Timeline for question in prime numbers
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Dec 14, 2010 at 22:23 | history | edited | Thomas Bloom | CC BY-SA 2.5 |
Edited to point out the mistake I made.
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| Dec 14, 2010 at 22:18 | comment | added | Thomas Bloom | Aah, forgive me. I made a simple error, which I've edited the answer to reflect. This answer is now irrelevant, and I leave it for historical interest. | |
| Dec 14, 2010 at 20:50 | comment | added | Asterios Gkantzounis | so what do u think is the answer thomas? | |
| Dec 14, 2010 at 20:33 | comment | added | Thomas Bloom | Sorry, that should be $x$ rather than $2x$. | |
| Dec 14, 2010 at 20:33 | comment | added | Thomas Bloom | Yes; or rather, the asymptotic formula that is a corollary of Theorem 7.11 (and letting $y=p_n$) -- clearly, for sufficiently large x, we can guarantee that $\Phi(x,y)\geq 0.1\frac{2x}{\log p_n}$ (say), and just take $N$, $N'$ greater than x with a difference of $2p_n$. | |
| Dec 14, 2010 at 17:45 | comment | added | Gerhard Paseman | Also, I am finding online theorem 7.11 credited to Buchstab, and I am not seeing how you derive your statement from that. Are you using equation 7.42 which has phi(x,y) defined as the number of numbers at most x which have only prime factors > y ? Gerhard "Ask Me About System Design" Paseman, 2010.12.14 | |
| Dec 14, 2010 at 17:39 | comment | added | Thomas Bloom | This seems to show the result for all sufficiently large intervals; this still contradicts your answer, however. I'll think about it overnight. | |
| Dec 14, 2010 at 17:27 | comment | added | Gerhard Paseman | I interpret the question as "all such intervals", not "exist infinitely many such intervals". Perhaps the original poster will clarify. Gerhard "Ask Me About System Design" Paseman, 2010.12.14 | |
| Dec 14, 2010 at 17:23 | history | answered | Thomas Bloom | CC BY-SA 2.5 |