Reply With Quote15799176637249321 | 4847*2^6001023+1
15799176637249320 = 2 ^ 3 x 3 ^ 2 x 5 x 7 x 47 x 1163 x 3137 x 36563
3593405260603031 | 10223*2^5774585+1
3593405260603030 = 2*5*31*3187*16573*219463
15799176637249321 | 4847*2^6001023+1
15799176637249320 = 2 ^ 3 x 3 ^ 2 x 5 x 7 x 47 x 1163 x 3137 x 36563
5720000 -> 5730000 had 5 factors:
13860049223357177 | 55459*2^5720278+1
13860049223357176 = 2^3*7*823*1361*1657*133351
52252110283617819719 | 10223*2^5725757+1
see previous post
2408494707706313 | 67607*2^5726411+1
2408494707706312=2^3*353*1597*9041*59069
1990620885055391 | 21181*2^5726972+1
1990620885055390=2*5*7*41*2351*4483*65809
363916348856369 | 24737*2^5727391+1
363916348856368=2^4*7*11*19^2*14891*54949
407337342926141 | 21181*2^5777180+1
407337342926140 = 2 ^ 2 x 5 x 11 x 31 x 41 ^ 2 x 59 ^ 3 x 173
Does anyone have smoother P-1?
Well you still have to have B1 >= 1681 (41^2) and B2 >= 205379 (59^3) to find that, so I don't really know if it counts as smooth.Originally posted by Nuri
407337342926141 | 21181*2^5777180+1
407337342926140 = 2 ^ 2 x 5 x 11 x 31 x 41 ^ 2 x 59 ^ 3 x 173
Does anyone have smoother P-1?
To be honest, I do not know much about the theory (or math) behind it, so I'll take your word for the concept of smoothness. (I previously thought it meant having many small prime factors for P-1).
But, there's something I do not get. I used B1=10000 and B2=65000 (That's the first k/n pair after I switched to lower settings of 1.25x and 160MB to keep up with PRP).
Smooth means that all the factors are small. What I don't remember is whether it's 59 or 59^3 you have to consider a factor.Originally posted by Nuri
To be honest, I do not know much about the theory (or math) behind it, so I'll take your word for the concept of smoothness. (I previously thought it meant having many small prime factors for P-1).
But, there's something I do not get. I used B1=10000 and B2=65000 (That's the first k/n pair after I switched to lower settings of 1.25x and 160MB to keep up with PRP).
As for the bounds it would be possible to find the factor if the factorer doesn't only try primes (and prime powers) between B1 and B2. Whether that or something I can't think of is the explanation, we have to hear Louie about, since he's probably the only one knowing that part of the code well enough.
Stage 2 does drag in some composite numbers between B1 and B2. So I think stage 1 caught the 59^2 and stage 2 picked up the other 59 in a composite number between B1 and B2.Originally posted by hc_grove
As for the bounds it would be possible to find the factor if the factorer doesn't only try primes (and prime powers) between B1 and B2. Whether that or something I can't think of is the explanation, we have to hear Louie about, since he's probably the only one knowing that part of the code well enough. [/B]
5740000 -> 5745000 gave me
9422557846421851 | 28433*2^5741185+1
9422557846421850=2*3^9*5^2*13*17*29*89*113*108287
249632915700529 | 10223*2^5743901+1
249632915700528=2^4*3^3*1399*2027*203773
but this is real bragging
A pretty smooth factor:
322163790109561 | 19249*2^5763038+1
322163790109560 = 2^3*3^3*5*7*107*113*1637*2153
Just found a factor in a range beings fast-checked (factor value 1.2)
1172536902962371 | 19249*2^5794718+1
It seems to me that we are having trouble keeping up with PRP, and I'm being closed in by priwo if I don't find some more factors fastI only have something like 1:15 of his factors in store so I have to keep finding P-1-factors to stay in the race.
I did it again
My newest factor:
220516583040629 | 28433*2^5840545+1
220516583040628 = 2^2*7*127*1879*2113*15619
Once again, this is a factor that Mystwalker will find in his sieving range
I found 233793054676939 | 4847*2^5884647+1 today through sieving.
Unfortunately, it turned out to be a duplicate of 173003399270327 | 4847*2^5884647+1.
I guess the chances of such a thing to happen is really low (i.e. ~1 out of every 5.200 factors => to find a factor through sieving that's in the active range is 200k/19m and ratio of duplicates is currently at 1.8%).
Last edited by Nuri; 03-22-2004 at 02:03 PM.
1185354900631669 | 55459*2^5843506+1
1185354900631668 = 2^2*3*13*5557*6949*196771
246188828265473 | 10223*2^5796845+1
246188828265472 = 2^10*53*97*2111*22153
171388807977727 | 21181*2^5797244+1
171388807977726 = 2*3^2*83*139*6247*132113
437621215726002191369 | 4847*2^5803263+1 <-- the new 2nd largest factor of the year
437621215726002191368=2^3*43*211*227*7127*19441*191693
2015485189907779 | 4847*2^5801991+1
2015485189907778=2*7*263*17921*3054449
5172107658035209 | 21181*2^5804804+1 <-- submitted about an hour before it would have gone to PRP
5172107658035208=2^3*3*11*16193*19759*61231
Last edited by dmbrubac; 03-30-2004 at 11:06 AM.
21143274634835342257 | 55459*2^5890846+1
21143274634835342256 = 2 ^ 4 x 3 x 23 x 31 x 3119 x 4021 x 4091 x 12041
It's been more than a week since the last one, but now I found:
16865935315528471 | 10223*2^5884637+1
284817113042325587201 | 55459*2^5874526+1
1927699125194094887 | 22699*2^5874598+1
16865935315528470 = 2*3*5*17*113*4483*5791*11273
(second one is larger than 2^64, I will factor that tomorrow).
1927699125194094886 = 2*653*709*1259*11467*144203
284 817113 042325 587200 = 2 ^ 8 x 5 ^ 2 x 37 x 43 x 89 x 547 x 2333 x 246277Originally posted by hc_grove
284817113042325587201 | 55459*2^5874526+1
(second one is larger than 2^64, I will factor that tomorrow).
And found one myself:
2918283220459681 | 55459*2^5868118+1
2918 283220 459680 = 2 ^ 5 x 3 x 5 x 29 x 509 x 6011 x 68521
Thanks. Now 2^8 that's something I haven't seen before.Originally posted by Mystwalker
284 817113 042325 587200 = 2 ^ 8 x 5 ^ 2 x 37 x 43 x 89 x 547 x 2333 x 246277
I had a 2^10 a few days back (see my previous post).
8057451392817109 | 10223*2^5869769+1
8057 451392 817108 = 2 ^ 2 x 3 x 29 x 31 x 59 x 71 ^ 2 x 347 x 7237
3796805718943647229 | 24737*2^5951647+1
3796805718943647228 = 2^2*3^2*7*13*41*47*193*22751*136973
3025213428395171034079 | 28433*2^6051073+1
3025213428395171034078 = 2*3*43*83*353*367*599*7907*230239
3492869317679191 | 24737*2^6053911+1
3492869317679190 = 2*3*5*613*3251*3389*17239
Just found my first one with self-compiled factorer:
260840209049520371 | 21181*2^6103172+1
260840209049520370 = 2 x 5 x 179 x 367 x 1217 x 4363 x 74779
not very smooth (even though I don't really understand that term) but that counts in favor for the program.
65922812594639417 | 33661*2^6055008+1
65922812594639416 = 2^3*13^2*23*103*1483*1949*7121
oh boy oh boy oh boy
just found a big one
35935660269812747397727 | 4847*2^6168951+1
It's definetly the largest for 4847 so far, not all that smooth (or is it?)
35935 660269 812747 397726 = 2 x 3 ^ 2 x 19 x 29 x 367 x 2707 x 4651 x 5273 x 148711
It's been three weeks, but I finally found a new factor:
501737316342481 | 33661*2^6280416+1
501737316342480 = 2^4*3^4*5*19*47*317*273521
Apart from the last factor this is the smoothest I've seen.
Ahh, only two days.
5274028739098589 | 4847*2^6281463+1
5274028739098588 = 2^2*7*167*877*16993*75683
235703760784331778069847 | 19249*2^6405242+1
235703760784331778069846 = 2 x 3 x 11 x 13 ^ 2 x 17 x 389 x 1427 x 2521 x 2791 x 318259
Very smooth - apart from the last 'bummer'...
6258055154249506859 | 4847*2^6508047+1
6258055154249506858 = 2 x 19 x 577 x 811 x 1151 x 14759 x 20717
622529950198081 | 55459*2^6821266+1
622529950198080 = 2^6*3*5*73*83*401*266897
Apart from the last factor this is extremely smooth.
But the main "feature" of this factor is that I used George's new factoring code to find it!
11529059469964763 | 28433*2^6989137+1
11529059469964762 = 2 x 11 x 67 x 167 x 1787 x 4817 x 5441
10340964718983403 | 55459*2^6990238+1
10340964718983402 = 2 x 3 x 7 x 41 x 83 x 401 x 457 x 394811
1405455930948721 | 55459*2^6990250+1
1405455930948720 = 2 ^ 4 x 3 x 5 x 7 x 31 x 47 x 7793 x 73679
524413893207156299 | 10223*2^6705785+1
524413893207156298 = 2 x 17 ^ 2 x 1091 x 1693 x 18287 x 26861
Would have been a B1 hit...
75249573592556119391 | 24737*2^6826063+1
75249573592556119390 = 2*5*43*557*1031*4111*4507*16447
Fairly smooth.
Would have been a stage 1 factor, but I used Georges new code.
So does George's new code not do a GCD at the end of stage 1??
Depends on the setting of the ini file. I switched it off, as I think most factors will be found in the B2 stage...Originally posted by garo
So does George's new code not do a GCD at the end of stage 1??
Well then it's a bit unfair to "blame" George's code for not finding it in stage 1. At least that's what hc_grove's post in particular and your post to some extent indicated.
I'd rather say these postings were a bit misleading.
Of course, George's new algorithm would also find it with GCD after stage 1.
P.S.: If I can't blame George's code, I can blame himself for giving the tip to switch off stage 1 GCD.
Ouch! Don't beat me again! Ouch!
I hope you can forgive us.
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