I have noticed a fact when verifying the Goldbach Conjecture. Let $n$ be an even number larger than 6, we can easily write $n=i+j$, where $i$ and $j$ are both prime numbers. Now let $i\le j$, and there's no prime number $i^{'}<i$ s.t. $n-i^{'}$ is prime. Usually $i$ is much smaller than $n$. By programming, I found some $n$ and $i$. Here are them:

$n(i):6(3), 12(5), 30(7), 98(19), 220(23), 308(31), 556(47), 992(73), 2642(103), 5372(139),7426(173), 43532(211), 54244(233), 63274(293), 113672(313), 128168(331), 194428(359), 194470(383) ,413572(389), 503222(523), 1077422(601), 3526958(727), 3807404(751), 10759922(829), 24106882(929), 27789878(997), 37998938(1039), 60119912(1093) ,113632822(1163), 187852862(1321), 335070838(1427), 419911924(1583) ,721013438(1789), 1847133842(1861).$

All the $n$ and $i$ are the smallest ones. I'm just curious about how fast $n$ increases when $i$ increases.(Under $2\times 10^9$,there's no $n$ for which $i$ is larger than $2000$) .Any comments will be appreciated.

  • $\begingroup$ Try the random model for the primes : $X_n = 1_{n \text{ is prime}}$ is a sequence of independent random variables with $P[X_n = 1] = \frac{1}{\log n},P[X_n = 0] = 1-\frac{1}{\log n}$ $\endgroup$ – reuns Oct 10 '17 at 11:42
  • $\begingroup$ Maybe (if such is your itention) you can add the tag (experimental-mathematics), and from my viewpoint always the tag (prime-numbers). Isn't required a response of this comment. $\endgroup$ – user243301 Oct 10 '17 at 11:54

COMMENT.-It is clear that your perspective is the same as solve the equation

$$2n-X=i \space \space \text{where }X\text{ and }\space \space i\text{ are prime and } n, i\space \space \text{are both minimum } $$

I think your problem is very difficult, more than the usual (difficult) problems about prime numbers. What you have on hand is only "brute force" with the limits of computation that this implies. Am I wrong?

  • $\begingroup$ Yes, I don't have any other ideas except the brute force with computer. This problem may just be a problem forever. Anyway, thanks. $\endgroup$ – Sinh Oct 10 '17 at 13:19
  • $\begingroup$ It was only a modest opinion. However the equation I wrote was after almost sending you a comment about why you do not consider cases like $8 (3), 12(5) $ or $24(5) $. With the equation well established there is no place for these misunderstandings that I almost commit. $\endgroup$ – Piquito Oct 10 '17 at 13:56
  • 3
    $\begingroup$ By the way, what do you think about this suggestive formulation of Golbach's conjecture I could never read in any book: "All integer greater or equal to $ 4 $ is equidistant of two primes"? ($2n=p+q\iff n-p=q-n$) $\endgroup$ – Piquito Oct 10 '17 at 14:09

i will never be a divisor of n. so for 210 for example, i can't be 2,3,5, or 7. If n is the even number, then n=2m and i < m, i is conditional on m as follows:

  • if m is not a multiple of three, take it's remainder on division by 3, i will have that same remainder, or be 3.

you could repeat this with any modular arithmetic mod a prime.


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