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RichD 2012-09-20 12:52

[QUOTE=RichD;312000]I'll take (3323^53-1)/(3323-1) C184

[SIZE=1]Mod note: [URL="http://www.mersenneforum.org/showthread.php?p=311744#post311744"]post 209[/URL][/SIZE][/QUOTE]

Done.

[CODE]prp77 factor: 46734885888095313337534966119007288522387634262813506482906421065181042932439
prp107 factor: 28188195379470209453078189735266116818099452752866115981501889434026866130147599277102446672981883202348479[/CODE]

jcrombie 2012-09-20 16:30

I'll take the two c186s,

(1487539^31-1)/(1487539-1) C186
(1583653^31-1)/(1583653-1) C186

[size=1]Mod note: [URL="http://www.mersenneforum.org/showthread.php?p=311744#post311744"]post 209[/URL][/size]

Mathew 2012-09-20 21:40

I would like to reserve:

(2917^53-1)/(2917-1) C181

[size=1]Mod note: [URL="http://www.mersenneforum.org/showthread.php?p=311744#post311744"]post 209[/URL][/size]

Dubslow 2012-09-22 19:11

I've decided that the ECM I cited is sufficiently boring as to be put on hold and interject with doing one of these. As such, I'll do [URL="http://www.mersenneforum.org/showthread.php?p=311744#post311744"](2957^53-1)/(2957-1) C181[/URL]. Carlos kindly provided me with the instructions to make a polynomial, but I'll want someone to verify that the other GGNFS params are at least somewhat optimal.
[code]n: 3050814277402253005216532590252942796387057798725128693786269369247133430781425434750331216918239346881571145099269367801565727743008617949625469906333072998219531813874210310000301
[STRIKE]m: 51110866676606486280448872612994249[/STRIKE]
Y0: -17284702968077945985948215290157
Y1: 1
c0: -2957
c6: 1
degree: 6
skew: 0.26395189193
type: snfs
rlim: 8312900
alim: 8312900
lpbr: 28
lpba: 28
mfbr: 55
mfba: 55
rlambda: 2.6
alambda: 2.6[/code]
I get the feeling this is on the border between 13e and 14e; which should it be? With lpb==28, I should be expecting 20-25M rels?

Batalov 2012-09-22 19:29

You have both m and Y1/Y0... anyway, m [STRIKE]will[/STRIKE] may be ignored but better keep it tidy and drop that line.
Also, c0: -[STRIKE]53[/STRIKE] 2957

pinhodecarlos 2012-09-22 19:33

c0 is wrong, it is -2957, therefore skew value is wrong, it is 0.263951892.

Edit: batalov posted at the same time.

Dubslow 2012-09-22 19:48

[QUOTE=Batalov;312452]You have both m and Y1/Y0... anyway, m [STRIKE]will[/STRIKE] may be ignored but better keep it tidy and drop that line.
Also, c0: -[STRIKE]53[/STRIKE] 2957[/QUOTE]

[QUOTE=pinhodecarlos;312453]c0 is wrong, it is -2957, therefore skew value is wrong, it is 0.263951892.

Edit: batalov posted at the same time.[/QUOTE]

Whoops, base!=53 :blush:

What about the siever?

wblipp 2012-09-25 20:14

Today I added a C157 to the list in [URL="http://www.mersenneforum.org/showthread.php?p=311744#post311744"]post 209[/URL]. It would use a quartic. This joins the C171 and C175 not yet claimed. The new number is

[code]
(2403293005054398937076590802836226619421^5-1)/(5*2403293005054398937076590802836226619420)[/code]

Dubslow 2012-09-25 21:58

[QUOTE=wblipp;312760]Today I added a C157 to the list in [URL="http://www.mersenneforum.org/showthread.php?p=311744#post311744"]post 209[/URL]. It would use a quartic. This joins the C171 and C175 not yet claimed. The new number is

[code]
(2403293005054398937076590802836226619421^5-1)/(5*2403293005054398937076590802836226619420)[/code][/QUOTE]

Whoops, scratch that, quartic, not quintic :razz:

Dubslow 2012-09-25 23:23

[QUOTE=Dubslow;312767]Whoops, scratch that, quartic, not quintic :razz:[/QUOTE]

Okay, on further thought, I believe this is the correct polynomial (and is quartic):
[code]c4: 2403293005054398937076590802836226619421
c0: -1
m: 2403293005054398937076590802836226619421 # This is the same as Y0 == -m and Y1=1, right?
skew: 7001670688
[/code]
If so, then I'll do this when my [URL="http://www.mersenneforum.org/showthread.php?p=311744#post311744"]current reservation[/URL] is done (and I'll then do the C17*s if they're still available).

[SIZE="1"](PS In case it's not clear, after some reading/thinking I'm pretty sure I can create polys now for most numbers of the form a^b+-1, and possibly other forms if I thought about those. I'd still like to be sure though :razz:)[/SIZE]

Edit: Is there a way to take advantage of the apparently-larger-than-usual algebraic factor? (Meaning usually it's just a-1, but in this case it looks like b*(a-1).)

wblipp 2012-09-25 23:59

No. The polynomial you want is x^4+x^3+x^2+x+1. The (x-1) factor of x^5-1 is already divided out.


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