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Andi47 2009-12-23 19:24

Question about 8,7,261+
 
I am currently factoring 8,7,261+ using this poly (SNFS-157 according to the reservation page):

[code]n: 12187326859715888775759007056979393308328466344262944112888775386220897145330520610542827371711778627006104470708598479594800395415247787536887
c4: 64
c2: -56
c0: 49
Y0: 680564733841876926926749214863536422912
Y1: 2183814375991796599109312252753832343
rlim: 3200000
alim: 3200000
skew: 1.41
lpbr: 27
lpba: 27
mfbr: 48
mfba: 48
rlambda: 2.3
alambda: 2.3
[/code]

Yield seems to be rather low - ~1 rel/Q:

[code]
gnfs-lasieve4I13e -a 8_7_261+.poly -o 8_7_261+_3M.3M2.out -f 3000000 -c 200000
Warning: lowering FB_bound to 2999999.
total yield: 213213, q=3200003 (0.16669 sec/rel)[/code]

Did I do something wrong with the polynomial? Is there a better poly available for 8,7,261+?

jyb 2009-12-23 19:31

12-5,205 has gone missing. It was reserved on the reservations page for a long time and isn't there anymore, so I assume it was completed.

frmky 2009-12-23 19:41

After polynomial division, you end up with a quadratic, so this can be made into a quartic or a sextic. Though neither is great, the sextic looks a bit more appropriate than the quartic at this size. Also, since 174 is divisible by 6 the algebraic poly is nicer.

x^6 - x^3 +1
7^29 x - 8^29

Just make sure you sieve on the algebraic side if you use the sextic and on the rational side if you use the quartic.

Andi47 2009-12-23 19:46

[QUOTE=frmky;199724]
Just make sure you sieve on the algebraic side if you use the sextic and on the [B]rational side[/B] if you use the quartic.[/QUOTE]

:doh!: that's what I missed!

Thanks!

R.D. Silverman 2009-12-24 22:55

[QUOTE=frmky;199724]After polynomial division, you end up with a quadratic, so this can be made into a quartic or a sextic. Though neither is great, the sextic looks a bit more appropriate than the quartic at this size. Also, since 174 is divisible by 6 the algebraic poly is nicer.

x^6 - x^3 +1
7^29 x - 8^29

Just make sure you sieve on the algebraic side if you use the sextic and on the rational side if you use the quartic.[/QUOTE]

This last advice is wrong. Sieve the rational side for both.

Consider, e.g. 2^1149-1. M = 2^129. A 'typical' lattice point
is ~(1M, 1M). The rational side norms are about 10^45. (2^129 * 10^6)
while the algebraic side norms are ~ (1M)^6 i.e. about 10^36 to 10^37.

The rational side norms are still larger, even with the sextic.

frmky 2009-12-24 23:20

[QUOTE=R.D. Silverman;199842]
The rational side norms are still larger, even with the sextic.[/QUOTE]
But for this particular number, 10^6 8^29 ~ 10^32, so the algebraic norm is larger. Hence my advice to sieve on the algebraic side. :smile:

xilman 2009-12-29 17:38

[QUOTE=xilman;199716]With luck, another update should occur before the end of the year.[/QUOTE]The update has just happened.

Paul

Andi47 2010-01-01 19:59

[QUOTE=frmky;199724]After polynomial division, you end up with a quadratic, so this can be made into a quartic or a sextic. Though neither is great, the sextic looks a bit more appropriate than the quartic at this size. Also, since 174 is divisible by 6 the algebraic poly is nicer.

x^6 - x^3 +1
7^29 x - 8^29

Just make sure you sieve on the algebraic side if you use the sextic and on the rational side if you use the quartic.[/QUOTE]

Did I get it right - the above is true when the exponent is divisible by 9 (N=a[SUP]9k[/SUP]+b[SUP]9k[/SUP])?

So you did (x[SUP]9[/SUP]+1)/(x[SUP]3[/SUP]+1) = x^6 - x^3 +1 with a rational poly a[SUP]k[/SUP]*x - b[SUP]k[/SUP] ?

Batalov 2010-01-01 20:35

[quote=Andi47;200552]Did I get it right - the above is true when the exponent is divisible by 9 (N=a[sup]9k[/sup]+b[sup]9k[/sup])?

So you did (x[sup]9[/sup]+1)/(x[sup]3[/sup]+1) = x^6 - x^3 +1 with a rational poly a[sup]k[/sup]*x - b[sup]k[/sup] ?[/quote]
Right except for the divisible by 9 part.
Exponents divisible by 3 work with small adjustments (they end up with the f(x) = m^2 x^6 - m n x^3 + n^2, where m and n are small powers of a and b).

EDIT: or if you meant: this particular form (m=n=1) happens for exponents divisible by 9 - that's definitely true. I meant that any of the sum/diff of cubes can be done with a sextic or a quartic whatever if faster.

Andi47 2010-01-10 15:25

7,3,279+, SNFS
 
I have factored 7,3,279+ by SNFS:

[CODE]n: 427660418727285466516569035385971081999943171976638480240422397222920746344190352291320635862513456252229501044835885164445260820304007
c6: 1
c3: -1
c0: 1
Y1: 157775382034845806615042743
Y0: -617673396283947
skew: 1.38
rlim: 3200000
alim: 3200000
lpbr: 27
lpba: 27
mfbr: 48
mfba: 48
rlambda: 2.3
alambda: 2.3
type: snfs
[/CODE]

[CODE]Sun Jan 10 15:36:16 2010 prp54 factor: 317807417090272466450602187275715723563234058722292659
Sun Jan 10 15:36:16 2010 prp82 factor: 1345659024080641602556743461663576664786123431161932751286151089081519314605086173
[/CODE]

xilman 2010-01-19 14:32

Update posted
 
Another update has been posted. There are 32 new factorizations and 801 composites left in the tables.

The ECMNet server has been updated.


Paul


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