Gratuitous OPN factors - Page 21

RichD · 2012-09-20, 12:52

Quote:

Originally Posted by RichD

I'll take (3323^53-1)/(3323-1) C184

Mod note: post 209

Done.

Code:

prp77 factor: 46734885888095313337534966119007288522387634262813506482906421065181042932439
prp107 factor: 28188195379470209453078189735266116818099452752866115981501889434026866130147599277102446672981883202348479

jcrombie · 2012-09-20, 16:30

I'll take the two c186s,

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

Mod note: post 209

Mathew · 2012-09-20, 21:40

I would like to reserve:

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

Mod note: post 209

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 (2957^53-1)/(2957-1) C181. 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
m: 51110866676606486280448872612994249
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

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 ~~will~~ may be ignored but better keep it tidy and drop that line.
Also, c0: -53 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:

Originally Posted by Batalov

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

Quote:

Originally Posted by pinhodecarlos

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

Edit: batalov posted at the same time.

Whoops, base!=53

What about the siever?

wblipp · 2012-09-25, 20:14

Today I added a C157 to the list in post 209. It would use a quartic. This joins the C171 and C175 not yet claimed. The new number is

Code:

(2403293005054398937076590802836226619421^5-1)/(5*2403293005054398937076590802836226619420)

Dubslow · 2012-09-25, 21:58

Quote:

Originally Posted by wblipp

Today I added a C157 to the list in post 209. It would use a quartic. This joins the C171 and C175 not yet claimed. The new number is

Code:

(2403293005054398937076590802836226619421^5-1)/(5*2403293005054398937076590802836226619420)

Whoops, scratch that, quartic, not quintic

Dubslow · 2012-09-25, 23:23

Quote:

Originally Posted by Dubslow

Whoops, scratch that, quartic, not quintic

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

If so, then I'll do this when my current reservation is done (and I'll then do the C17*s if they're still available).

(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

)

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.

2012-09-20, 16:30	#222
jcrombie "Jonathan" Jul 2010 In a tangled web... 5·43 Posts	I'll take the two c186s, (1487539^31-1)/(1487539-1) C186 (1583653^31-1)/(1583653-1) C186 Mod note: post 209 Last fiddled with by wblipp on 2012-09-21 at 04:40 Reason: Add back link

2012-09-20, 21:40	#223
Mathew Nov 2009 2×5²×7 Posts	I would like to reserve: (2917^53-1)/(2917-1) C181 Mod note: post 209 Last fiddled with by wblipp on 2012-09-21 at 04:41 Reason: Add back link

2012-09-22, 19:11	#224
Dubslow Basketry That Evening! "Bunslow the Bold" Jun 2011 40<A<43 -89<O<-88 7221₁₀ Posts	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 (2957^53-1)/(2957-1) C181. 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 ~~m: 51110866676606486280448872612994249~~ 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 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? Last fiddled with by wblipp on 2012-09-23 at 18:51 Reason: Add back link

2012-09-22, 19:33	#226
pinhodecarlos "Carlos Pinho" Oct 2011 Milton Keynes, UK 3·17·97 Posts	c0 is wrong, it is -2957, therefore skew value is wrong, it is 0.263951892. Edit: batalov posted at the same time. Last fiddled with by pinhodecarlos on 2012-09-22 at 19:34

2012-09-25, 20:14	#228
wblipp "William" May 2003 New Haven 2×7×13² Posts	Today I added a C157 to the list in post 209. It would use a quartic. This joins the C171 and C175 not yet claimed. The new number is Code: (2403293005054398937076590802836226619421^5-1)/(5*2403293005054398937076590802836226619420)

2012-09-22, 19:29	#225
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9477₁₀ Posts	You have both m and Y1/Y0... anyway, m ~~will~~ may be ignored but better keep it tidy and drop that line. Also, c0: -53 2957

2012-09-25, 23:59	#231
wblipp "William" May 2003 New Haven 2366₁₀ Posts	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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