Goldbach's Conjecture

[science_man_88]

Okay maybe not but I made a list of things last night that we know must happen if the GC is correct.

1) all even>3 = prime + prime
2)all numbers>1 must be equidistant from 2 primes including distance=0
3)distance is opposite parity to number except number=2


but these lead to a conundrum and namely that allowing distance=0 means technically anyone could claim all numbers >1 are prime.

so again:

1) all even>7 = prime + prime
2)all numbers>3 must be equidistant from 2 primes
3) distance is opposite parity to number.

anyone care to add any more?

[3.14159]

The first even example where Goldbach's Conjecture is correct is actually 4:

4 = 2 + 2, followed by 6 = 3 + 3, etc.

This has been tested to about 10¹⁵ if I remember correctly.

Originally Posted by science_man_88

2)all numbers>3 must be equidistant from 2 primes

Have you found any counterexamples less than 10⁶? Are there any less than 10⁶?

[science_man_88]

Originally Posted by 3.14159

The first even example where Goldbach's Conjecture is correct is actually 4:

4 = 2 + 2, followed by 6 = 3 + 3, etc.

This has been tested to about 10¹⁵ if I remember correctly.



Have you found any counterexamples less than 10⁶? Are there any less than 10⁶?

I changed the rule for a reason Pi because to include those 2 I need distance = 0 but then anyone could claim that all numbers > 1 are prime with the assertion

#n; distance = 0 = #n,#n #n must be prime.

[3.14159]

Originally Posted by science_man_88

I changed the rule for a reason Pi because to include those 2 I need distance = 0 but then anyone could claim that all numbers > 1 are prime with the assertion

You could have used, "All numbers > 4 must be equidistant from two primes", while, "All even numbers greater than 2 must be the sum of two primes."

The only way these two could be invalidated is if you do not count p*2 as being valid.

[science_man_88]

Originally Posted by CRGreathouse

Notice that c and n need to be of opposite parity... that'll save a few cycles.

yeah I took that code and made it into:

for(c=2,500,if(c%2==0,forstep(n=1,c-3,2,if(isprime(c-n) && isprime(c+n),print(c","n);break())),forstep(n=0,c-3,2,if(isprime(c-n) && isprime(c+n),print(c","n);break()))))

I'll double check it works and then I'd have to check everything to satisfy anyone.

[CRGreathouse]

One step better:

forstep(c=2,500,2,forstep(n=1,c-3,2,if(isprime(c-n) && isprime(c+n),print(c","n);break));forstep(n=0,c-3,2,if(isprime(c-n) && isprime(c+n),print(c","n);break)))

The speed improvement of changing from the for/if combination to forstep is tiny, but it improves readability greatly by reducing the number of nested loops. Also I fixed a syntax error, see the red semicolon.

[science_man_88]

Originally Posted by CRGreathouse

One step better:

forstep(c=2,500,2,forstep(n=1,c-3,2,if(isprime(c-n) && isprime(c+n),print(c","n);break));forstep(n=0,c-3,2,if(isprime(c-n) && isprime(c+n),print(c","n);break)))

The speed improvement of changing from the for/if combination to forstep is tiny, but it improves readability greatly by reducing the number of nested loops. Also I fixed a syntax error, see the red semicolon.

funny the code I posted worked fine for me and in fact with one small alteration according to my timing mine with 1 alteration is faster than yours by factor 4. I'll try it on yours to see what i get.

[science_man_88]

oh wait your code checks only even c my idea checks both so mine must do about twice the work in .25* time. I can't use my alteration on yours as my alteration speeds up checks for odd c only and yours doesn't check any odd c's

[CRGreathouse]

Oh, I see what you're doing. Ignore my post, I translated it wrong. I could fix it but it's too much of a pain to do so efficiently.

[science_man_88]

Originally Posted by CRGreathouse

Oh, I see what you're doing. Ignore my post, I translated it wrong. I could fix it but it's too much of a pain to do so efficiently.

I think what you missed at first was the c+n part as it shows something above c hence c can't be the even number that are sum of 2 primes as c+n would have to sum with a negative number to get c.

[science_man_88]

(16:31) gp > (s^2-2)/(2^p-1)
%89 = (1.442695040888963407359924681*p^-1 - 0.5000000000000000000000000000 + 0.05776226504666210911810267679*p + 0.E-30*p^2 - 0.0004625342388735131662761855314*p^3 - 2.445106030 E-32*p^4 + 0.000005291094502550969612465167461*p^5 - 1.041948592 E-34*p^6 - 0.0000
0006355305751691600116677266267*p^7 + 1.248167584 E-35*p^8 + 0.0000000007710671219120984642037884552*p^9 - 1.611085877 E-37*p^10 - 9.376883229129214727836136500 E-12*p^11 + 2.649812298 E-39*p^12 + 1.140957404492010740074329500 E-13*p^13 - 2.484199030 E-41*p^14
 + O(p^15))*s^2 + (-2.885390081777926814719849362*p^-1 + 1.000000000000000000000000000 - 0.1155245300933242182362053536*p + 0.E-30*p^2 + 0.0009250684777470263325523710627*p^3 + 4.890212060 E-32*p^4 - 0.00001058218900510193922493033492*p^5 + 2.083897185 E-34*p^
6 + 0.0000001271061150338320023335453253*p^7 - 2.496335169 E-35*p^8 - 0.000000001542134243824196928407576910*p^9 + 3.222171754 E-37*p^10 + 1.875376645825842945567227300 E-11*p^11 - 5.299624596 E-39*p^12 - 2.281914808984021480148658999 E-13*p^13 + 4.96839805 E-
41*p^14 + O(p^15))

if only it was easy to calculate if it was integer lol.