[QUOTE=Batalov;233484][FONT=Arial Narrow]> echo "(10^9270-1)/3^4/7" | ecm 5e4
Input number is (10^9270-1)/3^4/7 (9268 digits)
Using B1=50000, B2=15446350, polynomial x^2, sigma=3682263502
Step 1 took 99069ms
********** Factor found in step 1: 9573726981993820200551709701677247363792903115583020462044881149461982541836124423370428786937847788404630054782651343281856841362377745746844353951111544591359795066465415084994921961147188454685295861838434800628490443
Found composite factor of 220 digits: ...
[/QUOTE]

I didn't get a chance to post a reply to this when I read it
this morning; and see that there haven't been further developments
since? Finding a composite factor of this size (and in step 1, no less)
would appear to suggest that the composite is highly composite (so
not one of those "semi-primes"). If the c220 were a product of just
three prime factors p*q*r, then the elliptic curve for this sigma mod
this c220 would give curves mod p, mod q and mod r each with order
that's B1,B2-smooth for the limits used. Even aside from the un-likelyhood
of simultaneously smooth curves for the two largest primes, the largest
prime can't be very large --- less than p60, say (for this B1!). Best
case (not a likely case ...) of three p60's doesn't get above 200 digits;
and I'd expect to see at least four prime factors. That makes the
smallest prime factor less than (10^219)^(1/4) = 10^55. So either
four factors near p56 (and four times more likely to find one), or a
substantially smaller smallest prime factor. But four factors with the
smallest around p50 doesn't work well either; with the elliptic curve
mod the three larger prime factors ... that's three curves, mod p2,
mod p3 and mod p4 all with the same sigma ... all B1,B2-smooth for
this tiny B1. That's starting to suggest at least five factors, maybe
more. Even if there's no particular need to have the factors of this
c220, I'd be interested to hear how this c220 shows up as an ecm
factor. -Bruce

[QUOTE=bdodson;233512]I didn't get a chance to post a reply to this when I read it
this morning; and see that there haven't been further developments
since? Finding a composite factor of this size (and in step 1, no less)
would appear to suggest that the composite is highly composite (so
not one of those "semi-primes"). If the c220 were a product of just
three prime factors p*q*r, then the elliptic curve for this sigma mod
this c220 would give curves mod p, mod q and mod r each with order
that's B1,B2-smooth for the limits used. Even aside from the un-likelyhood
of simultaneously smooth curves for the two largest primes, the largest
prime can't be very large --- less than p60, say (for this B1!). Best
case (not a likely case ...) of three p60's doesn't get above 200 digits;
and I'd expect to see at least four prime factors. That makes the
smallest prime factor less than (10^219)^(1/4) = 10^55. So either
four factors near p56 (and four times more likely to find one), or a
substantially smaller smallest prime factor. But four factors with the
smallest around p50 doesn't work well either; with the elliptic curve
mod the three larger prime factors ... that's three curves, mod p2,
mod p3 and mod p4 all with the same sigma ... all B1,B2-smooth for
this tiny B1. That's starting to suggest at least five factors, maybe
more. Even if there's no particular need to have the factors of this
c220, I'd be interested to hear how this c220 shows up as an ecm
factor. -Bruce[/QUOTE]

Try the following: Let N = (2^420-1) * (3^420-1) * (5^420-1) * .....
(or similar)

run a very small number of Pollard Rho iterations...... You might be
surprised at what pops out.

I would not be surprised that the C220 factor above has 20 or more
factors. Indeed. I expect it.

[QUOTE=Batalov;233484][FONT=Arial Narrow]> echo "(10^9270-1)/3^4/7" | ecm 5e4
Input number is (10^9270-1)/3^4/7 (9268 digits)
Using B1=50000, B2=15446350, polynomial x^2, sigma=3682263502
Step 1 took 99069ms
********** Factor found in step 1: 9573726981993820200551709701677247363792903115583020462044881149461982541836124423370428786937847788404630054782651343281856841362377745746844353951111544591359795066465415084994921961147188454685295861838434800628490443
Found composite factor of 220 digits: 9573726981993820200551709701677247363792903115583020462044881149461982541836124423370428786937847788404630054782651343281856841362377745746844353951111544591359795066465415084994921961147188454685295861838434800628490443
Composite cofactor ((10^9270-1)/3^4/7)/9573726981993820200551709701677247363792903115583020462044881149461982541836124423370428786937847788404630054782651343281856841362377745746844353951111544591359795066465415084994921961147188454685295861838434800628490443 has 9048 digits
[/FONT]
[FONT=Arial Narrow]:wblipp:[/FONT]
[FONT=Arial Narrow]:rolleyes:[/FONT][/QUOTE]
9270=2*3^2*5*103
So 10^9270-1 has many algebraic factors.
9573726981...<220> = 11 · 13 · 19 · 31 · 37 · 41 · 211 · 241 · 271 · 619 · 1031 · 1237 · 2161 · 7211 · 9091 · 29611 · 46351 · 52579 · 181693 · 238681 · 333667 · 380071 · 497491 · 704521 · 1358983 · 1405333 · 2906161 · 3762091 · 7034077 · 43604227 · 44092859 · 102860539 · 569836171 · 984385009 · 2013681931<10> · 5905014721<10> · 8985695684401<13> · 301917502615411<15> · 664958944842271489<18>
the maximum prime factor of this c220 is only p18, which is a factor of 10^927-1.
If you really want to find a new factor of 10^9270-1, change the input to 10^3090-10^1545+1 would be better.
As I know, B1=25e4 or B1=1e6 should be used.
There are many p3x not yet be found for the numbers 10^n-1, n>5000.
and there are many p2x not yet be found for the numbers 10^n-1, n>20000.
Makoto Kamada's site [url]http://homepage2.nifty.com/m_kamada/math/11111.htm[/url]
contains the factorization of 10^n-1 and it is update rather quick.

It was meant to be a joke, of course. What happened was that I had a list of a hundred composites and ran it through "ecm 100 < s1"; all produced factors of 3, 7, ..., 203, and such, but this one immediately spat out a c95 that really stood out in the output. Then I randomly poked at it for a while until I got c145s with B1=100 and then poked until I got the first composite over 200 digits. Just for fun.

Let me just say what that list was. I was looking for N+-1's that are differences of cubes for various (k*10^n+c)/d near-repunit forms that happen to be known PRPs. This in turn was a followup to the following fortuitous but at the first sight bleak factorization:

[COLOR=darkred]n+2 = 10^4297+2 = 2*3*284887*140235709*806209146522749*[U]p4268[/U][/COLOR]

After proving the p4268 with Primo, I got a PFGW proof for a 12891-digit number in 20 seconds (because that number is (n[SUP]3[/SUP]+8)/3+1):

[FONT=Arial Narrow]./pfgw -f0 -lBlarge.txt -hhelp12981.txt -t -q"[COLOR=red][COLOR=darkred](10^12891+11)/3"[/COLOR][/COLOR][/FONT]
[FONT=Arial Narrow]Primality testing (10^12891+11)/3 [N-1, Brillhart-Lehmer-Selfridge][/FONT]
[FONT=Arial Narrow]Running N-1 test using base 5[/FONT]
[FONT=Arial Narrow]Generic modular reduction using generic reduction FFT length 4K on A 42823-bit [/FONT]
[FONT=Arial Narrow]number[/FONT]
[FONT=Arial Narrow]Calling Brillhart-Lehmer-Selfridge with factored part 33.34%[/FONT]
[FONT=Arial Narrow][COLOR=darkred](10^12891+11)/3 is prime[/COLOR]! (20.4607s+0.0008s)[/FONT]
[FONT=Arial Narrow]___________[/FONT]

Getting back to the highly composite (10^9270-1). Of course I should have (and I have now) excluded all 10^n+-1 forms from the list -- they are too well factored. For them, I have to look up e.g. [URL="http://hpcgi2.nifty.com/m_kamada/f/tm.cgi?p=93#N9270"]these pages[/URL] rather than factor. But for 10^n+-2, no one appears to have factored them at any significant rate.

[QUOTE=wreck;233515]...There are many p3x not yet be found for the numbers 10^n-1, n>5000.
and there are many p2x not yet be found for the numbers 10^n-1, n>20000.
Makoto Kamada's site [URL]http://homepage2.nifty.com/m_kamada/math/11111.htm[/URL]
contains the factorization of 10^n-1 and it is update rather quick.[/QUOTE]
Absoultely agreed, the Phin10 pages are great for that.

Actually I've just had a gratuitous* factor there, too:
n=[URL="http://hpcgi2.nifty.com/m_kamada/f/phin10.cgi?p=93#N9209"][COLOR=#810081]9209[/COLOR][/URL]: c9209(1111111111......) = 198052390085540279237423977869613 * c9176
# ECM B1=1000000, sigma=6868161888639481 (a Prime95-assisted p33)
_______

[SIZE=1]P.P.S. (8*10^27811-71)/9 is the PRP that could be proven by the full factorization of (10^9270-1), but it is far, far from that.[/SIZE]

The Brent composite 199^107-1 had one previously known factor - 14767. The remaining C240 has been factored by ECM as P42 * P198.

[code]199^107-1

P42: 992931754118562627932662314175010297485573
P198: 326890508695138894487801194152836252409085904065582941000001593942976603208376299025081602428984723037347397832187099688401483991047730145439635054493541626201825220861997383182071823813006656408011[/code]

The Brent composite 971^79-1 had no previously known factors. The C234 has been factored as P48 * P52 * P66 * P69. This is the fortieth SNFS factorization of an OPN composite by RSALS and the fiftieth composite factored in this collaboration (the other ten were factored during ECM pre-testing).

ECM to t50 by yoyo@home
SNFS sieving by RSALS
Post Processing by Pace aka Zetaflux

[code]971^79-1
P48: 136309342083874417275129175096146908735062759529
P52: 6009530879395480205205381605292602410168586385034487
P66: 252232436346723484873060190204903427401880263864479522437465965587
P69: 487955593714913947955270698730032820301175748237253170472034877300449
[/code]

[QUOTE=wblipp;234394]The Brent composite 971^79-1 had no previously known factors. The C234 has been factored as P48 * P52 * P66 * P69.[/QUOTE]
How many dependencies did that take?

Capt.Obvious (a.k.a. reality in this case) reports that t50 was way low.

[QUOTE=Batalov;234403]Capt.Obvious (a.k.a. reality in this case) reports that t50 was way low.[/QUOTE]

Irrelevant in this case. SNFS would need to split the final two factors
anyway.

However, I am curious as to why time was spent on this number.
What made it important? Have the Brent tables been extended to all
bases less than 1000? What is special about base = 971?

Serge: as you may remember, a while ago, I started feeding the RSALS grid with integers only after said integers had survived t50 (for SNFS tasks, i.e. 2/9 of SNFS difficulty 225) or t45 (i.e. 2/7 of GNFS 16x difficulty) :smile:
As shown by [URL]http://boinc.unsads.com/rsals/crunching.php[/URL], 971^79 - 1 is the first integer in months (several dozens of integers), which could, retrospectively, be qualified an ECM miss.
t52 might have found both the p48 and the p52, leaving the p66 * p69 for GNFS. But if it had found only one, SNFS would have remained the most efficient method to factor the number.

One ECM miss in several dozens doesn't look such a terrible ratio that we're going to demand William to kill the OP quota on yoyo@home in order to fully respect the "2/9" rule of thumb :wink:
Especially, a full t55 for all integers of SNFS difficulty >= 245, is too expensive for OP's quota on yoyo@home. Maybe 2t50 (8000 @ B1=43e6 + 2500-3000 @ B1=11e7) for all tasks of SNFS difficulty above, say, 238, could be manageable, however ?