20170909, 17:27  #34  
Jun 2012
Boulder, CO
2^{3}·5·7 Posts 
Quote:
Code:
n: 126995944778384428343817467782763835965797446765537889426743251291836133619348289537505627894595882345088764593485439135004224536319773689870707463091317828498596530820195257967005647404285565398502129893378166401 # 2^1100+1^1100, difficulty: 264.91, skewness: 1.00, alpha: 1.45 # cost: 2.54145e+19, est. time: 12102.14 GHz days (not accurate yet!) skew: 1.000 c4: 1 c3: 1 c2: 1 c1: 1 c0: 1 Y1: 1 Y0: 1684996666696914987166688442938726917102321526408785780068975640576 m: 1684996666696914987166688442938726917102321526408785780068975640576 type: snfs 

20170911, 23:08  #35 
Dec 2011
11×13 Posts 
Speaking of fishing lessons...
There is another unfactored 11smooth number with 213 digits: c213.of.2^3150+1L jcrombie's Web Site reports a degree 8 polynomial with SNFS difficulty 252.865. (I'm hoping to understand how that is obtained, so I might apply the method to numbers with larger bases.) I define two functions, applicable to base 2 Cunninghams, , where n is an odd multiple of 2 By recursively dividing out the algebraic factors, I obtain an algebraic expression for the Aurifeuillean L primitive of 2^3150+1: The result is a 217digit number, which has been partially factored to: 12601 * c213 (as shown above). The algebraic expression for the L primitive seems too complex to use in SNFS. A long division of (1) would result in not fewer than 105 terms. Obviously, since (1) computes the primitive, nothing will divide it. However, it may be that multiplication by some of the algebraic factors which were removed might result in a simpler algebraic expression. Indeed, the Aurifeuillean Lprimitive of 2^450+1 is: The result is a 37digit number, which has been fully factored. Multiplying (1) by (2), we obtain an expression that when fully expanded should have fewer terms: The result is a 253digit number, whose logarithm is approximately 252.865. Setting a=2, I can rewrite (3) in a standard algebaic form: Setting , I can rewrite (4) in an even simpler form: Now, with the help of Mathematica, I expand the numerator and denominator: Mathematica's PolynomialRemainder function reports the remainder is 0. The PolynomialQuotient function provides the longdivision as: Now, the path becomes less clear, and I could use some guidance. (Fishing lessons.) I factor out from (7): Next, I substitute . How do we complete this? Can someone provide the detailed algebra to obtain an SNFS polynomial? Thanks in advance. Last fiddled with by rcv on 20170911 at 23:41 
20170912, 20:46  #36 
Jun 2012
5×599 Posts 
@rcv  do you intend to factor the C213 from 2^3150+1L?
I ran some test sieving on it and it's a tough composite to crack  on the order of 400500,000 corehours to sieve. The octic demands its toll! As to your algebra problem, (10) seems almost intractable but I'm no expert. Good luck! 
20170912, 22:00  #37  
Dec 2011
10001111_{2} Posts 
Quote:
As far as my equation (10), this is in the form (as I recall, with palindromic coefficients) by which each pair of terms, , which have identical coefficients, can be combined to a single term, perhaps in another variable and perhaps with a different coefficient. I have approached this several times, but each time I have run into a dead end. I have been working at this from the point of view of the old high school algebra question, if you are given , then calculate . So, if , then Last fiddled with by rcv on 20170912 at 22:04 

20170913, 06:41  #38 
Dec 2011
10001111_{2} Posts 
Let Using the method outlined in my previous post, we combine pairs of terms from the right hand factor of (10) and write them in terms of : Summing, and replacing the right hand factor, we get Now, I substitute Recall that Let Is there any chance this is correct? Last fiddled with by rcv on 20170913 at 06:44 
20170917, 04:13  #39 
Dec 2011
11·13 Posts 
With respect to my previous series of posts, I think I have it figured out. I believe the algebra shown above is basically correct for the algebraic side. Disregard the last few numbered equations which were an irrational stab at the rational side. I will plan to post something about the corresponding rational side when I have the time to write something cogent.
Meanwhile, a new ElevenSmooth 49digit factor of M(29700) has appeared in factordb. (Not my work.) p49 = 26400155219669229975537170269019760942301 
20170917, 11:25  #40  
Jun 2012
5663_{8} Posts 
Quote:
Ryan is also attempting the lowest difficulty SNFS composites as well, with factors reported here. There do not appear to be many feasible SNFS factorization jobs left in that project. 

20170925, 19:46  #41 
Jun 2012
Boulder, CO
2^{3}·5·7 Posts 
For those keeping score at home, also got a p60 ECM hit on this c564 cofactor of 2^2800 + 1:

20171012, 17:41  #42 
Jun 2012
5×599 Posts 
2 more ECM hits by RyanP
Ryan continues to run ECM on the ElevenSmooth project. Two recent results.
M12096 composite cofactor C996 yielded an ECM hit Code:
Using B1=2900000000, B2=81523616554398, polynomial Dickson(30), sigma=3904585649 ... Found probable prime factor of 58 digits: 7899984695355088860500288590353045075963122543602810243969 M23100L cofactor C701 also got an ECM hit Code:
Using B1=7600000000, B2=322813090700118, polynomial Dickson(30), sigma=3343323451 ... Found probable prime factor of 56 digits: 48760786864644474856730499321644513230245978254290690801 Last fiddled with by swellman on 20171012 at 17:47 
20171013, 23:36  #43 
Dec 2011
10001111_{2} Posts 
Thanks for the info! Subsequent to Ryan's p60 hit on 2^2800+1 (two posts up), someone reported a p58+p447 split to factordb.com. I presume that was also Ryan?
Last fiddled with by rcv on 20171013 at 23:39 
20171014, 01:20  #44  
Jun 2012
2995_{10} Posts 
Quote:
Also found by Ryan was a p57 prime factor of a C852 cofactor of 2^5400+1, leaving a C795. Results reported to factordb. Last fiddled with by swellman on 20171014 at 01:33 

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