Timeline for Let $X$ be a positive integer. Then $\pi{(X+\ln^2{X})}-\pi{(X-\ln^2{X})}>\ln{X}$?
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| May 10 at 15:05 | comment | added | Terry Tao | Actually, now that I look at it more carefully, for realistic values of $X$ it simply isn't worth it to try to chase the small $2e^{-\gamma}$ gain, and just set $w=1$ in the above analysis, i.e., brute force search. The same calculation suggests that a given interval has a $X^{-2h(-1/2)} = X^{-0.30685}$ chance of working, which is a reasonably feasible success rate for the size of $X$ in your answer. | |
| May 10 at 14:41 | comment | added | Terry Tao | For this specific problem one han heuristically allow any value of w as long as $X/W \ll X^{0.20386}$ ( though for small values of $w$ the RHS may not be fully accurate and one would gave to perform the analysis in my answer more carefully). | |
| May 10 at 8:40 | comment | added | Bogdan Grechuk | Thank you for the answer. Yes, I understood some time ago that the described effect does not help finding new counterexamples. All I wanted is to check the theory by checking that intervals $(Y-\ln^2 Y,Y+\ln^2 Y)$ with random $Y$ of the form $Y=WN$ indeed contain on average just about $e^\gamma \ln Y$ primes, instead of $2 \ln Y$. However, I failed even this, and your comment clarifies why: we would need to look at numbers with about $10^{100}$ digits(!) to decrease the effect of extra candidates below 5%. I will see whether increasing $w$ help, although I am not sure what $w$ is optimal. | |
| May 10 at 2:49 | comment | added | Terry Tao | One could reduce this loss by increasing $w$ to be closer to $\log X$ than the proposed $\log X/\log\log X$ (while keeping $W$ significantly smaller than $X$). But regardless of how one optimizes in $w$, given that each interval only had approximately a $X^{-0.20386}$ chance of producing a counterexample even if we could neglect all lower order terms, I withdraw the claim that this would be a feasible alternate approach to locate a counterexample. | |
| May 10 at 2:48 | comment | added | Terry Tao | I did some calculations. The extra candidates $WN \pm pq$ where $p,q > w$ and $pq \leq \log^2 X$ are roughly $2\frac{\log\log\log X}{\log\log X}$ as frequent as the $WN \pm p$ candidates, which tracks with your 1.5 multiplier. So one would have to take quite an enormous value of $X$ to start seeing a significant win; this is also related to how the original Cramer conjecture $p_{n+1}-p_n \leq \log^2 p_n$ has not yet been contradicted even though we expect eventually this to fail by a factor of $2e^\gamma \sim 1.1129$. It just takes too long for the $1/\log\log X$ type terms to decay! | |
| May 9 at 13:49 | comment | added | Bogdan Grechuk | I tried X being a random 1000-digit(!) number, so that $w=297.4$, W is the product of all primes up to $297$, N is random is the suggested range, and $Y=WN$. Then the number of primes in $(Y-\ln^2 Y, Y+\ln^2 Y)$ happen to be about $2.005 \ln Y$, so no effect in comparison to selecting Y just at random. The problem is that the number of candidates was 1.5 times more than stated. Extra candidates are not of the form $WN+p$. I wonder why there are so many of them for so large Y and what Y I should select so that candidates of the form $WN+p$ start truly dominate. | |
| May 4 at 16:08 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| May 4 at 15:57 | history | edited | Terry Tao | CC BY-SA 4.0 |
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| May 4 at 15:51 | history | answered | Terry Tao | CC BY-SA 4.0 |