[QUOTE=frmky;134644]The factorization of 7,271- is complete. The sieving for this one went faster than we expected, and therefore it was very oversieved. I initially filtered only the relations I had here in California to see the size of the matrix created. Here's that filtering log:



Greg[/QUOTE]


Nice result......

I am about half way through the LA for 2,1101+, half-way through the
sieving for 2,1104+ and will then do 2,1538M. Sam Wagstaff does not show
anyone doing 2,799+. Is anyone doing it or may I reserve it?

[quote=R.D. Silverman;134697]Nice result......

I am about half way through the LA for 2,1101+, half-way through the
sieving for 2,1104+ and will then do 2,1538M. Sam Wagstaff does not show
anyone doing 2,799+. Is anyone doing it or may I reserve it?[/quote]

I'm doing it. Sieving is 95% done. Sorry, I never reserved it with Sam Wagstaff.

[QUOTE=bsquared;134698]I'm doing it. Sieving is 95% done. Sorry, I never reserved it with Sam Wagstaff.[/QUOTE]


Thanks for the info.

[QUOTE=bsquared;134698]I'm doing it. Sieving is 95% done. Sorry, I never reserved it with Sam Wagstaff.[/QUOTE]

I like your Avatar....

The student, of course, is 100% correct. The question did not
say "compute the value of x". The teacher needs to be more careful......

[quote=R.D. Silverman;134791]I like your Avatar....

[/quote]

Thanks! I ran across that and several others like it in a mass e-mail awhile back (attached).

[quote=R.D. Silverman;134791]I like your Avatar....

The student, of course, is 100% correct. The question did not
say "compute the value of x". The teacher needs to be more careful......[/quote]

Agreed. Nonetheless, something tells me that the teacher did not award any creativity points to the student in this case...

- ben.

More that didn't fit in previous post...

I'd like to reserve 2,1598L for SNFS. SNFS seems to be the way to go, especially since I now have experience with SNFS on this size number (namely, 2,799+), while I have none with GNFS on a C171.

I'm still playing around with parameterization, using this poly

[CODE]
n: 344682277495113483576737243789457319844973285487058684789549245491555989770819433641425904625087602042261741581803359854653090524174719942048090727106451863564668263059909
skew: 1
c6: 2
c3: -2
c0: 1
Y1: -1
Y0: 10889035741470030830827987437816582766592
[/CODE]

- ben.

p.s. I've sent email to Sam Wagstaff about both this and 2,799+.
p.p.s Is this the best place to post such a reservation, or should I start a new thread?

[QUOTE=bsquared;134817]I'd like to reserve 2,1598L for SNFS.
[/QUOTE]

Numbers like this always frustrate me. I keep thinking there surely must be a way to use an algebraic factor like, for this one,

65536x^736+65536x^713 + 32768x^690 - 16384x^644 - 16384x^621 - 8192x^598 + 4096x^552 + 4096x^529 + 2048x^506 - 1024x^460 - 1024x^437 - 512x^414 + 256x^368 - 128x^322 - 128x^299 - 64x^276 + 32x^230 + 32x^207 + 16x^184 - 8x^138 - 8x^115 - 4x^92 + 2x^46 + 2x^23 + 1

Degree 32 (or 16) polys are useless, though.

[QUOTE=bsquared;134817]
p.p.s Is this the best place to post such a reservation, or should I start a new thread?[/QUOTE]

They don't need to be reserved here at all, just with Sam.

Greg

[quote=frmky;134833]Numbers like this always frustrate me. I keep thinking there surely must be a way to use an algebraic factor...

Greg[/quote]

Yeah, I know how you feel. I had pari print out the first 90 some splittings, and stared at that same one for a while.

something like
for(n=2,99,print(n,",",1598\n,",",factor(2^(1598%n)*x^n+1)))

[QUOTE=bdodson;132309]The NFSNET resources are mostly Richard's and Greg's; the two 10,257's (minus and plus) at difficulty 257.

<snip>

[/QUOTE]

Did you do ECM pre-testing on 10,257-? While the C241 had 3 factors,
and would have required NFS to finish in any case, should the smaller
factor (a p54) be considered an ECM miss?

[QUOTE=R.D. Silverman;135175]Did you do ECM pre-testing on 10,257-? While the C241 had 3 factors,
and would have required NFS to finish in any case, should the smaller
factor (a p54) be considered an ECM miss?[/QUOTE]

Probaby so. From Bruce,

[QUOTE]I have the
pre-test recorded as 7*t50 (which refers to the effort, not the limits;
almost all of the curves were either B1=110M (p55-optimal) or
B1=260M (p60-optimal)). That's supposed to be somewhere in
the range between t55 and 2*t55; which means that the chances
of finding a known p55 are between 63.2% and 86.4%, midrange
might be 1-exp(-1.5) = 77.68%, so I spent an effort sufficient
to find 3 of 4 known p55's, leaving 1 of 4 to be found by sieving.
There've certainly been lots of p54-p56's found; and among recent
sieving candidates I found a p53 (on-schedule) and a p57 (early).
[/QUOTE]

Greg