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Old 2019-12-14, 16:46   #2
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Dec 2008
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Default Merit and CSG via Riemann's R(x) - a matter of belief?

_I don't think so. But I'm probably going to need some feedback. None of the papers I've read so far discussed the topic at hand.

_While calculating the merit of a gap via g/log(p), and the Cramér-Shanks-Granville (CSG) ratio accordingly as g/log²(p), the first few values of the latter expression don't fit in the picture. Even when the right-hand bounding prime p'=p+g is used, for p=7 and p'=11, CSG=g/log²(p')=0.69566, a value that is first superseded at p=2010733. And the situation is worse for the two smaller maximal gaps. You know what I mean, we can't compare CSG ratios like that for small p.
_Recall that CSG is always M²/g, multiplied by \varphi(q) when arithmetic progressions p+iq are considered. (Ah, there's the phi I was looking for!) If the calculation for CSG is altered, it's directly from the calculation for the merit M.
_For my musings, consistency is key, thus I resorted to calculating the merit M=R(p')-R(p) in terms of the formula
R(x)=1+\sum_{n=1}^\infty \frac{log^nx}{n\hspace{1}n!\hspace{1}\zeta(n+1)
_and I was quite happy with it for some time now. The average probability of a random number x being prime, by this reasoning, is a bit smaller than 1/log(x), namely R'(x)=\frac{1}{\log x}\sum_{n=1}^\infty \frac{\mu(n)}{n\hspace{1}x^{1-\frac{1}{n}}}

_Yet something kept bugging me. There is another term (well, actually, two terms) in the smooth part of the famous Riemann prime counting formula, which gives a strictly increasing function for x>1 that fits perfectly between the stairs of \pi(x) from the very beginning.
Ri(x)=R(x)+\frac{\arctan\frac{\pi}{\log x}}{\pi}-\frac{1}{\log x}
_But now [Ri(3)-Ri(2)]²/1=0.91808, a value that is only challenged by Nyman's gap with CSG=0.92064 for all primes<264. Things are getting more troublesome with prime gaps in arithmetic progression. The comparison in the attached table shows that it's not quite right to simply take M=Ri(p')-Ri(p). (I've just noticed that I used ln instead of log there, just don't get confused by that:)
_M=Ri(p'+½)-Ri(p+½) is good for ordinary gaps (q=2), but not for arithmetic progressions.

_The most appropriate and consistent way I could find of dealing with the measure of the gaps is to take the sum of the derivatives of Ri(x) at all integers x=p+iq for 1\lei\lek where k=g/q. Ri'(x) would then serve as the probability à la Cramér.
_Cross-check: Ri(x) ~ \sum_{i=2}^x Ri'(i)

_Better yet: Ri(x)-Ri(c) < \sum_{i=2}^x Ri'(i) < Ri(x+1)-Ri(c) where c=1.5920763885...

_What follows is that we have to distinguish the values of q mod 4.
_When q mod 4=0, M=\sum_{i=1}^k Ri'(p+iq)
_When q mod 4=2, M=\sum_{i=1}^{2k} Ri'(p+\frac{iq}{2})

_So the question goes to the reader: Is this getting out of hand?
Attached Files
File Type: pdf Merit_CSG comparison.pdf (40.4 KB, 247 views)
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