### Monday, April 17, 2006

## Screwy little prime results

I have started looking at my prime counting function looking for some kind of prime result that will break the current impasse, but so far just have screwy little prime results, where I had some troubles with the details.

Like I'm currently using a minor tidbit result that given a prime x, if x-1 has 3 as a factor, then it must be true that if f is a factor of x-2, f<x^{1/3}, or f>sqrt(x), or it turns out, f may equal the largest prime less than sqrt(x).

That's what I call a screwy little prime result. It is true out to positive infinity.

That's the result I've been using.

The special case where f equals the largest prime less than sqrt(x) happens if p is the largest prime less than sqrt(x), and floor(x/p) is prime, while floor((x-3)/p) is not.

It is only then that x-2 wil have p as a factor.

So that's what I have so far. I'm looking to get something bigger, but in the meantime, the tidbit result is fun to play with, and you can wonder about how it's proven!

I do have the proof.

Oh yeah, it's possible to prove it with previously known prime counting functions! You can use Meissel's for instance, but I think it's a bit easier with mine.

Can anyone prove it? Or do you think maybe what I said is false, and someone wants to find a counterexample?

Just try. I have the proof, so knock yourselves out.

Oh goody, I figured out a way to simplify it!!!

If x mod p equals 2 then you have the special case, otherwise you do not.

So I can toss all that about floor(x/p) and floor((x-3)/p), which makes it less ugly!

It's almost not screwy and kind of a nifty result now.

Can anyone prove it?

I can.

[A reply to someone who found a counterexample, namely x = 67 and f = 5.]

Yuck. You're right.

Maybe I should give the argument—which clearly is not a proof as it doesn't work—and move from there.

Ok, here's the latest fix.

If x is prime and x-1 = 0 mod 3

then either x-2 has only two prime factors, is prime, or its factors are less than x

I was still doing the silly mistake.

The problem is that the prime p can be any prime in the interval from x^{1/3} to sqrt(x), and not just the largest prime up to sqrt(x).

I don't know if it's worth mentioning now at this point, but hey, at least I think that's it.

Like I'm currently using a minor tidbit result that given a prime x, if x-1 has 3 as a factor, then it must be true that if f is a factor of x-2, f<x^{1/3}, or f>sqrt(x), or it turns out, f may equal the largest prime less than sqrt(x).

That's what I call a screwy little prime result. It is true out to positive infinity.

That's the result I've been using.

The special case where f equals the largest prime less than sqrt(x) happens if p is the largest prime less than sqrt(x), and floor(x/p) is prime, while floor((x-3)/p) is not.

It is only then that x-2 wil have p as a factor.

So that's what I have so far. I'm looking to get something bigger, but in the meantime, the tidbit result is fun to play with, and you can wonder about how it's proven!

I do have the proof.

Oh yeah, it's possible to prove it with previously known prime counting functions! You can use Meissel's for instance, but I think it's a bit easier with mine.

Can anyone prove it? Or do you think maybe what I said is false, and someone wants to find a counterexample?

Just try. I have the proof, so knock yourselves out.

Oh goody, I figured out a way to simplify it!!!

If x mod p equals 2 then you have the special case, otherwise you do not.

So I can toss all that about floor(x/p) and floor((x-3)/p), which makes it less ugly!

It's almost not screwy and kind of a nifty result now.

Can anyone prove it?

I can.

[A reply to someone who found a counterexample, namely x = 67 and f = 5.]

Yuck. You're right.

Maybe I should give the argument—which clearly is not a proof as it doesn't work—and move from there.

Ok, here's the latest fix.

If x is prime and x-1 = 0 mod 3

then either x-2 has only two prime factors, is prime, or its factors are less than x

^{1/3}or greater than sqrt(x).I was still doing the silly mistake.

The problem is that the prime p can be any prime in the interval from x^{1/3} to sqrt(x), and not just the largest prime up to sqrt(x).

I don't know if it's worth mentioning now at this point, but hey, at least I think that's it.