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Old 2013-10-06, 02:18   #18
TheCount
 
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Sep 2013
Perth, Au.

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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^n-1.
How to go about that?

I tried by finding the weight of the term k*n-1.
I could just average sieve runs for the first 100,000 k's but sieve programs don't deal with the term k*n-1.
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*n-1, chance is about 1/log (k*n-1)

To find the probability for a fixed k over a given range we integrate from n = A to n = B.
Integrating 1/log (k*n-1) gives li(k*n-1)/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*n-1 is:
(li(k*B-1)-li(k*A-1))/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*n-1 is (li(k*B-1)-li(k*A-1))/k = (li(20*110000-1)-li(20*100001-1))/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.
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