Goldbach's Conjecture

[science_man_88]

also since even/2 can equal odd or even is it a fair statement to say that Goldbach's conjecture implies that both odd and even number>3 have 2 primes at equal spacing from them ? that's what I'd try to prove. as essentially that's what I find it boiling down to.

[science_man_88]

the attached file is me trying to figure more out, I see one weakness in one part of the drawing (representation if you will) and it's in the ones that have a 6 gap. however this shouldn't hinder a proof of any kind. . Basically I found that you can always find a combination ((6n+1,6n-1),(6n+1,6n+1),(6n-1,6n-1)) that you can find at equal lengths from any position on my line drawing.never mind even compressed I can't upload the file lol

[science_man_88]

cropped the image down to the bare essentials.

now it's just the probability of each turning up 2 primes lol by the way all these can be extended the reason the 6 gap ones are what i see fault in is that because they contain both gaps shown 4,2 combo or 2,4 combo that they end up leaving gaps of 6 necessarily to expand them.

[3.14159]

also since even/2 can equal odd or even is it a fair statement to say that Goldbach's conjecture implies that both odd and even number>3 have 2 primes at equal spacing from them ? that's what I'd try to prove. as essentially that's what I find it boiling down to.

So you're stating that using some integer b, a ± b is a prime number, when a is greater than 3?

4 ± 1 = 3, 5.
5 ± 2: 3, 7.
6 ± 1: 5, 7.
7 ± 4: 3, 11.

As for your attaching it to the Goldbach conjecture, (p₁ + p₂ = 2n), the decompositions may have more than one decomposition, which may affect your findings:

Ex: 164200
164200 = 17 + 164183
164200 = 53 + 164147
164200 = 83 + 164117
164200 = 107 + 164093
164200 = 149 + 164051
164200 = 227 + 163973
Etc.

I'm not sure if the different decompositions affect your findings. If you wish, you could run some tests to see whether or not they do.

[science_man_88]

Originally Posted by 3.14159

So you're stating that using some integer b, a ± b is a prime number, when a is greater than 3?

4 ± 1 = 3, 5.
5 ± 2: 3, 7.
6 ± 1: 5, 7.
7 ± 4: 3, 11.

As for your attaching it to the Goldbach conjecture, (p₁ + p₂ = 2n), the decompositions may have more than one decomposition, which may affect your findings:

Ex: 164200
164200 = 17 + 164183
164200 = 53 + 164147
164200 = 83 + 164117
164200 = 107 + 164093
164200 = 149 + 164051
164200 = 227 + 163973
Etc.

I'm not sure if the different decompositions affect your findings. If you wish, you could run some tests to see whether or not they do.

I realized my equation works on the basis of 6 if that, But it got me drawing the photo above. which basically shows my point better as you can go equal distance from the original sizes (not always just one jump) but since all of my gaps are small compared to the larger numbers you can extend them (almost all to the next number either 6n-1 or 6n+1) on both sides so you can always jump to a number that may be prime however as the numbers get bigger you have more and more chances to have 2 primes line up. you know eventually on the low side you'd have to hit a prime and on the high side unless the gap between primes has grown to x-3 or x-2 then you can safely say you will come across some primes going higher as well.

oh and by the way 3.14159 your first decomposition fits my formula lol. .5 * 164200 -17 = n then add .5*164200 and you get the other one. EDIT: on further review all your examples fit my calculation lol.

[science_man_88]

according to my pari code i was able to find 8028 matches for my formula that had primes for both. I only checked primes under .5*164200.

edit: I just ran the test to 164200 and I got 15031 matches but some are the others reversed.


a=0;forprime(x=2,164200,if(isprime(floor((.5*164200-x)+(.5*164200))),print(x","floor((.5*164200-x)+(.5*164200))));a=a+1);print(a)


there's my pari code.

I know floor() is probably wrecking it but I can't get it to work without it. anyone with an idea to help ?

I put the a=a+1 in a different place and got 2444 probably more accurate lol.

[CRGreathouse]

Originally Posted by science_man_88

I put the a=a+1 in a different place and got 2444 probably more accurate lol.

That's the right answer. My code:

t=0;forprime(p=2,164200/2,t+=isprime(164200-p));t

gives 1222 (and runs in 30 ms).

[science_man_88]

Originally Posted by CRGreathouse

That's the right answer. My code:

t=0;forprime(p=2,164200/2,t+=isprime(164200-p));t

gives 1222 (and runs in 30 ms).

so my formula comes up with all of them apparently ? cool

see CRGreathouse mine wasn't just to count the amount it was to show that the 2 primes are equal distance from .5*x and since it found all of them i could (stupidly) conclude that this works for all but I'm willing to try another should I run it or should someone else ?
the code that worked moved the a=a+1 in one set of parentheses. (sorry I realize yours seems to do the trick lol)

[science_man_88]

outside one thing (a proof of one thing) I could give a possible proof for Goldbach's conjecture.

1) take e.g 6, half of 6 is 3, 3+3 = 6 (both same distance away from center), 2+4=6(both same distance away from center), and so is 1+5, all of these are the same distance from the center so any prime breakdown should fit this pattern,

2)take e.g 4 show it has primes at equal distance from it, take 5 as the odd (and prime) example it too has primes at equal distance from it. so both are centers of a number (in fact both have to be the center of higher even numbers).

So assuming all odd and even numbers greater than 3 can be shown to fit number 2 above (part of what I don't have proofs for) every even number greater than 3 will have the property #1 and hence .5 times the number would be subject to #2. without a formula to prove #2 though I can only assume this to be a proof.

edit:assuming a distance of 0 is acceptable # 2 can work for 2 and 3 as well.

[3.14159]

see CRGreathouse mine wasn't just to count the amount it was to show that the 2 primes are equal distance from .5*x and since it found all of them i could (stupidly) conclude that this works for all but I'm willing to try another should I run it or should someone else ?

Don't fall for the pesky strong law of small numbers. Just because a few of them work, doesn't mean it works forever. Do some tests involving the larger integers.

Try doing some tests on.. 3595987409275436787584502226.

An example of a decomposition with that integer:

150323 + 3595987409275436787584351903.

Or try some tests using the integer 387682744505309687545236851526113965095246317784.

[CRGreathouse]

Originally Posted by 3.14159

Don't fall for the pesky strong law of small numbers. Just because a few of them work, doesn't mean it works forever. Do some tests involving the larger integers.

Why test larger numbers? Just prove the (very minor) result.

Originally Posted by 3.14159

Try doing some tests on.. 3595987409275436787584502226.

Since that will have (in all likelihood) over 4 x 10^23 Goldbach partitions, I doubt there's time to list them all...