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Old 2020-04-21, 12:35   #78
kriesel
 
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Quote:
Originally Posted by kriesel View Post
80-81 under way.
Done. Typical of the 8 processes is
Code:
M(60651732991) has 0 factors in range k = [9966119042880, 19932270758400], passes 14-15
Performed 47147307808 trial divides
Clocks = 05:46:53.712
mfactor-base-2w -m 60651732991 -bmin 80 -bmax 81 -passmin 14 -passmax 15 done at Tue 04/21/2020  7:23:50.14
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Old 2020-04-23, 00:20   #79
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Thanks kriesel. Now we have to wait for some revolutionary hardware to do the PRP/LL test of M(60651732991)
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Old 2020-05-12, 05:35   #80
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Default Another idea

Code:
{forprime(p=3,30000000000,if(
Mod(Mod(1,p+17)*x+2,x^2-3)^lift(Mod(2,p+1)^p)==1&&
Mod(Mod(1,p+257)*x+2,x^2-3)^lift(Mod(2,p+1)^p)==1&&
Mod(Mod(1,p+65537)*x+2,x^2-3)^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])]
Produces only one non-2^p-1 Mersenne exponent. The following produces two:

Code:
{forprime(p=3,30000000000,if(
Mod(Mod(1,p+17)*x+2,x^2-3)^lift(Mod(2,p+1)^p)==1&&
Mod(Mod(1,p+65)*x+2,x^2-3)^lift(Mod(2,p+1)^p)==1&&
Mod(Mod(1,p+65537)*x+2,x^2-3)^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 2020-05-12 at 05:44
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Old 2020-07-25, 13:11   #81
kriesel
 
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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 dual-E5-2690 gave the following range of run times (less than 3 days, ~0.97% range max-min). 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(95-85) 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 64-bit 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 14-14
Performed 377347101769 trial divides
Clocks = 67:47:56.301
mfactor-base-2w -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 9-9
Performed 377346866607 trial divides
Clocks = 67:08:56.768
mfactor-base-2w -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 2020-07-25 at 14:02
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Old 2020-07-25, 16:03   #82
kriesel
 
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Quote:
Originally Posted by kriesel View Post
on dual-E5-2690
actually dual Xeon E5-2697v2, which are 12-core +HT each. So one could do TF n bits to n+1 on 8 cores, and n+1 to n+2 on 16 cores, simultaneously, and still leave considerable capacity for prime95.
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Old 2020-08-18, 03:58   #83
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Can someone with big RAM please test
Code:
p=60651732991;n=2^p-1;V=Vec(lift(lift(Mod(Mod(1,n)*x,x^2-2*x-2)^2^20)));print([p,(V[1]+V[2])%p==0])
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Old 2020-08-19, 01:24   #84
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How much RAM is big RAM?
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Old 2020-08-19, 02:14   #85
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Quote:
Originally Posted by lavalamp View Post
How much RAM is big RAM?
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^2-2*x-2 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_{i-1}^2-2.

Last fiddled with by paulunderwood on 2020-08-19 at 02:18
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Old 2020-08-19, 02:18   #86
R. Gerbicz
 
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Quote:
Originally Posted by paulunderwood View Post
Can someone with big RAM please test
Code:
p=60651732991;n=2^p-1;V=Vec(lift(lift(Mod(Mod(1,n)*x,x^2-2*x-2)^2^20)));print([p,(V[1]+V[2])%p==0])
8 Mb memory is more than enough, what Pari Gp starts on my system:
Code:
parisize = 8000000, primelimit = 500000
? p=60651732991;e=20;V=Vec(lift(Mod(x,x^2-2*x-2)^2^e));
print([p,(V[1]+V[2])==0])
? [60651732991, 0]
? ##
  ***   last result computed in 0 ms.
?
not speaking about the running time. Notice that you can prove bounds for the coefficients, for this smallish iteration there will be no "overflow", so never ever calculate/store that silly n value.
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Old 2020-08-19, 02:30   #87
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Thanks Robert. I was actually seeking
Code:
? print([p,(V[1]+V[2])%p==0])
[60651732991, 1]
No surprises there.
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Old 2020-08-19, 02:39   #88
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Well I just tried it with 24GB allocated to it and it couldn't complete the first 2^p-1 instruction.

Edit: attached screenshot.
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Last fiddled with by lavalamp on 2020-08-19 at 02:41
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