Conjecture
> > In every case but one, n was indeed a Carmichael number,
> > but for n = 507529 = 11*29*37*43, the conditions are satisfied
> a) To discover 1 non-Carmichael: 1 GHz-year (MO)
> b) To discover 6 more: less than 1 GHz-hour (DB)
> c) To recover all 7: less than 1 GHz-minute (MO)
> Which is the more meritorious?
> I incline to think : (a), by MO, at beginning of this process.
For c), I benefited from your Chinese hint: hereby gratefully acknowledged.
I think the best thing about a) was that it was truly exhaustive, in that it assumed no restriction on the number of factors of n; and as we have seen, the bulk of the time needed is for checking out semiprimes (even with your CRT improvement).
We can be certain that 507529 is a for Neil's OEIS entry, when we get round to creating it.
But a is currently not certain, as b) assumed no semiprime solution exists.
The algorithm in c) is almost linear in n_max, which is remarkable, isn't it.
I have run it up to n_max=10^10 so far, and found 89 non-Carmichaels.
The core of the lemma is one line of pari-GP code, but it's unproven. It was found "by inspection", after wrestling for some time with the following deep paper by Richard Pinch:
[I can't remember, did you flag it in an earlier posting?]
The algorithm seems to work, and certainly gives those solutions, and only those, which your code gives, where they overlap.
I wonder if you can co-discover it?
(My turn to set a Puzzle:-)
- --- In firstname.lastname@example.org,
"mikeoakes2" <mikeoakes2@...> wrote:
> > My revised count up to 2*10^10 is 116.http://physics.open.ac.uk/~dbroadhu/cert/dbmo116.out
> My (original) count up to 2*10^10 was 105.
> So it must have missed 11, i.e. a bigger proportion.
gives my 116, in the format [n, factors, number of solutions]
With n < 2*10^10, the record-holder for the number of solutions is
[2214495361, [13, 17, 23, 29, 83, 181], 147407]
which googles quite nicely, linking to