There are remarkable patterns of density in the graph http://upload.wikimedia.org/wikipedia/commons/7/7c/Goldbach-1000000.png plotting the number of representations of even numbers up to a million as the sum of two primes. Has anyone measured these patterns? The referenced graph was uploaded in 2006. Given the enormous data bases of primes, it would seem to be a matter of simple checking of sums to generate the graph, with a complexity that grows not much more than linearly with the size of the even number. Thus, with the growth of computer power, some advance on a million might have been achieved over the past few years. Has this been done?
4 Answers
That link reveals that a certain heuristic but extremely plausible asymptotic formula is highly accurate at least up to $10^{10}$. That information tells us what a graph over a larger range would look like. So one knows certain patterns are there even if the graph is not drawn. (Maybe someone has drawn it, I don't know).
Roughly, the expected number of pairs is about $0.66\frac{n}{{\log}^2(n)}$ for a large number which is twice a prime. Multiply that by $\prod_{p|n}\frac{p-1}{p-2}$ to get the estimate for any even n (the product over odd prime divisors of $n$).
This explains these patterns (do you see others?): There should be a lower half very roughly hitting at $10^6$ from 3460 to 6700 and then an upper half from 6920 to 13400 about twice the lower part. The lower part is for 2 or 4 mod 6 and the upper for mutiples of 6. Numbers which are or are not multiples of 5 and/or 7 should create 4 bands in each half (I can't see much beyond that). Those patterns (including with more primes taken into account)continue very faithfully as far as calculated.
added information
The name "Goldbach's Comet" is probably not useful in finding extended computations. "Goldbach partitions" might do better. As far as I can tell the state of the art is
MR1850627 (2002g:11142) Richstein, Jörg . Computing the number of Goldbach partitions up to $5\cdot 10^8$. Algorithmic number theory (Leiden, 2000), 475--490, Lecture Notes in Comput. Sci., 1838, Springer, Berlin, 2000.
That paper does include a graph for the range $500500000 \le n \le 500660160$. The author no doubt has the data and could generate other graphs, but may not see any reason to. There are a number of observations in that paper about patterns. You might be better served by looking at color coded plots over short ranges (say according to congruence class mod 30 or 210). for example a plot over a modest range colored mod 6 shows that the top is all of the $0 \mod 6$ and nothing else (of course) BUT it also shows that the very bottom boundry is heavily in favor of $2 \mod 6$ leading me to pose this question.
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$\begingroup$ Thank you for the search term "Goldbach comet", given in your edit to Gerry Myerson's answer, which does indeed lead to many sites via Google. However, I have not been able to find among them plots of the comet even as far as the original reference's $10^6$: they are mostly relatively old sites where the plot doesn't reach $10^5$. $\endgroup$ Commented Feb 2, 2011 at 21:27
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$\begingroup$ I just fixed the link. What are you looking for in a plot with a larger range? It will look "the same" only more. $\endgroup$ Commented Feb 2, 2011 at 21:49
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$\begingroup$ We would expect it to look similar over a larger range, from the Hardy--Littlewood formula. But, as far as I know, that formula is still only a conjecture. Moreover, the formula was proposed only as an asymptotically correct description, although it may be more precise than just that. The definition of the bands in the comet improves greatly as the graph is extended. By eye-balling pictures with very clear, regular, and measurable patterns, one can get ideas. I don't have any prior view on what I would expect to see in an extended graph. $\endgroup$ Commented Feb 3, 2011 at 12:10
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$\begingroup$ Ok I expanded the answer $\endgroup$ Commented Feb 3, 2011 at 21:20
Yes. See http://www.ieeta.pt/~tos/goldbach.html
EDIT: The graph is sometimes referred to as the Goldbach Comet, and you'll find numerous links by typing that phrase into your favorite search engine. http://en.wikipedia.org/wiki/Goldbach's_comet for example covers some of the same ground that Aaron does in his answer.
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$\begingroup$ Thanks for the link. Although the site doesn't consider patterns in the referenced graph, it indicates (without reference) that enough data have been collected to extend the graph to an abscissa of $10^{10}$ ─ and perhaps more. $\endgroup$ Commented Feb 1, 2011 at 16:13
Permit me to add two nice images of Goldbach's Comet: Richard Tobin's color-coded image,
"Multiples of 3 are green, multiples of 5 are red, and multiples of 7 are blue. These are combined so that multiples of 3 and 5 are yellow, multiples of 3 and 7 are cyan, multiples of 5 and 7 are magenta, and multiples of all three are light grey. Numbers that are a power of 2 times a prime are orange."
and Andrew Morris's density-based plot,
"I used color to indicate density — from multiple points landing on the same pixel. This occurs because there are 10,000,000 numbers on the x axis but the image is only 2,560 pixels wide. The color variation is automatically scaled so that each part of the spectrum from red (only one point landed on the pixel) to purple (very many points landed on the pixel) is used equally."
It is probably hard to give an exact function for the actual number of Goldbach partition. The best way to go about it is to consider the upper and lower limit cases. If we take 2k to be a Goldbach number with k>1, the number of primes in the interval [1, 2k] can be shown to be given by (1) π(2k)=2k(1/2-r).
Here r is the ratio of composite odd numbers to the Goldbach number 2k. Just to illustrate how it works: if 2k = 26, then r = 4/26 and π(26) = 9.
By (1) above we note that (2) π(2k)<=k. Since the pairs of Goldbach partition are in the interval [1,2k], the upper limit for the function, R(2k), for the number of Goldbach partitions is given by (3) R(2k)<=k/2. The Goldbach partition equations are given by: (4) p2=k+√(k^2- L) (5) p1=k-√(k^2- L The product of the two equations give (6) L =p1*p2. The sum of the two equations give the result (7) 2k =p1*p2, and the difference is given by (8) p2- p1 = 2√(k^2-L). These results mean that each Goldbach partition is associated to some semiprime p1*p2 of the form (9) p1*(2k-p1) for semiprimes the interval [1,k^2].
The number of Goldbach partitions is equal to the number of these Goldbach partition semiprimes.
Now how do we formulate a function for the number of Goldbach partitions? From the above discussion of the upper limit of R(2k), the way forward is to divided 2k by a partition number d. (10) If d = d2- d1, we can write d as
(11) d = 2√((d2+d1)^2)/4-d1*d2) = 2√(k^2- m), where k is some Goldbach integer k>=2 and m is some elastic rational number with lower limit governed by the upper limit of R(2k). Thus R(2k) can take the form (11) R(2k) = k/√k^2- m). Now since R(2k)<=k/2 it also means k^2- m>=4. But since k>=2, it means that m>=0. These results fix the lower limit of R(2k) as R(2k)>=1, meaning that the minimum number of Goldbach partitions is one and this proves Goldbach conjecture The facility I am using to make this post does not have an equation editor. I request some editing.
https://www.academia.edu/38024182/Goldbach_strong_conjecture_and_its_proof
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1$\begingroup$ Suggestions: break into short paragraphs; use TeX encoding; use complete sentences, and check grammar and punctuation. $\endgroup$ Commented Jul 20, 2019 at 18:30
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1$\begingroup$ Then after you do all that, check the mathematics. This elementary faffing around is not going to prove the conjecture. $\endgroup$ Commented Jul 20, 2019 at 22:27
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$\begingroup$ The math is fine. Just to explain to you. Consider 2k = 20. Then k^2 = 100. The Goldbach partition semiprimes less than 100 of are13*7 and 3*17. The number of partitions is also equal to the number of semiprimes for Goldbach partition in the interval [1,2k]. $\endgroup$– SamuelCommented Jul 21, 2019 at 6:06
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$\begingroup$ The math is garbage. Try submitting it to a journal, and see what happens. $\endgroup$ Commented Jul 21, 2019 at 9:24