20200421, 12:35  #78 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
31·149 Posts 
Done. Typical of the 8 processes is
Code:
M(60651732991) has 0 factors in range k = [9966119042880, 19932270758400], passes 1415 Performed 47147307808 trial divides Clocks = 05:46:53.712 mfactorbase2w m 60651732991 bmin 80 bmax 81 passmin 14 passmax 15 done at Tue 04/21/2020 7:23:50.14 
20200423, 00:20  #79 
Sep 2002
Database er0rr
2·1,723 Posts 
Thanks kriesel. Now we have to wait for some revolutionary hardware to do the PRP/LL test of M(60651732991)

20200512, 05:35  #80 
Sep 2002
Database er0rr
2×1,723 Posts 
Another idea
Code:
{forprime(p=3,30000000000,if( Mod(Mod(1,p+17)*x+2,x^23)^lift(Mod(2,p+1)^p)==1&& Mod(Mod(1,p+257)*x+2,x^23)^lift(Mod(2,p+1)^p)==1&& Mod(Mod(1,p+65537)*x+2,x^23)^lift(Mod(2,p+1)^p)==1, print([p,factor(p+1)])))} [3, Mat([2, 2])][2147483647, Mat([2, 31])] [7, Mat([2, 3])] [31, Mat([2, 5])] [127, Mat([2, 7])] [4423, [2, 3; 7, 1; 79, 1]] [8191, Mat([2, 13])] [131071, Mat([2, 17])] [524287, Mat([2, 19])] [2147483647, Mat([2, 31])] Code:
{forprime(p=3,30000000000,if( Mod(Mod(1,p+17)*x+2,x^23)^lift(Mod(2,p+1)^p)==1&& Mod(Mod(1,p+65)*x+2,x^23)^lift(Mod(2,p+1)^p)==1&& Mod(Mod(1,p+65537)*x+2,x^23)^lift(Mod(2,p+1)^p)==1, print([p,factor(p+1)])))} [3, Mat([2, 2])] [7, Mat([2, 3])] [31, Mat([2, 5])] [127, Mat([2, 7])] [607, [2, 5; 19, 1]] [4423, [2, 3; 7, 1; 79, 1]] [8191, Mat([2, 13])] [131071, Mat([2, 17])] [524287, Mat([2, 19])] [2147483647, Mat([2, 31])] Last fiddled with by paulunderwood on 20200512 at 05:44 
20200725, 13:11  #81 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
1001000001011_{2} Posts 
M60651732991 has been run with no factor found through 85 bits TF using Ernst Mayer's Mfactor program.
Sixteen processes running a pass each on dualE52690 gave the following range of run times (less than 3 days, ~0.97% range maxmin). Impact on prime95 progress using all physical cores is about a 10% slowdown during that time. Apparently hyperthreading is pretty effective for Mfactor runs. I estimate from a large extrapolation of https://www.mersenneforum.org/showpo...7&postcount=12 that the TF optimal stopping point for this exponent on cpu is around 95 bits. An estimated bit level duration is 2^{(9585)} times 67.5 hours = 7.9 years for the last bit level, 15.8 years cumulative, well out of reach. The odds of finding a factor in those 10 bit levels are around 11%. A 64bit mfaktc version on a modern gpu would be good for accelerating this, but none exists. (Estimated gpu TF limit 99 bits, ~15% odds of factor with an NVIDIA RTX 2080 Super.) Code:
M(60651732991) has 0 factors in range k = [159458035376640, 318916087089600], passes 1414 Performed 377347101769 trial divides Clocks = 67:47:56.301 mfactorbase2w m 60651732991 bmin 84 bmax 85 passmin 14 passmax 14 done at Sat 07/25/2020 4:05:44.19 M(60651732991) has 0 factors in range k = [159458035376640, 318916087089600], passes 99 Performed 377346866607 trial divides Clocks = 67:08:56.768 mfactorbase2w m 60651732991 bmin 84 bmax 85 passmin 9 passmax 9 done at Sat 07/25/2020 3:26:40.17 Last fiddled with by kriesel on 20200725 at 14:02 
20200725, 16:03  #82 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
11013_{8} Posts 

20200818, 03:58  #83 
Sep 2002
Database er0rr
2×1,723 Posts 
Can someone with big RAM please test
Code:
p=60651732991;n=2^p1;V=Vec(lift(lift(Mod(Mod(1,n)*x,x^22*x2)^2^20)));print([p,(V[1]+V[2])%p==0]) 
20200819, 01:24  #84 
Oct 2007
London, UK
2^{2}·3·109 Posts 
How much RAM is big RAM?

20200819, 02:14  #85 
Sep 2002
Database er0rr
D76_{16} Posts 
60 Billion bits can be stored in ~8 GB. Then there are the arithmetic operations. I guess 128GB RAM is enough. Pari uses FFT. So 20 squaring iterations over x^22*x2 should take a few minutes. But I am pretty sure the answer is known now. I am looking for similarities to p=607, but differences from other divisors of S_k where S_0 is 4 and S_i = S_{i1}^22.
Last fiddled with by paulunderwood on 20200819 at 02:18 
20200819, 02:18  #86  
"Robert Gerbicz"
Oct 2005
Hungary
17·83 Posts 
Quote:
Code:
parisize = 8000000, primelimit = 500000 ? p=60651732991;e=20;V=Vec(lift(Mod(x,x^22*x2)^2^e)); print([p,(V[1]+V[2])==0]) ? [60651732991, 0] ? ## *** last result computed in 0 ms. ? 

20200819, 02:30  #87 
Sep 2002
Database er0rr
D76_{16} Posts 
Thanks Robert. I was actually seeking
Code:
? print([p,(V[1]+V[2])%p==0]) [60651732991, 1] 
20200819, 02:39  #88 
Oct 2007
London, UK
2^{2}·3·109 Posts 
Well I just tried it with 24GB allocated to it and it couldn't complete the first 2^p1 instruction.
Edit: attached screenshot. Last fiddled with by lavalamp on 20200819 at 02:41 
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