Sunday, April 16, 2006
A twin primes conjecture
From my prime counting function I can easily derive the following rules for twin primes, which I'll call a conjecture, as, um, I might be wrong:
Given a prime x where x > 3, if x-1 is divisible by 3, and for every prime p less than x^{1/3} but greater than 3 it is true that
x > 2 mod p
then x-2 must be prime.
I think that should work. It may be known anyway, but the key thing here is that you have this condition on primes less than x^{1/3} which is one of those interesting things, I think.
Now if only I could prove that there must exist an infinite number of x's out to positive infinity where all those conditions are met.
Oh well. Here's an example of it in action.
Let x=139. 138 is divisible by 3, so that condition is met.
The only prime less than 138^{1/3} and greater than 3 is 5, and
139 = 4 mod 5
so the last condition is met, and yup, 137 is prime.
A single counterexample ends the conjecture, of course.
The conjecture follows from my prime counting function easily enough, with a little bit of work, but I did it all in my head, so hey, I might be wrong. I don't think so, but I just thought it up in an effort to give this thing a little boost.
Can any of you re-derive it?
[A reply to someone who noticed that x = 233 is a counterexample.]
Yeah. Thanks. You get a counterexample in the following situation.
If p is the largest prime less than sqrt(x), and floor(x/p) is prime, then you have a counterexample.
With 223, the largest prime less than sqrt(223) is 13 and
floor(223/13) = 17.
That's the only way. The explanation for why I missed that is simple enough.
Oh, the full rules then are, if x is prime and
x-1 = 0 mod 3
and now the new rule—and with p the largest prime less than sqrt(x), if floor(x/p) is not prime, if x-2 is coprime to all primes less than x^{1/3} then x-2 is prime.
That is the absolute set of conditions which I can prove, so it's not a conjecture.
I did one dumb thing in the derivation which is why I missed that other rule.
Besides, now I think it's uglier. I liked it simple!!!
Given a prime x where x > 3, if x-1 is divisible by 3, and for every prime p less than x^{1/3} but greater than 3 it is true that
x > 2 mod p
then x-2 must be prime.
I think that should work. It may be known anyway, but the key thing here is that you have this condition on primes less than x^{1/3} which is one of those interesting things, I think.
Now if only I could prove that there must exist an infinite number of x's out to positive infinity where all those conditions are met.
Oh well. Here's an example of it in action.
Let x=139. 138 is divisible by 3, so that condition is met.
The only prime less than 138^{1/3} and greater than 3 is 5, and
139 = 4 mod 5
so the last condition is met, and yup, 137 is prime.
A single counterexample ends the conjecture, of course.
The conjecture follows from my prime counting function easily enough, with a little bit of work, but I did it all in my head, so hey, I might be wrong. I don't think so, but I just thought it up in an effort to give this thing a little boost.
Can any of you re-derive it?
[A reply to someone who noticed that x = 233 is a counterexample.]
Yeah. Thanks. You get a counterexample in the following situation.
If p is the largest prime less than sqrt(x), and floor(x/p) is prime, then you have a counterexample.
With 223, the largest prime less than sqrt(223) is 13 and
floor(223/13) = 17.
That's the only way. The explanation for why I missed that is simple enough.
Oh, the full rules then are, if x is prime and
x-1 = 0 mod 3
and now the new rule—and with p the largest prime less than sqrt(x), if floor(x/p) is not prime, if x-2 is coprime to all primes less than x^{1/3} then x-2 is prime.
That is the absolute set of conditions which I can prove, so it's not a conjecture.
I did one dumb thing in the derivation which is why I missed that other rule.
Besides, now I think it's uglier. I liked it simple!!!
# posted by James Harris @ 23:52 
