20211129, 23:56  #12 
May 2007
Kansas; USA
295F_{16} Posts 

20211130, 03:16  #13  
Feb 2017
Nowhere
5×1,069 Posts 
Quote:
Gleitkomma = floating point; Rechner = computer; Heizgerät = heating device The word for computer sounds very close to "reckoner." I like that! I'm not sure how the results are being found. I suspect that in some cases, a length is chosen, blocks of that length ending in 1, 3, 7, or 9 are extracted from the decimal digits of M_{p}, the integers represented by those blocks are checked for small factors, and those without small factors are subjected to a PRP test. 

20211130, 03:55  #14  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
25D7_{16} Posts 
Quote:
I used my own because it was way before that. (The sharp eyes will notice that this thread is from '16; but actually there was one even before that; then Mike recondensed it to the top thread of this thread.) 

20211130, 16:22  #15 
Feb 2017
Nowhere
5×1,069 Posts 
It occurred to me to wonder how likely it might be that an Ldigit block of decimal digits of M_{p} is prime. I'll assume that L is large enough to make it interesting, but small compared to the number D of decimal digits in M_{p}.
As we all know, D = 1 + floor(p*log(2)/log(10)). For 0 < L D, the number of Ldigit blocks is D  L + 1. This will be much less than the number 10^{L} of Ldigit blocks of decimal digits, for any L of interest here. About all I can do is invoke the Assumption of Ignorance, and treat the Ldigit blocks as "random" WRT whether they represent prime or composite numbers. That would give about 1/(L*log(10)) of the blocks being prime. I would also guess that the question doesn't really depend on whether M_{p} is prime, or even if p is prime. That is (keeping the above assumptions on L), Ldigit blocks of decimal digits of 2^{n}  1 are just as "random" for composite n as for prime exponents, and just as "random" for composite M_{p} as for Mersenne primes. If anything is known about the likelihood of Ldigit blocks being prime, I'd appreciate hearing about it. 
20211201, 18:57  #16 
Einyen
Dec 2003
Denmark
3,253 Posts 
I tried searching M51 again but only from the beginning and the ending (like pi PRPs (and the famous pi end PRPs ))
Code:
51,82589933,24862048,47,AH,2^82589933/10^24862001 51,82589933,24862048,98,AH,2^82589933/10^24861950 51,82589933,24862048,169,AH,2^82589933/10^24861879 51,82589933,24862048,6644,AH,2^82589933/10^24855404 51,82589933,24862048,7815,AH,2^82589933/10^24854233 51,82589933,24862048,24853,AH,2^82589933/10^24837195 51,82589933,24862048,4,AH,(2^825899331)%10^4 51,82589933,24862048,6,AH,(2^825899331)%10^6 51,82589933,24862048,73,AH,(2^825899331)%10^73 51,82589933,24862048,139,AH,(2^825899331)%10^139 51,82589933,24862048,558,AH,(2^825899331)%10^558 51,82589933,24862048,759,AH,(2^825899331)%10^759 51,82589933,24862048,1271,AH,(2^825899331)%10^1271 51,82589933,24862048,34652,AH,(2^825899331)%10^34652 51,82589933,24862048,151020,AH,(2^825899331)%10^151020 Last fiddled with by ATH on 20211215 at 04:23 
20211201, 20:29  #17  
Feb 2017
Nowhere
5·1,069 Posts 
Quote:
Code:
lift(Mod(2,10^4)^82589933  1) = 2591 lift(Mod(2,10^6)^82589933  1) = 902591 lift(Mod(2,10^73)^82589933  1) = 4472526640076912114355308311969487633766457823695074037951210325217902591 lift(Mod(2,10^139)^82589933  1) = 7100570481952136608062107557947958297531595208807192693676521782184472526640076912114355308311969487633766457823695074037951210325217902591 Also, lift(Mod(2,10^k)^82589933  1) for k = 558, 759, 1271 

20211201, 23:53  #18 
Einyen
Dec 2003
Denmark
3,253 Posts 
Thanks, I'm an idiot. I used my old pi sieving code and modified it to work on the end part as well, but did a very poor job of it. Luckily I only wasted a few hours on 2 cores.
I do not know why I did not check manually, normally I double check and triple check when I can, I guess I'm getting old and sloppy. 