Timeline for Are there infinitely many natural numbers not covered by one of these 7 polynomials?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 30, 2014 at 22:53 | vote | accept | joebloggs | ||
| Apr 30, 2014 at 21:50 | comment | added | Christian Elsholtz | To go a step away from the prime triple conjecture. $f_1$ and $f_2$ allow that $30n+11$ contains prime factors $p\equiv 1,11,19,29 \mod 30$. $f_3,f_5,f_6,f_7$ together are more restrictive, All possible prime factors are forbidden, For $f_4$: $30n-11$ may contain prime factors $p\equiv 1, 7, 19, 13\mod 30$. (Note: 1,11,19,29 and 1,7,19,13 are both multiplicative subgroups mod 30). Hence there is one prime condition, and 2 half-prime conditions, (sieve dimension 2), which is still undoable. (For sieve dimension 3/2 there is sometimes hope, Iwaniec half dimensional sieve). | |
| Apr 30, 2014 at 21:27 | comment | added | joebloggs | Great response @Max! | |
| Apr 30, 2014 at 21:25 | comment | added | Jeremy Rouse | This factoring trick is called "completing the rectangle". | |
| Apr 30, 2014 at 21:19 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
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| Apr 30, 2014 at 21:18 | comment | added | Lucia | Excellent! And for the general problem of $axy+bx+cy$, one multiplies through by $a$ and gets $(ax+c)(ay+b)-bc$, so that the numbers missed by such polynomials are again closely related to shifted primes. | |
| Apr 30, 2014 at 21:12 | history | answered | Max Alekseyev | CC BY-SA 3.0 |