Timeline for Why are Goldbach laggards biased towards $2 \bmod 6$?
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Mar 9, 2023 at 17:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Nov 9, 2022 at 16:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
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Feb 12, 2022 at 14:53 | history | edited | LSpice | CC BY-SA 4.0 |
While this is on the front page, uniformising number (TeX vs. non-TeX) formatting; `\mod` -> `\bmod`; links to comments
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Feb 12, 2022 at 13:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
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Oct 20, 2020 at 10:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Sep 20, 2020 at 9:14 | answer | added | user140242 | timeline score: 1 | |
Apr 7, 2014 at 12:29 | comment | added | Joël | Oh, I didn't notice this whole discussion was old. | |
Apr 7, 2014 at 10:34 | comment | added | GH from MO | @Joel: Thank you, but note that my comment with the link is more than 2 years old. Links come and go! | |
Apr 6, 2014 at 22:12 | comment | added | Joël | @GHfromMO The link you give seems to be broken. I think web.math.princeton.edu/sarnak/MazurLtrMay08.PDF works instead. | |
Apr 6, 2014 at 14:38 | comment | added | Ricardo Andrade | The last edit to this question (on April 6, 2014) was a bit extensive, nevertheless I approved the suggested edit by user48851. I checked the correctness of the edit using this text file which lists the number of ways that $2n$ can be written as a sum of two primes for $n\leq 20000$. That file is linked from the corresponding OEIS sequence A045917. | |
S Apr 6, 2014 at 14:27 | history | suggested | Edward Porcella | CC BY-SA 3.0 |
962 does not belong--no need to note it as an exception to each number being a power of 2 times a prime, and a 24 is not an exception since $a_2 = 68$, which is $2 mod 6$8
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Apr 6, 2014 at 13:27 | review | Suggested edits | |||
S Apr 6, 2014 at 14:27 | |||||
Apr 6, 2014 at 11:59 | review | Suggested edits | |||
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Feb 4, 2011 at 23:54 | comment | added | GH from MO | This is also a lovely read: math.princeton.edu/sarnak/MazurLtrMay08.PDF | |
Feb 4, 2011 at 7:38 | comment | added | Aaron Meyerowitz | I looked at g(a) and g(b) for a and b twice a prime (to remove the effect of other prime divisors) then it does seem much more balanced (but not totally) if we let a and b each be the jth member of their congruence class mod 6. | |
Feb 4, 2011 at 7:11 | comment | added | Aaron Meyerowitz | @Gerry (and Gjergi and David) Yes! that does explain a lot. Of course actually (if I haven't made an error) we would have $1194 \le N \le 1216$ so that the number of primes congruent 1 mod 6 would be between 94 and 96. So it is as many 50 vs no more than 47 or 48. Still that may be enough. And the effect is more pronounced mod 3 than mod p for larger primes because there is no mixing of classes. | |
Feb 4, 2011 at 6:05 | comment | added | Gerry Myerson | Let the minutes record that Gjergji's comment wasn't yet visible to me when I posted mine. | |
Feb 4, 2011 at 6:03 | comment | added | Gerry Myerson | To expand on David Speyer's comment: suppose that up to $N$ there are 100 primes congruent 2 mod 3, and only 80 congruent 1 mod 3. Then if $N$ is 4 mod 6 it could have as many as 50 Goldbach representations; if 0 mod 6, it could have as many as 80; if 2 mod 6, it can't have more than 40. This could well be why more laggards are 2 mod 6, and no laggards (except 24) are 0 mod 6. You have to go up to 608,981,813,029 before you get more primes 1 than 2 mod 3, which seems to be beyond the range of the Goldbach tables, so testing the hypothesis won't be easy. | |
Feb 4, 2011 at 6:00 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
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Feb 4, 2011 at 5:55 | comment | added | Gjergji Zaimi | If primes were as randomly distributed as I think they are, then since for $n\equiv 1\pmod{3}$ you are picking pairs from $\binom{P(2n;1,3)}{2}$, for $n\equiv 2\pmod{3}$ you are picking pairs from $\binom{P(2n;2,3)}{2}$, and for $n\equiv 0\pmod{3}$ you pick pairs from $P(2n;1,3)\times P(2n;2,3)$, you will likely have $g(3n)>g(3n+2)>g(3n+1)$. Now combining this with the standard growth conjectures on $g(n)$ (meaning, it is roughly increasing :)) this gives some "explanation". | |
Feb 4, 2011 at 5:55 | comment | added | Gjergji Zaimi | Let's denote by $\pi(x;a,b)$ the number of primes that are $a \pmod{b}$ and $\le x$ (and by $P(x;a,b)$ the set of such primes). The Chebyshev bias that David Speyer mentions says that $\pi(x;1,3)<\pi(x;2,3)$ happens with a very high logarithmic density. This is the only heuristic I've seen to explain the fractal patterns in Goldman's partition function. Though it is very likely that $\pi(x;1,3)<\pi(x;2,3)$, it is almost never likely that $2\pi(x;1,3)<\pi(x;2,3)$. | |
Feb 4, 2011 at 5:33 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
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Feb 4, 2011 at 5:27 | history | edited | Aaron Meyerowitz | CC BY-SA 2.5 |
updated information
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Feb 4, 2011 at 0:21 | comment | added | David E Speyer | Numerical data suggests that, for almost all $x$, the number of primes less than $x$ which are $2 \mod 3$ is greater than those which are $1 \mod 3$. The term to search on is "prime race"; here is a friendly introduction mathdl.maa.org/images/upload_library/22/Ford/granville1.pdf Is this effect strong enough to explain your observations? | |
Feb 3, 2011 at 23:28 | comment | added | Gerry Myerson | Multiplication by 2 yields the sequence at oeis.org/A000954 but that site doesn't discuss the problem raised here. | |
Feb 3, 2011 at 23:20 | history | asked | Aaron Meyerowitz | CC BY-SA 2.5 |