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20181211, 23:30  #12  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
4,457 Posts 
Quote:


20181211, 23:37  #13  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
4,457 Posts 
Quote:


20181211, 23:41  #14 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
2^{2}·2,161 Posts 
Your former range and other clues (based upon at least 1 being interpreted) point to 5 exponents. Your updated range limits that to <5.
Last fiddled with by Uncwilly on 20181211 at 23:42 
20181212, 00:00  #15 
Sep 2003
5024_{8} Posts 
Methodology for my estimate:
Magnify the graph about 5x in a Paint program. Under magnification, we see that the lines that draw the squarebox plot points are not always sharp, but rather they are often blurred and smeared into two or even three adjacent lines of varying intensity. This reflects the fact that the exact center of the desired plot point usually does not map exactly onto an exact pixel point, so the plotting program uses blurring to compensate. So we can use the relative paleness/brightness of the components of the blurred/smeared lines to assign a fractional vertical pixel position to both the top horizontal line and the bottom horizontal line of each squarebox plot point, and then average them to find the center. For the relative vertical pixel positions, with the zero reference point at the bottom horizontal bar of the square representing M46, I get: Code:
bottom top center Mn log(Mn) M46 0.2 9.1 4.45 42643801 17.56839 M47 0.3 10.0 5.15 43112609 17.57933 M48 20.2 30.0 25.1 57885161 17.87397 M49 37.05 46.7 41.875 74207281 18.12237 M50 39.8 49.4 44.6 77232917 18.16234 M51 44.3 54.0 49.15 xxxxxxxx 18.2296 to 18.2301 Last fiddled with by GP2 on 20181212 at 00:04 
20181212, 00:26  #16  
P90 years forever!
Aug 2002
Yeehaw, FL
7171_{10} Posts 
Quote:
I'd also like to see your graph along with a line with the slope predicted at https://primes.utm.edu/notes/faq/NextMersenne.html. "we should get approximately a straight line with slope 1/egamma (about 0.56145948)" 

20181212, 00:34  #17  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10001110100111_{2} Posts 
I think it (that discussion) has started (since post like #10 and #14 )
It reminds me of that anecdote where someone heard about the prime and the only question that they had was "but what was the sign of the penulitmate iteration?" Numbers, shnumbers... Ideas! Found it  Quote:


20181212, 00:52  #18  
P90 years forever!
Aug 2002
Yeehaw, FL
71×101 Posts 
Quote:
We have a theory that Mersenne primes "follow" a logarithmic line with a specific slope. We have data on the first 51 Mersennes. If someone with a good probability/statistics background could weigh in with an analysis of how well the 51 data points are matching the theory, then I and others here could hopefully learn a thing or two. 

20181212, 02:43  #19  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9,127 Posts 
Quote:
I later googled a free version of Wagstaff (1983) (below) and with all due respect the UTM simplified explanation has more holes than the original paper. I don't think the https://primes.utm.edu/notes/faq/NextMersenne.html quite matches what the paper does. 

20181212, 03:56  #20  
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
4,457 Posts 
Quote:
Zoom first, then screen copy paste, and there's little or no antialiasing in the paint program input. A spreadsheet using pixel locations of X marks the center gives method error estimates provided by the knowns. You seem to be using the already known Mn values above, rather than computing estimates from the pixel locations for comparison to the knowns. What did I miss? 

20181212, 05:38  #21  
Sep 2003
2^{2}·3·5·43 Posts 
Quote:
This time, after zooming to the maximum in the PDF and saving the image, I then zoomed some more in the paint program to get nice big pixels. It is this second process that makes the antialiasing more readily visible. I suspect that if I redid the calculations with the new pixel counts it would give a more accurate estimate, to an extra decimal place. Perhaps I'll try that tomorrow. What estimate do you get? 

20181212, 06:09  #22  
Sep 2003
2^{2}×3×5×43 Posts 
Quote:
If A = 2p and B = sqrt of the Mersenne number, then you can try to estimate the total number of factors a Mersenne number of a given size would be expected to have. I tried to do that in a post a few months ago, maybe someone who actually understands the math can correct it. However, thanks to user TJAOI and his very systematic "by k" factoring, we can be quite certain that we know all factors of bitlength 65 or less of all Mersenne numbers with prime exponent less than 1 billion. As I posted last year in the "User TJAOI thread", empirically we can see that he does not miss any factors, except for one very minor glitch in 2014 that he very quickly corrected. So we can use our factor data to check if the actual exact number of factors smaller than B = 2^65 matches what Wagstaff's heuristic predicts, for various ranges of p. If our factor data is in fact consistent with the Poisson process implied by Wagstaff's analysis, then we should expect that the predicted number of Mersenne primes is also consistent with it. I really think someone here with the required math knowledge could do the analysis and maybe make a paper out of it. 

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