20200517, 07:39  #1 
"Jeppe"
Jan 2016
Denmark
2·71 Posts 
Mersenne prime in a Cunningham chain
Often when a Mersenne prime M = 2^p  1 is found, we check also the numbers M  2 (twin), (M + 2)/3 (Wagstaff) and M*(M + 1)/2 + 1 (right perfect) to make sure M does not have a "companion" prime.
In this message, I suggest we check one companion extra, namely 2*M  1. Because M is prime, it would be possible to do a deterministic test of 2*M  1, even if the canonical form of 2*M  1does not reveal it: 2*M  1 = 2^{p+1}  3 Because of simple modulo 3 considerations, M cannot be a Sophie Germain prime or a safe prime, because 3 always divides 2*M + 1 and (M  1)/2 (except for trivial initial examples), so we do not need to check them. And (M + 1)/2 is a power of two. So 2*M  1 is the only relevant candidate for a Cunningham chain. Here is the status so far: Code:
p: factor of 2^{p+1}  3  2: prime 3: prime 5: prime 7: 11 13: prime 17: 11 19: prime 31: 9241 61: 37 89: 1823 107: 11 127: 11 521: 193 607: 11 1279: 98807 2203: 47 2281: 131 3217: 11 4253: 23 4423: 313 9689: composite 9941: 419303 11213: composite 19937: 11 21701: composite 23209: 37 44497: 11 86243: 6553 110503: 313 132049: composite 216091: 23 756839: 757 859433: 193 1257787: 11 1398269: 96997 2976221: 1451 3021377: 11 6972593: 103869349 13466917: 11 20996011: 86183 24036583: 59 25964951: 76382783 30402457: 11 32582657: 11 37156667: 11 42643801: 38334482051 43112609: 59 57885161: 193 74207281: unknown! 77232917: 11 82589933: 37463 It would also be interesting to find factors in the cases where the table above says only "composite". I have submitted all the relevant numbers to factordb.com. /JeppeSN 
20200517, 10:26  #2 
"Jeppe"
Jan 2016
Denmark
10001110_{2} Posts 
Found a factor of one of them:
Code:
p: factor of 2^{p+1}  3  2: prime 3: prime 5: prime 7: 11 13: prime 17: 11 19: prime 31: 9241 61: 37 89: 1823 107: 11 127: 11 521: 193 607: 11 1279: 98807 2203: 47 2281: 131 3217: 11 4253: 23 4423: 313 9689: 289935200473 9941: 419303 11213: composite 19937: 11 21701: composite 23209: 37 44497: 11 86243: 6553 110503: 313 132049: composite 216091: 23 756839: 757 859433: 193 1257787: 11 1398269: 96997 2976221: 1451 3021377: 11 6972593: 103869349 13466917: 11 20996011: 86183 24036583: 59 25964951: 76382783 30402457: 11 32582657: 11 37156667: 11 42643801: 38334482051 43112609: 59 57885161: 193 74207281: unknown! 77232917: 11 82589933: 37463 
20200517, 17:36  #3 
"Oliver"
Sep 2017
Porta Westfalica, DE
371_{8} Posts 
2^112143 has a factor: 1796673824281091446021443177301
2^217023 has a factor: 22313687645759 
20200517, 18:02  #4  
"Jeppe"
Jan 2016
Denmark
2·71 Posts 
Quote:
Only one composite without known factor, and one for which it is not known if it is prime or not. Updating: Code:
p: factor of 2^{p+1}  3  2: prime 3: prime 5: prime 7: 11 13: prime 17: 11 19: prime 31: 9241 61: 37 89: 1823 107: 11 127: 11 521: 193 607: 11 1279: 98807 2203: 47 2281: 131 3217: 11 4253: 23 4423: 313 9689: 289935200473 9941: 419303 11213: 1796673824281091446021443177301 19937: 11 21701: 22313687645759 23209: 37 44497: 11 86243: 6553 110503: 313 132049: composite 216091: 23 756839: 757 859433: 193 1257787: 11 1398269: 96997 2976221: 1451 3021377: 11 6972593: 103869349 13466917: 11 20996011: 86183 24036583: 59 25964951: 76382783 30402457: 11 32582657: 11 37156667: 11 42643801: 38334482051 43112609: 59 57885161: 193 74207281: unknown! 77232917: 11 82589933: 37463 Last fiddled with by JeppeSN on 20200517 at 18:03 

20200517, 18:23  #5 
"Oliver"
Sep 2017
Porta Westfalica, DE
3×83 Posts 

20200517, 19:54  #6 
Einyen
Dec 2003
Denmark
2·5·17^{2} Posts 
The 5 chains are only length 2.
The next steps 2^{p+2}  7 are composite and the first 4 of them are square numbers: p=2: 2^{p+2}  7 = 3^{2} p=3: 2^{p+2}  7 = 5^{2} p=5: 2^{p+2}  7 = 11^{2} p=13: 2^{p+2}  7 = 181^{2} p=19: 2^{p+2}  7 = 5*419429 
20200517, 20:59  #7 
"Jeppe"
Jan 2016
Denmark
2·71 Posts 
Updated with the latest factor from kruoli.
The only pending is 2^74207282  3. The same one that causes trouble with right perfect primes. For the largest factors, there is no guarantee they are minimal. Code:
p: factor of 2^{p+1}  3  2: prime 3: prime 5: prime 7: 11 13: prime 17: 11 19: prime 31: 9241 61: 37 89: 1823 107: 11 127: 11 521: 193 607: 11 1279: 98807 2203: 47 2281: 131 3217: 11 4253: 23 4423: 313 9689: 289935200473 9941: 419303 11213: 1796673824281091446021443177301 19937: 11 21701: 22313687645759 23209: 37 44497: 11 86243: 6553 110503: 313 132049: 593818027643 216091: 23 756839: 757 859433: 193 1257787: 11 1398269: 96997 2976221: 1451 3021377: 11 6972593: 103869349 13466917: 11 20996011: 86183 24036583: 59 25964951: 76382783 30402457: 11 32582657: 11 37156667: 11 42643801: 38334482051 43112609: 59 57885161: 193 74207281: unknown! 77232917: 11 82589933: 37463 
20200517, 21:07  #8  
"Jeppe"
Jan 2016
Denmark
10001110_{2} Posts 
Quote:
Edit: It has been proved that 2^{p+2}  7 is never going to be square again. /JeppeSN Last fiddled with by JeppeSN on 20200517 at 21:36 

20200518, 02:39  #9  
Nov 2003
2^{2}×5×373 Posts 
Quote:


20200518, 03:56  #10 
Einyen
Dec 2003
Denmark
2·5·17^{2} Posts 
Maybe some of us like "numerology". The whole GIMPS project is really "numerology" as well as filling out the Cunningham Tables. Unless we find so many Mersenne Primes that the theory of the distribution of Mersenne Primes needs to be altered.

20200518, 12:58  #11  
Feb 2017
Nowhere
3,331 Posts 
Quote:
2^{n}  7 = x^{2} is called Ramanujan's square equation. He posed the question of whether it has any solutions for n > 15 in 1913. So I guess you're asking, did he understand the difference between math and numerology? 

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