# An equation of prime counting function

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1. Mar 26, 2015

### secondprime

I have encountered the below problem-
Given,
$z(z-1)$ has all prime < $\sqrt{z} <n$ , Prove(or disprove)-
$π(z)-w(z-1)-A= π(2z-1)- π(z)$ where A={0 ,1}, π (z) is the prime counting function, π(2z-1)- π(z) is the number of primes in between z and (2z-1), $\omega(z-1)$ is the number of distinct prime factors of (z-1).

I think the eq: is true and the proof I propose is -

Proof:
1.a. Consider (z+k) and (z-k) where 0<k<z, if both are prime or composite at the same time then, $π(z)= π(2z-1)- π(z)$
is true .

If one is prime and another is not, then the equation is not true but I claim both have to be prime when one of them is prime and k | (z-1) .

let, the assumption is false , so (z-k) is prime and (z+k) is composite.

then (z+k)=cd where (z,k)=1(if (z,k)>1 then (z-k) is not prime.)

c,d does not divide z,k since k<z and (z,k)=1

so, at least c | (z-1) implies c| k(means c divides z since z=cd+k ) and leads to a contradiction since all primes (consecutive) of n are in z, (z-1) .So, assumption is true.

1.b. Consider (z+k) and (z-k)

where 0<k<z,

if both are prime or composite at the same time then, then the equation is true, if one is prime and another is not, then the equation is not true.

Say, (z+k) is prime an (z-k) is not, then z-k=(z-1)-(k-1) is also composite, this implies (z-1)+(k-1) is also composite.

Mod note: Removed link to external site.

Thus it can be shown, that there is a “one –to -one” relation between every composite number before z and a composite after z. So, the equation π(z)= π(2z-1)- π(z) holds.

2. But the primes of (z-1) are less than z, unbalance the equation (using above argument). So, w(z-1) is subtracted.

$π(z)-w(z-1)= π(2z-1)- π(z)$

3. The prime 2 cannot be considered above way so it is ignored in the numbers of prime so,

$π(z)-w(z-1)-A= π(2z-1)- π(z)$, where A={0,1}

"Could you verify my proof, and/or point out any errors or ways I can improve it?"

Last edited by a moderator: Mar 26, 2015
2. Mar 31, 2015

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

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