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Old 2011-03-06, 00:42   #12
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Dec 2008
you know...around...

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Slightly modified to make it sound more sophisticated :

Originally Posted by mart_r View Post
Let q(n,p) denote the parameter such that the actual probability of primality for a number n, when trial divided up to p, equals \frac{\log n}{q(n,p)*\log p}.
Applying SOE for an appropriate range in the vicinity of a given n and juxtaposing the values for n = \infty, the values q(n,p) for the fixed n fluctuate around the other, q(\infty,p), as p increases (...)
I wrote "actual probability" in contrast to the "theoretical probability" which is q(\infty,p)
The "appropriate range" should not be too small (<~n1/4) to avoid a ragged graph, but not too big either (>~n/10) to keep n "within the focus".

I'll put this into another variable: r(n,p) = q(n,p)-q(\infty,p).

For 1<p<n^{(1/e)}, r(n,p) has a
- local minimum of about -0.018 at p\approx n^{0.362} and a
- local maximum of about 0.0021 at p\approx n^{0.288}. Furthermore, for 1<p<n^{0.288}, a
- local minimum of about -0.00014 at p\approx n^{0.237}, preceded by a
- local maximum of about 0.00003 at p\approx n^{0.202} in the range 1<p<n^{0.237}.
Enough yet? The preceding minimum should be somewhere around p\approx n^{0.164}.

These are numbers taken from 109 sample values between 1016-5*108 and 1016+5*108 and might not be accurate to the last digit if this pattern continues that way for n --> \infty.

I still can't make head nor tail of this phenomenon, but I would guess these numbers are connected to something which is already known, i.e. can be calculated from well-known constants. But for now, I've got to get some sleep.
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Old 2020-07-20, 13:37   #13
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Dec 2008
you know...around...

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If there's a single person other than me out there who was stumped by this problem, let me solve it by referring to the Buchstab function. This paper, on page 11, contains the very answers to my question:
(A. Y. Cheer and D. A. Goldston: A differential delay equation arising from the sieve of Eratosthenes)

- local minimum at p ~ n^(1/c2) = n^0.3618962566...
- local maximum p ~ n^(1/c3) = n^0.288206...
and so on and so forth.

If it was a message to myself in the past, I'd also suggest looking into the parity problem.

P.S.: sorry for the gravedigging, but I had to post it here, I remember that I linked to this thread at least on one occasion.

Last fiddled with by mart_r on 2020-07-20 at 13:43 Reason: + link description
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