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2021-11-29, 23:56   #12
gd_barnes

May 2007
Kansas; USA

5·13·163 Posts

Quote:
 Originally Posted by xilman Clue: PL == Paul Leyland.
GB == (yours truly)

Direct translation appears to be:
Floating point computer heater project

2021-11-30, 03:16   #13
Dr Sardonicus

Feb 2017
Nowhere

3·1,787 Posts

Quote:
 Originally Posted by gd_barnes GB == (yours truly) Direct translation appears to be: Floating point computer heater project
I get

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 Mp, the integers represented by those blocks are checked for small factors, and those without small factors are subjected to a PRP test.

2021-11-30, 03:55   #14
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

2×37×131 Posts

Quote:
 Originally Posted by Dr Sardonicus 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 Mp, the integers represented by those blocks are checked for small factors, and those without small factors are subjected to a PRP test.
Mark R. actually wrote a sieve for that.
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 re-condensed it to the top thread of this thread.)

 2021-11-30, 16:22 #15 Dr Sardonicus     Feb 2017 Nowhere 14F116 Posts It occurred to me to wonder how likely it might be that an L-digit block of decimal digits of Mp 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 Mp. As we all know, D = 1 + floor(p*log(2)/log(10)). For 0 < L $\le$ D, the number of L-digit blocks is D - L + 1. This will be much less than the number 10L of L-digit blocks of decimal digits, for any L of interest here. About all I can do is invoke the Assumption of Ignorance, and treat the L-digit 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 Mp is prime, or even if p is prime. That is (keeping the above assumptions on L), L-digit blocks of decimal digits of 2n - 1 are just as "random" for composite n as for prime exponents, and just as "random" for composite Mp as for Mersenne primes. If anything is known about the likelihood of L-digit blocks being prime, I'd appreciate hearing about it.
 2021-12-01, 18:57 #16 ATH 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,(2^82589933-1)%10^4 51,82589933,24862048,6,AH,(2^82589933-1)%10^6 51,82589933,24862048,73,AH,(2^82589933-1)%10^73 51,82589933,24862048,139,AH,(2^82589933-1)%10^139 51,82589933,24862048,558,AH,(2^82589933-1)%10^558 51,82589933,24862048,759,AH,(2^82589933-1)%10^759 51,82589933,24862048,1271,AH,(2^82589933-1)%10^1271 51,82589933,24862048,34652,AH,(2^82589933-1)%10^34652 51,82589933,24862048,151020,AH,(2^82589933-1)%10^151020 No others below 150K digits from the beginning. Last fiddled with by ATH on 2021-12-15 at 04:23
2021-12-01, 20:29   #17
Dr Sardonicus

Feb 2017
Nowhere

3·1,787 Posts

Quote:
 Originally Posted by ATH I tried searching M51 again but only from the beginning and the ending (like pi PRPs (and the famous pi end PRPs )) So far no end PRPs:
Really?

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

2021-12-01, 23:53   #18
ATH
Einyen

Dec 2003
Denmark

3,253 Posts

Quote:
 Originally Posted by Dr Sardonicus Really?
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.

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