Current status

[hlaiho]

Originally Posted by Wacky

Does anyone have any "easy" suggestions?

I suggest for example the following numbers:
2,1582L, 2,1586L, 2,1606M, 2,2046L and 11,235-.

Heikki

[R.D. Silverman]

Originally Posted by hlaiho

I suggest for example the following numbers:
2,1582L, 2,1586L, 2,1606M, 2,2046L and 11,235-.

Heikki

What polynomials would you suggest for 2,1586L and 1606M???

11,235- would be very difficult. A quartic is distinctly sub-optimal.

2046L would be easier with GNFS.

None of these numbers is "wanted".

[R.D. Silverman]

Originally Posted by R.D. Silverman

What polynomials would you suggest for 2,1586L and 1606M???

11,235- would be very difficult. A quartic is distinctly sub-optimal.

2046L would be easier with GNFS.

None of these numbers is "wanted".

Please. I assume you suggested these numbers for a reason.
I assume therefore that you have good polynomials for 2,1586L and
2,1606M. I'd like to know what they are.

[BotXXX]

Originally Posted by R.D. Silverman

Why is it that GIMPS can attract 10's of thousands (or more)?

Quite simple, GIMPS is around longer, is more known (also because of usage of Prime95 for stability testing of hardware), gets more attention in public (with the find of new prime number), has statistics, but most of all it has an client that is easy/simple (looking at people with hardly any computer knowledge).

--
But good news that there was a considerable speed up of gathering the relations.

[hlaiho]

Originally Posted by R.D. Silverman

Please. I assume you suggested these numbers for a reason.
I assume therefore that you have good polynomials for 2,1586L and
2,1606M. I'd like to know what they are.

I don't have any [good polynomials] for these numbers. I just suggested these
numbers because 1586 is divisible by 13 and 1606 is divisible by 11.

And I suggested 11,235-, because CWI has done earlier 12,235- with SNFS.

Heikki

[R.D. Silverman]

Originally Posted by hlaiho

I don't have any [good polynomials] for these numbers. I just suggested these
numbers because 1586 is divisible by 13 and 1606 is divisible by 11.

And I suggested 11,235-, because CWI has done earlier 12,235- with SNFS.

Heikki

Before you suggest "easy" numbers to do, please be certain that the
numbers are indeed easy.

12,235- was a LARGE effort. 11,235- will be a LARGE effort.

Please explain why you think that just because 1586 is divisible by 13
that the number will be easy?? Ditto for 1606/11?????
AFAIK, doing NFS with 12th and 10th degree polynomials is currently
impractical. Your claims suggest that you know something that I don't.
Teach me.

[em99010pepe]

Originally Posted by R.D. Silverman

If I had more sievers it would speed up sieving elapsed time

My siever is quite a bit faster than the one used by NFSNET, but I only
have ~15% of their computing power.

My home machine is free, I can always add more machines (AMD 2200+, AMD X2 3800+, a few Opterons). Send me a PM telling me how to help you. I'm tired of running the usual distributed projects.

Carlos

[R.D. Silverman]

Originally Posted by em99010pepe

My home machine is free, I can always add more machines (AMD 2200+, AMD X2 3800+, a few Opterons). Send me a PM telling me how to help you. I'm tired of running the usual distributed projects.

Carlos

There is no way for outsiders to get data to me other than putting it on
a CD and snail-mailing it.

[bdodson]

Originally Posted by R.D. Silverman

Please explain why you think that just because 1586 is divisible by 13
that the number will be easy?? Ditto for 1606/11?????
AFAIK, doing NFS with 12th and 10th degree polynomials is currently
impractical. Your claims suggest that you know something that I don't.
Teach me.

I'm not sure what the person you're asking is thinking about, but
if I recall correctly, Peter was able to get degree 5 from a factor
of 11, and degree 6 from 13. Ah, here's one I did - 2,671-
the root was 2^61 + (2^61)^(-1), polyn
pm5:=x^5-x^4-4*x^3+3*x^2+3*x-1; built from the cyclotomic
polyn you're thinking of. Same for 13.

bd

[frmky]

Originally Posted by R.D. Silverman

11,235- will be a LARGE effort.

Is a 196-digit quartic really THAT bad? Even with a good poly like x^4+x^3+x^2+x+1? I realize that a quartic is not optimal, but with good choice of factor base sizes, I didn't think it'd be too bad considering I've done a 185-digit with a quintic on a single computer with GGNFS.

Originally Posted by R.D. Silverman

I assume therefore that you have good polynomials for 2,1586L and 2,1606M.

Likewise, would a sextic be too difficult for 2,1586L? 2 x^6 - 2 x^3 + 1 with difficulty of 239 digits would rival the effort for M811, but would the sextic make it much harder than the quintic used then?

Greg

[xilman]

Originally Posted by bdodson

I'm not sure what the person you're asking is thinking about, but
if I recall correctly, Peter was able to get degree 5 from a factor
of 11, and degree 6 from 13. Ah, here's one I did - 2,671-
the root was 2^61 + (2^61)^(-1), polyn
pm5:=x^5-x^4-4*x^3+3*x^2+3*x-1; built from the cyclotomic
polyn you're thinking of. Same for 13.

bd

Hi Bruce, please let us know how to get a reciprocal polynomial for those two numbers so that we can then use the degree-halving trick to reduce them to a quintic and a sextic.

I can't yet see how to do it. Perhaps I need to think about it more.


Paul