[QUOTE=lorgix;233143]Funny example;
M2371703 has a factor: 172106762886153056494817 The funny part is that k= 2*2*2*2*1049*1811*34469*34631, so 2kp+1 could have been found with B2= 34631[/QUOTE] Even faster with B1=34631 and no stage 2. This 78 bit prime factor can be found in ~95 seconds on one core of a modern CPU. Pretty incredible. :smile: 
[QUOTE=MiniGeek;233151]Even faster with B1=34631 and no stage 2. This 78 bit prime factor can be found in ~95 seconds on one core of a modern CPU. Pretty incredible. :smile:[/QUOTE]
Yes, didn't think it through. I'm doing a large number of P1 in that exponent area with B1~ double the above, and no stage 2. Might add on stage 2 later. But as low as the limits are on some exponents in the area I think this might be a fairly effective approach. On the other end we have M234473 with its [SIZE=3]2kp+1=[/SIZE][SIZE=2][SIZE=3]750020570278149061867959407 k= 41*234,473*161,503,549*241,537,413,779 Good luck finding that with P1! :big grin: Would be nice to hear other more or less extreme examples, maybe that has/deserves a thread of its own. [/SIZE][/SIZE] 
My smallest factor yet;
M2428781 has a factor: 5404906126626057367 k=3*7*89*7583*78509 Evaded previous TF by 1.2bits. Another one; M2428661 has a factor: 1082894272153874474993 k=2*2*2*13553*22307*92177 
More cases where the largest prime factor of k is relatively small:
M[B]2430457[/B] has a factor: 113034265381620576607487  77bit k= 17*541*4987*14303*[B]35447[/B] M[B]4538137[/B] has a factor: 326333958725131675555063  79bit k= 3*7*109*15271*20509*[B]50153[/B] Is my intuition right, or are cases like these actually quite common? 
<silverman> Read my paper with Wagstaff.</silverman>.

Hi,
[QUOTE=lorgix;233295]Is my intuition right, or are cases like these actually quite common?[/QUOTE] I think so, some of "my" P1 factors, sorted by factor size: M49915309 has a factor: 2085962683046854861393 (70.82 Bits; k = 20895019231944 = 2 * 2 * 2 * 3 * 11 * 13 * 37 * 227 * 347 * [B]2089[/B]) M50739071 has a factor: 10474816683392115991831 (73.14 Bits; k = 103222393285365 = 3 * 3 * 5 * 19 * 181 * 227 * 769 * [B]3821[/B]) M51027377 has a factor: 77850684812475802805663 (76.04 Bits; k = 762832516479103 = 7 * 17 * 139 * 191 * 239 * 257 * [B]3931[/B]) M51679921 has a factor: 451068222670482355938121 (78.57 Bits; k = 4364056812997860 = 2 * 2 * 3 * 5 * 11 * 307 * 953 * 2957 * [B]7643[/B]) M53196851 has a factor: 129375114189794147350126111 (86.74 Bits; k = 1216003501690298805 = 3 * 5 * 11 * 17 * 71 * 79 * 2113 * 3571 * [B]10243[/B]) M51007903 has a factor: 416373044176966390884620641 (88.42 Bits; k = 4081456202747311440 = 2 * 2 * 2 * 2 * 3 * 3 * 3 * 5 * 13 * 59 * 89 * 563 * 4441 * [B]11071[/B]) M51094921 has a factor: 620135167283167713317314151 (89.00 Bits; k = 6068461944418778075 = 5 * 5 * 271 * 2851 * 3371 * 8429 * [B]11057[/B]) M51139447 has a factor: 1873562419055481575542379948831 (100.56 Bits; k = 18318172457510946251945 = 5 * 7 * 19 * 23 * 73 * 359 * 18457 * 35597 * [B]69557[/B]) Oliver 
But notice the latest factors my computer found using ECM:
M400087 has a factor: 286218557414155282359049 (k = 2 ^ 2 x 3 x 41737 x [B]714185251333[/B]) M400277 has a factor: 2081610233687632912124807 (k = 11 x 107 x [B]2209186189633807[/B]) These factors could not have been discovered using P1. 
[QUOTE=TheJudger;233535]Hi,
I think so, some of "my" P1 factors, sorted by factor size: M49915309 has a factor: 2085962683046854861393 (70.82 Bits; k = 20895019231944 = 2 * 2 * 2 * 3 * 11 * 13 * 37 * 227 * 347 * [B]2089[/B]) M50739071 has a factor: 10474816683392115991831 (73.14 Bits; k = 103222393285365 = 3 * 3 * 5 * 19 * 181 * 227 * 769 * [B]3821[/B]) M51027377 has a factor: 77850684812475802805663 (76.04 Bits; k = 762832516479103 = 7 * 17 * 139 * 191 * 239 * 257 * [B]3931[/B]) M51679921 has a factor: 451068222670482355938121 (78.57 Bits; k = 4364056812997860 = 2 * 2 * 3 * 5 * 11 * 307 * 953 * 2957 * [B]7643[/B]) M53196851 has a factor: 129375114189794147350126111 (86.74 Bits; k = 1216003501690298805 = 3 * 5 * 11 * 17 * 71 * 79 * 2113 * 3571 * [B]10243[/B]) M51007903 has a factor: 416373044176966390884620641 (88.42 Bits; k = 4081456202747311440 = 2 * 2 * 2 * 2 * 3 * 3 * 3 * 5 * 13 * 59 * 89 * 563 * 4441 * [B]11071[/B]) M51094921 has a factor: 620135167283167713317314151 (89.00 Bits; k = 6068461944418778075 = 5 * 5 * 271 * 2851 * 3371 * 8429 * [B]11057[/B]) M51139447 has a factor: 1873562419055481575542379948831 (100.56 Bits; k = 18318172457510946251945 = 5 * 7 * 19 * 23 * 73 * 359 * 18457 * 35597 * [B]69557[/B]) Oliver[/QUOTE] [QUOTE=alpertron;233537]But notice the latest factors my computer found using ECM: M400087 has a factor: 286218557414155282359049 (k = 2 ^ 2 x 3 x 41737 x [B]714185251333[/B]) M400277 has a factor: 2081610233687632912124807 (k = 11 x 107 x [B]2209186189633807[/B]) These factors could not have been discovered using P1.[/QUOTE] Yes, both extremes exist. But I'm curious about how common cases like these are. A 3D (exponent size, number of factors of k, sizes of factors of k) graph or a distribution table would be extremely nice. But I doubt that's at hand. Any thought or info/ideas? 
not found by me but still
M52000043 has a factor : 104000087 (k=2) thats an extreme... 
[QUOTE=firejuggler;233541]not found by me but still
M52000043 has a factor : 104000087 (k=2) thats an extreme...[/QUOTE] Yeah, I've seen a few of those... Not that big before though. A little sample; these should be the known ones in the range. M33623 has 67247 M34283 has 68567 M34319 has 68639 M34439 has 68879 M34631 has 69263 M34883 has 69767 M35099 has 70199 M35111 has 70223 M35291 has 70583 M35831 has 71663 M35999 has 71999 And it goes on.... I think it's safe to say that k=2 is common for small p. But just [I]how[/I] does the distribution look? I kinda feel like this info should be available [I]somewhere[/I]. P.S. Smallest p for which k=2 11, 23, 83, 131, 179, 191, 239, 251, 359, 419, 431, 443 After looking those up manually I realize that this is [URL="http://www.research.att.com/%7Enjas/sequences/A002515"]A002515.[/URL] 
These "k=2" you are talking about are really k=1. Factors are of the form 2kp+1. In other words, they're mp+1, with m always even. With these factors, m=2 and k=1, since the factor is equal to 2*p+1.
I'd bet that the factors of the k's break down, on average, like the factors of any natural number of about their size. And that the chance of any given k producing a factor is related to the equation given at [url]http://www.mersenne.org/various/math.php[/url]: "(how_far_factored1) / (exponent times [URL="http://www.utm.edu/research/primes/glossary/Gamma.html"]Euler's constant[/URL] (0.577...))". I don't know what that means precisely as far as how many k's will be smooth to suchandsuch bounds, but the GIMPS Math page says "The chance of finding a factor and the factoring cost both vary with different B1 and B2 values. Dickman's function (see Knuth's Art of Computer Programming Vol. 2) is used to determine the probability of finding a factor, that is k is B1smooth or B1smooth with just one factor between B1 and B2. The program tries many values of B1 and if there is sufficient available memory several values of B2, selecting the B1 and B2 values that maximize the formula above." 
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