Timeline for Semi-primes represented by quadratic polynomials
Current License: CC BY-SA 4.0
8 events
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| Nov 13, 2018 at 22:28 | comment | added | Delmastro | @rlo Oh, it's great to have your input here! Thank you for your comment, it is very useful (indeed, I thought that $P_2$ were numbers with exactly two prime factors, not up to; my bad). In my case, prime or semi-prime are both fine, but other composites do not really work. In fact, the existence of a single prime, or a single pair with $(p_1,p_2)=-1$ would suffice, but I guess that's too much to ask... Anyway, thank you again! | |
| Nov 13, 2018 at 18:52 | comment | added | rlo | I unfortunately have to agree with Stanley that this is hard. In fact, though it's been a while since I've thought about the problem, I believe I only proved that there are infinitely many values G(n) with at most two prime factors, not exactly two, so if you really need semiprimes, you might be out of luck. OTOH, if you want something weaker (e.g., that there's some factorization of G(n) = d_1*d_2, not necessarily into primes, with (d_1,d_2) of each sign), then there might be some games you could play, but this would have very a different feel. | |
| Nov 12, 2018 at 15:04 | history | edited | Delmastro | CC BY-SA 4.0 |
added 242 characters in body
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| S Nov 12, 2018 at 1:28 | history | suggested | Josiah Park | CC BY-SA 4.0 |
removed / and added details from reference
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| Nov 12, 2018 at 1:28 | comment | added | Stanley Yao Xiao | The short answer is that probably nothing is known, but your expectation should be correct. One of the issues is that the current way to prove that a quadratic $f$ represents semi-primes infinitely often uses a sieve, which doesn't quite pick up the correct order of magnitude of semi-primes of order 2. | |
| Nov 12, 2018 at 0:59 | review | Suggested edits | |||
| S Nov 12, 2018 at 1:28 | |||||
| Nov 12, 2018 at 0:50 | review | First posts | |||
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| Nov 12, 2018 at 0:46 | history | asked | Delmastro | CC BY-SA 4.0 |