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NOTE I have completely reworded this because I made a complete hash of it the first time, it got worse as I added to it. I apologize to anyone who might have been confused, and hope that this will be seen as a better attempt. Also, I'm not a (proper) mathematician so my terminology can be a little unconventional in places.

I have been investigating the Goldbach Strong Conjecture and have come up against a problem for which I can't find any references.

My approach is to prove that for any number in any given range of numbers there will be at least one pair of primes equidistant from it. I am attempting to do this (and have obviously not succeeded) by looking at the longest possible sequence of composites either side of a given number, regardless of whether that sequence is practically possible or not within the given range.

For an even number, the odd composites and primes will be an odd +ve or -ve offset from it. E.g. 60 is 1 away from 61 and 3 away from 63 on one side while it is also 1 away from 59 and 3 away from 57 on the other.

An odd number will conversely be an even offset from an odd composite or a prime.

I treat the odd and even cases separately.

To give this some meaning, I will try to explain what the following images represent.

Image 1 shows 0 and +-1 (mod 3)

Image 2 shows 0 , +-1 and +-2 (mod 5),

Image 3 shows 0, +-1, +-2 and +-3 (mod 7)

Take the number 60, it is 0 (mod 3) and so it is represented by the frame of Image 1 that has a single red dot. That's because the set of odd numbers that are divisible by 3 either side of 60 can be represented by a single series of period 6 (it's 6 because the prime in question is 3, and we are only interested in odd composites).

Now take the number 62. It is 2 (mod 3) which is the same as -1 (mod 3). The odd numbers that are divisible by 3 either side of 62 cannot be represented by a single series of period 6, but can be represented by 2 such series out of phase by 1 with one another, as shown in the frame of Image 1 with 2 red dots.

The number 64 is 1 (mod 3) and will be affected by the same pair of series as 62, hence the +-1 used to denote the pair of red dots.

Similar logic follows for mod 5, except in this case we have either a single series of period 10, 2 series of period 10 that are out of phase by 1, or 2 series of period 10 that are out of phase by 2 (or 3 depending on how you look at it) as depicted by the 3 frames of Image 2.

3 Mod 0,+-1 Image 1

5 Mod 0,+-1,+-2 Image 2

enter image description here Image 3

Odd Sequence Image 4

Now for Image 4 and the reason behind my question.

I wrote a program to test the different permutations of the first n odd primes and find the longest possible sequences of composites either side of an odd or even number.

Image 4 is a visual representation of the result I got for even numbers for the first 3 odd primes. My program shows that the longest sequence occurs for even numbers that are +-1 (mod 3) and +-1 (mod 5) and +-3 (mod 7). This is shown in the first 3 rows of Image 4. Because this result is for even numbers, the red dots in the bottom row of Image 4 are showing the odd columns for which a red dot appears in at least one of the first 3 rows. The black dot is the result. This represents the first guaranteed Goldbach Pair (for a number below $11^{2}$) - well maybe that's not 100% obvious but I have established that it's true.

In case you are wondering if I've lost the plot, I will make some practical sense of the result shown.

We can use the chinese remainder theorem to find even number solutions to the modular equations represented in Image 4 (just for the sake of this example):

  • x=+-1 (mod 3)
  • x=+-1 (mod 5)
  • x=+-3 (mod 7)

One such solution (chosen because it is within the range $7^{2}/2<n<11^{2}/2$) is 46

Now testing for equidistant primes either side of 46 we find:

 {45,47} 45 is composite (Image 4 - row 4 - 1st red dot) 
 {43,49) 49 is composite (Image 4 - row 4 - 2nd red dot) 
 {41,51} 51 is composite (Image 4 - row 4 - 3rd red dot)
 {39,53} 39 is composite (Image 4 - row 4 - 4th red dot)
 {37,55} 55 is composite (Image 4 - row 4 - 5th red dot)
 {35,57} both composite  (Image 4 - row 4 - 6th red dot)
 {33,59} 33 is composite (Image 4 - row 4 - 7th red dot)
 {31,61} Bulls-eye, a Goldbach pair at 46 +-15 (Image 4 - row 4 - black dot).

This does not prove anything in itself, but it does I hope provide some meaning to Image 4.

I was able to compute that for the first 11 odd primes, the greatest position of the black square using odd columns is

3,9,15,33,75,87,105,165,213,255,339 Series 1

I also checked the even columns for the first 10 odd primes, and got

6,12,24,42,60,90,150,180,216,312 Series 2

NOTE - Image 4 is not appropriate for odd numbers because a different set of modular equations is required: x = +-1 (mod 3); x= +-2 (mod 5) ; x = +-1 (mod 7). An appropriate solution for which is 43 which is +-24 from the nearest pair of equidistant primes {19,67} as predicted by the third element in Series 2

In case anyone is interested, one set of modular equations associated with 339 (the last member I have for Series 1) are

x=+-1 (mod 3); +-2 (mod 5); +-2 (mod 7); +-5 (mod 11); +-5 (mod 13); +-2 (mod 17); +-5 (mod 19); +-4 (mod 23); +-11 (mod 29); +-14 (mod 31); +-18 (mod 37)

I could not find any results on OEIS despite trying all variations of Series 1 and Series 2 that I could think of.

My question is, does anyone have any useful references, a formula (he said, not expecting an answer because it would probably prove Goldbach's strong conjecture) or an extension to the sequence(s) Series 1 and Series 2?

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I have no idea what you mean by "a potential composite number". –  Gerry Myerson Oct 8 '12 at 11:43
    
@Gerry Myerson - Sorry to throw that one in unexplained. I really meant that it was either a composite or the prime that gave rise to it. In contrast to the black-filled square that is definitely a prime. Sometimes there is a fine line between too much detail and not enough and maybe I got on the wrong side of it. –  Alan Gee Oct 8 '12 at 14:45
    
@Gerry Myerson - Actually the black-filled square represents a Goldbach Pair, oops. –  Alan Gee Oct 8 '12 at 15:03
    
OK, I'm almost beginning to understand what you have written. In image 4, in the 1st row, you've colored 1, 2, 4, 5, 7, 8, and so on --- the numbers that are $\pm1$ modulo 3. In the second row, 1, 4, 6, 9, and so on --- $\pm1$ modulo 5. In the 3rd row, 3, 4, 10, 11, --- $\pm3$ modulo 7. And the smallest number that isn't in any of these residue classes is 15. And no other choice of residue classes gets past 15. I still don't understand what you mean by "odd columns" and/or "odd numbered columns". –  Gerry Myerson Oct 8 '12 at 22:52
    
I mean 1st 3rd 5th ... With the evens being 2nd 4th 6th ... –  Alan Gee Oct 8 '12 at 23:23

1 Answer 1

It looks like this is related to A082467, Least $k>0$ such that $n-k$ and $n+k$ are both primes. A082467(46) = 15 corresponds to the pair $(46-15, 46+15)=(31,61)$ that you mention.

You may also be interested in A129302, the worst-case numbers (or best case, if you're hoping for a counterexample).

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Thanks Charles. I haven't established quite how relevant this is yet, but I can see that it is not totally unconnected at least. I will mark this up if I get nothing better, but I'm not sure if it's too late to award you the bounty. –  Alan Gee Oct 18 '12 at 15:06
    
A082467 is certainly of interest if only for confirmation of some of the numbers in my series. –  Alan Gee Oct 18 '12 at 15:14

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