[QUOTE=Dubslow;296677]We appear to have a 2^5*7 guide/driver/thingy, i3072; I'll start running 3-4M on C133 shortly.[/QUOTE]
1000@3M; starting 1000@11M.

"Thus conscience does make cowards of us all;
And thus the native hue of resolution
Is sicklied o'er with the pale cast of thought,
And enterprises of great pith and moment
With this regard their currents turn awry,
And lose the name of action."

[QUOTE=Batalov;296707]"Thus conscience does make cowards of us all;
And thus the native hue of resolution
Is sicklied o'er with the pale cast of thought,
And enterprises of great pith and moment
With this regard their currents turn awry,
And lose the name of action."[/QUOTE]

...:huh:?
:unsure:

I did ~300@3e6, no factor. Switching to 11e6.

[QUOTE=Dubslow;296677]We appear to have a 2^5*7 guide/driver/thingy, i3072; I'll start running 3-4M on C133 shortly.[/QUOTE]Shoot....and it was going so well with just 2^3, in fact it even lost a couple more digits. Well, as long as we don't pick up a 3, the 2^5 * 7 should be pretty gentle, though it is an "up" guide.....

1000@11M on the c133@i3072. Not knowing how much others have actually ran, I'll throw in another 400@43M. Presumably the number is about ready for NFS in any case.

4k@11M. Edit: can't do NFS on my home machine, would take a week or two.

[QUOTE=Dubslow;296677]We appear to have a 2^5*7 guide/driver/thingy, i3072; I'll start running 3-4M on C133 shortly.[/QUOTE]

Once again, I will think that
following 2[sup]5[/sup]*7 thing is being rather quite easy to escape from

2[SUP]5[/SUP] always gives away the thing as follows
1+2+4+8+16+32 = 63, thus thereby retaining the 7 as such
but 7 gives away 1+7 = 8 = 2[sup]3[/sup]

But if the remaining cofactor factors into product of two primes of the form 1 (mod 4) / single prime of the form 3 (mod 8), then it will mutate, i.e. the power of 2 will increase beyond 5, there is good chance to lose the 7 within the subsequent iterations, if and only if
1. There is being no prime factor of the form 6 (mod 7)
2. Power of 2 is not being 2 (mod 3) at once
3. If the power of prime p is being k, then (p[sup]k+1[/sup]-1)/(p-1) is not being congruent to 0 (mod 7) at all

If an iteration factors as into 2[sup]5[/sup]*7*(prime of the form 1 mod 4), then within the subsequent iteration, the power of 2, will automatically come down to 4 itself (i.e. 2[sup]4[/sup]*7*...)

1+2+4+8+16+32 = 63 = 0 (mod 3), as long as the 2[sup]5[/sup]*7 thing is being on hold, it is not being possible to develop a 3, over on the subsequent iterations, either

This is not a 2[SUP]2[/SUP]*7 driver, whereby 1+2+4 = 7 preserves the 7 always; 1+7 = 8 = 2[sup]3[/sup] > 2[sup]2[/sup] thereby always retain the 2[sup]2[/sup] factor, as such as both.

Edit:-> Developed all the aliquot sequences being starting with the powers of 2 to, through 2[sup]300[/sup] throughout, and then, thereby thus storing out them within into the factoring database itself

1000@11e6, no factor.

2250 curves @ B1=1e7, B2=46842680440 on line 3072. NF

Holy cow, finished my 1K@11M; will move it to higher bounds, but this number is more than ready for NFS.