Stronger Goldbach-related conjecture

  • Thread starter Thread starter rokirovka
  • Start date Start date
  • Tags Tags
    Conjecture
rokirovka
Messages
3
Reaction score
1
For any positive integer n>1, define gap(n) as

gap(n) = |n-p| = |n-q| for p,q the two primes closest to n such that p+q = 2n.

The Goldbach Conjecture is equivalent to the existence of p and q, and thus the existence of gap(n), for all n>1. (Or all n>2 if p and q are required to be odd primes, which does not affect the substance of the conjecture.)

But I make the following conjecture, which is stronger than the mere existence of gap(n) for all n>1:

For all n>1, the maximum value of gap(n)/n is gap(22)/22 = 9/22 = .40909... (for primes p=13 and q=31).

Other high values of gap(n)/n are

gap(8)/8 = 3/8 = .375 (p=5, q=11)
gap(46)/46 = 15/46 = .326... (p=31, q=61)
gap(28)/28 = 9/28 = .321... (p=19, q=37)
gap(32)/32 = 9/32 = .28125 (p=23, q=41)
gap(58)/58 = 15/58 = .259... (p=43, q=73)
gap(4)/4 = 1/4 = .25 (p=3, q=5)

Other n for which gap(n)/n > .2 include [in descending order of gap(n)/n] n=49, 25, 38, 146 [gap(146)=33 for p=113, q=179], 9, 68, 55, 14, 24, 74.
 
Physics news on Phys.org
Hello, rokirovka,
I don't know much about the subject, but I would imagine that a weaker version of your conjecture, like
If gap(n) exists, then it is not bigger than 9n/22​
would be more useful, in terms of bounding the solution space.
 
And surely it is much more likely that a large gap(n)/n exists for some n, than it is that gap(n) does not exist at all for some n.

The former is a lack of primes p,q that sum to 2n within a certain large range from about 3n/5 to 7n/5. The latter is a lack of primes p,q that sum to 2n within the entire range from 1 to 2n.

So if your weaker version of the conjecture is true, that would imply with overwhelming likelihood the truth of the stronger version as well. Of course "overwhelming likelihood" is not an acceptable form of mathematical proof. But proving your weaker version could be the right way to begin to tackle the problem.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top