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-   -   Reserved for MF - Sequence 4788 (https://www.mersenneforum.org/showthread.php?t=11615)

Gimarel 2012-04-29 19:37

I had a lucky hit with ecm and got a p48.

Dubslow 2012-04-29 21:38

The C123 has been ECMd through t35; I've now started 2K@12M.

jrk 2012-04-29 21:55

[QUOTE=Dubslow;297903]The C123 has been ECMd through t35; I've now started 2K@12M.[/QUOTE]

p39: 718068333241085723448923908459684536389

Dubslow 2012-04-29 22:38

We appear to have picked up a 2^2*17 guide/thingy; we're still (slowly) dropping.

science_man_88 2012-04-29 22:59

[QUOTE=Dubslow;297914]We appear to have picked up a 2^2*17 guide/thingy; we're still (slowly) dropping.[/QUOTE]

to requote from: [QUOTE="http://www.mersenneforum.org/showpost.php?p=174259&postcount=20"]Define a guide to be [TEX]2^a[/TEX], together with a subset of the prime factors of [TEX]\sigma (2^a)[/TEX]. A driver is defined as a number [TEX]2^a v[/TEX]with a>0, v odd, [TEX]v| \sigma (2^a)[/TEX] and [TEX]2^{a-1} | \sigma (v)[/TEX] . The last requirement is included so that the power of the prime 2 will tend to persist at least as well as it does for the driver 2 itself, for which the condition is trivially satisfied.[/QUOTE] 2^2*17 does not work for this.

axn 2012-04-29 23:09

[QUOTE=Dubslow;297914]We appear to have picked up a 2^2*17 guide/thingy; we're still (slowly) dropping.[/QUOTE]

Appearances can be deceiving. aka Coincidence :smile:

Dubslow 2012-04-30 00:26

[QUOTE=science_man_88;297916]to requote from: 2^2*17 does not work for this.[/QUOTE]

[QUOTE=axn;297918]Appearances can be deceiving. aka Coincidence :smile:[/QUOTE]

Hence the "thingy"; I guess I really meant that the 17 is persisting across lines (Edit: Just like [URL="http://www.mersenneforum.org/showpost.php?p=174250&postcount=18"]schickel[/URL]), and that we are (incidentally) still going down. :razz: (Now that I've seen a concise definition of driver/guide, I won't misuse them again. :smile:)

Edit2:
Okay, a question and reality check, the latter first: [tex]\sigma\left(2^\alpha\right)/2=2^\alpha- 1[/tex], correct? (Derf thanks sm88)
And, with that definition, a driver requires that [tex]2^{\alpha-1}|\sigma(v)[/tex], but, if v is prime, then [tex]\sigma(v)[/tex] is 1, and then that condition doesn't hold, so 2^2*3 isn't a driver? What am I doing wrong?

science_man_88 2012-04-30 01:23

[QUOTE=Dubslow;297927]Hence the "thingy"; I guess I really meant that the 17 is persisting across lines (Edit: Just like [URL="http://www.mersenneforum.org/showpost.php?p=174250&postcount=18"]schickel[/URL]), and that we are (incidentally) still going down. :razz: (Now that I've seen a concise definition of driver/guide, I won't misuse them again. :smile:)

Edit2:
Okay, a question and reality check, the latter first: [tex]\sigma\left(2^\alpha\right)=2^\alpha- 1[/tex], correct?
And, with that definition, a driver requires that [tex]2^{\alpha-1}|\sigma(v)[/tex], but, if v is prime, then [tex]\sigma(v)[/tex] is 1, and then that condition doesn't hold, so 2^2*3 isn't a driver? What am I doing wrong?[/QUOTE]

[CODE](22:05)>sigma(2^3)
%5 = 15
(22:17)>sigma(2^4)
%6 = 31
(22:18)>sigma(2^5)
%7 = 63
(22:18)>sigma(2^2)
%8 = 7
(22:18)>sigma(2^1)
%9 = 3[/CODE]

sigma is a sum of divisors but it can also be used as a sum of proper divisors and I believe you use it both ways, the sum of proper divisors of a prime is 1 the sum of divisors = prime+1. the sigma Pari uses is the sum of divisors, proper divisors don't include the number itself so we have to subtract that.

Dubslow 2012-04-30 01:28

Derf, I managed to screw up the edit. So what I really meant was that [tex]\sigma(2^\alpha)=2^{\alpha+1}-1[/tex]. Then for 2^2*3, alpha=2, and 2^1=2 | sigma(3)=4.

Also, I'm about halfway through bringing the C129 up to t40, so I'll let yafu run it up to t39.69 (a bit low perhaps) and start NFS. Somebody should probably run at least a few curves at a higher bound.

Dubslow 2012-04-30 04:01

[QUOTE=Dubslow;297933]
Also, I'm about halfway through bringing the C129 up to t40, so I'll let yafu run it up to t39.69 (a bit low perhaps) and start NFS. Somebody should probably run at least a few curves at a higher bound.[/QUOTE]

Poly select started. (It will take a total of ~13 hours.)

jrk 2012-04-30 17:12

I did 5000 curves @ B1=4e6, B2=11415205630 on line 3140. NF


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