Timeline for Every prime number > 19 divides one plus the product of two smaller primes?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2015 at 14:37 | comment | added | joro | @EricNaslund for large n, isn't yours asymptotically the same? | |
| Dec 29, 2011 at 10:31 | comment | added | Eric Naslund | @joro: On the probabilistic grounds, we should be integrating the density and looking at $$\frac{1}{p}\left(\int_2^p \frac{1}{\log t}dt\right)^2=\frac{1}{p}\text{li}(p)^2.$$ Indeed, in the case plugging in $p=1000003$ we find the much closer approximation $$\frac{1}{p}\text{li}(p)^2=6182.307\dots$$ | |
| Dec 28, 2011 at 11:34 | comment | added | joro | @Mark if someone finds the pari program correct, probably you can do up to $10^10$ with it in about 10 hours (very rough estimate). The only change needed is replacing "10^9" with "10^10" | |
| Dec 28, 2011 at 11:07 | comment | added | user6976 | @joro: I used Maple which is very slow. I do have pari installed on my computer but never learned how to use it, unfortunately. Thank you for checking the conjecture. I think that the analytic tools work better when the density argument gives truthful answer. So there is a hope. | |
| Dec 28, 2011 at 8:40 | history | answered | joro | CC BY-SA 3.0 |