20131004, 09:01  #12  
Sep 2013
Perth, Au.
2·7^{2} Posts 
Quote:
To determine the weight for a random number I can just sieve "k*n1" for different k values (first 10,000 say) and find the average. srsieve does not support integer sequences of this type though. I note R916 fell pretty quickly, and with a double hit. 

20131004, 11:00  #13  
"Mark"
Apr 2003
Between here and the
1929_{16} Posts 
Quote:


20131004, 12:19  #14  
May 2007
Kansas; USA
5^{2}·421 Posts 
Quote:
On the Pdepth, it does sieve to P=1e6 and then uses a multiplier to estimate a sieve to P=1e12...and to me it's more than a "rough guess" on the number of tests at P=1e12. I've found it to be very accurate. Last fiddled with by gd_barnes on 20131004 at 12:20 

20131004, 14:25  #15  
"Mark"
Apr 2003
Between here and the
3·19·113 Posts 
Quote:
Last fiddled with by rogue on 20131004 at 14:26 

20131004, 18:22  #16 
Sep 2013
Perth, Au.
2×7^{2} Posts 
OK, so the "n per 25000 at 1e12" value in the Unproven Conjectures table is derived by calling:
> srsieve n 100001 N 110000 P 1e6 "k*b^n+/1" For each k, multiplying the results by 1.25, and adding them all together. The point is that P=1e6 gives a more accurate weight than P=511, so I will use that from now on. 
20131004, 20:04  #17  
May 2007
Kansas; USA
5^{2}·421 Posts 
Quote:
Last fiddled with by gd_barnes on 20131004 at 20:05 

20131006, 02:18  #18 
Sep 2013
Perth, Au.
2·7^{2} Posts 
Part of the reason I like prime finding is the fun of learning some new maths and number theory.
In the Proth/Riesel probability equation I am working on the weight w was the Nash weight divided by 1751.542 giving a coefficient of 1.00 for the average k of base 2. The trouble with that is it assumes base 2 has the same density of primes as a randomly chosen set of odd numbers of the same magnitude. So I want to find weight of the a randomly chosen set of odd numbers of the same magnitude as k*b^n1. How to go about that? I tried by finding the weight of the term k*n1. I could just average sieve runs for the first 100,000 k's but sieve programs don't deal with the term k*n1. Instead I used Prime Number Theory which says that the chance of a random integer x being prime is about 1/log x For x = k*n1, chance is about 1/log (k*n1) To find the probability for a fixed k over a given range we integrate from n = A to n = B. Integrating 1/log (k*n1) gives li(k*n1)/k, where li() is the logarithmic integral. So the number of primes, or fraction thereof, expected in a given range n = A to B for k*n1 is: (li(k*B1)li(k*A1))/k There is a logarithmic integral calculator here: http://keisan.casio.com/exec/system/1180573428 Now I use this for R620, where b=620 and k=20. The Unproven Conjectures table has the P=1e6 weights in the n per 25000 at 1e12 column if you divide by 1.25. For R620 the P=1e6 weight is 1007/1.25=805.6 The weight of the term 20*n1 is (li(k*B1)li(k*A1))/k = (li(20*1100001)li(20*1000011))/20 = 686.885 So w = 805.6/686.885 = 1.1728 This compares to my old version where the Nash weight was 1855, so w = 1855/1751.542 = 1.0591 So the probability increases from 11.42% to 12.64% using this new w. This is still too low compared to 19.5%. I'll have to keep thinking on it. 
20131006, 17:09  #19 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{2}×1,481 Posts 
TheCount's calculations fail for b != 2 as base 2 is assumed in the calculation of 1751.
I have spent today delving though the paper linked in http://www.mersenneforum.org/showpos...3&postcount=34 and have implemented the method it uses in the spreadsheet. It produces the same results as the original method. I also extended both methods to support ranges of n based on integration. At some point in the future I might add ranges of k as well. In the attached spreadsheet the first tab is the original method. The second is the new method and also includes a proth weight calculation. It wouldn't surprise me if either the new or old methods have regions where they give better/worse estimates than the other. If you find they are different for certain input then please let me know. I have been unable to find an example. I left it set up for R620 using Gary's data sieved to 5T. Last fiddled with by henryzz on 20131006 at 17:10 
20131013, 09:08  #20 
Sep 2013
Perth, Au.
2·7^{2} Posts 
Thanks for that post henryzz. I will read the paper you linked to see what a real Mathematician thinks.
I noted with the integrated probability equation that if you make the reasonable assumptions: ln k << B ln b, and ln k << A ln b, (where w is the weight, b is the base, k is the k, testing range from n=A to n=B) and use some logarithmic identities that the equation: #Primes = w * (ln(ln k + B*ln b)  ln(ln k + A*ln b))/ln b becomes #Primes = w * log_{b}B/A Quote a bit simpler! Also I used the term k*b*n1 for the weight of a random set of odd numbers of the same magnitude and I got 18.23% for R620. Much closer to 19.5%. I'll compare all the 1 k's for the old and new methods in your spreadsheet and see what I can find. 
20131014, 14:12  #21 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2^{2}×1,481 Posts 
In this kind of spreadsheet formula simplification of that form isn't necessary. I would prefer the accuracy myself.

20131017, 12:44  #22 
Sep 2013
Perth, Au.
2×7^{2} Posts 
Well, I completed checking the probability formulas for the 1k's. What I found is that:
 henryzz new method is the same as the old. Same inputs and results.  these methods require the range to be presieved. Only 21% of the 1k's are sieved.  my new method don't require presieving and is 98.9% correlated on the Riesel side and 97.0% correlated on the Sierpinski side.  I don't know which method is more accurate. The weight (w) I used in the spreadsheet column O, attached, is: w = k*b*P_{1e6} / (li(110,000*k*b)li(100,001*k*b)) Where li(x) ∼ x/ln(x) + x/ln(x)² + 2x/ln(x)³ + 6x/ln(x)⁴ + 24x/ln(x)⁵ + ... CRUS  1k's.zip In the early 1970s, following a counting session, the Count would laugh maniacally, "AH AH AH AH AH!" 
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