Goldbach's Conjecture

[science_man_88]

Originally Posted by CRGreathouse

Did you have something particular in mind?

You could get a small speedup by installing (possibly as a second OS rather than replacing your primary) some form of Linux, installing gp2c, running the script through gp2c, and hand-editing the C code to type-specialize, as well as to replace general functions with custom specific functions. But for this problem, most of the time is just spent proving primality, so there's no a whole lot to be done.

is .5*c+/- n in there maybe we can find a way to speed it up with that info.

[CRGreathouse]

Originally Posted by science_man_88

is .5*c+/- n in there maybe we can find a way to speed it up with that info.

As always, I'm open to ideas.

[science_man_88]

Originally Posted by CRGreathouse

As always, I'm open to ideas.

i can't even get my original idea to work right lol:

forstep(c=2,100,[2],for(n=1,.5*c,if(isprime(.5*c+n) && isprime(.5*c-n),print(c))))

gives a isprime error of not integer value in arithmetic function.

[3.14159]

i can't even get my original idea to work right lol:

My original idea of the prime test let all the pseudoprimes pass.

I somehow construed the Fermat test so that 11 is now occasionally marked off as a composite.

Well, a * b = 11. a =?; b = ?

Ideas: a = 5.5, b = 2.

Resolved: Set base to 2.

[science_man_88]

t=0;forstep(n=2,400000,[2],forprime(p=2,n/2,t+=isprime(n-p);if(t>0,break())))

this is a alteration of your quick code and performs to 400,000 in (94 ms I think is what i got) the other code took 1907 for a one time test if I looked up my stats properly. so that's about 2000% as fast.

the limitation is the precomputed primes here though.

I love this i tested repeatedly at 100,000 and 1,000,000 and the 100,000 always got 93-94 ms the 1000000 went as low as 891 and never to 940 so in the linear to sub-linear range(I say this because 94 overwhelmed 93 in the 100,000 test) and technically if we get around the precomputed prime thing in under about 1900% the time it takes now we can still have it faster than the other one.

[science_man_88]

Originally Posted by 3.14159

My original idea of the prime test let all the pseudoprimes pass.

I somehow construed the Fermat test so that 11 is now occasionally marked off as a composite.

Well, a * b = 11. a =?; b = ?

Ideas: a = 5.5, b = 2.

Resolved: Set base to 2.

maybe you should get a test mode where they can pick the base and how far to test up.

or maybe put a for loop around it to test one base at a time up to the given limit.

[3.14159]

maybe you should get a test mode where they can pick the base and how far to test up.

This would only be useful for the Miller-Rabin test, as the Fermat test is weak to Carmichael numbers, which satisfy the test in every base except those that are its factors or multiples/powers of its factors.

[3.14159]

Also: If 6n+1, 12n+1, and 18n+1 are prime, (6n+1)(12n+1)(18n+1) = Carmichael number.

My best guess:

(6n+1)(12n+1) = 72n^2 + 6n + 12n + 1 = 72n^2 + 18n + 1

(18n+1)(72n^2 + 18n + 1) = 1296n^3 + 72n^2 + 324n^2 + 18n + 18n + 1

That = 1296n^3 + 396n^2 + 36n + 1

Code!:

for(n=10^3,3*10^3,(1296n^3 + 396n^2 + 36n + 1),if(numdiv(1296n^3 + 396n^2 + 36n + 1) == 8,print(n))

Refined to:

b(n) = {
for(n=10^3,3*10^3,(1296n^3 + 396n^2 + 36n + 1, 
if(numdiv(1296n^3 + 396n^2 + 36n + 1) == 8,print(n)))
);
}

I need a bit of assistance here..

(13:20) gp > b(n) = {
for(n = 10^3, 3*10^3,(1296*n^3 + 396*n^2 + 36*n + 1,
if(numdiv(1296*n^3 + 396*n^2 + 36*n + 1) == 8, print(n)));
}

***   syntax error, unexpected ')', expecting KPARROW or ',': ...*n^2+36*n+1)==8,print(n)));

[science_man_88]

b(n)=for(n=10^3,3*10^3,if(numdiv(1296*n^3+396*n^2+36*n+1)==8,print(n)));

I got this to do nothing but no syntax errors lol

[3.14159]

I got this to do nothing but no syntax errors lol

That's something of an improvement. Keep working at it until PARI recognizes the command and obeys it.

[CRGreathouse]

Originally Posted by science_man_88

t=0;forstep(n=2,400000,[2],forprime(p=2,n/2,t+=isprime(n-p);if(t>0,break())))

this is a alteration of your quick code and performs to 400,000 in (94 ms I think is what i got) the other code took 1907 for a one time test if I looked up my stats properly. so that's about 2000% as fast.

This code shows that there is some Goldbach partition for some even number up to 400000, not that every even number up to 400000 has a Goldbach partition. It's faster by doing far less.