Thread: Can there EVER be more primes from n to 2n than from 0 to n?

Thank you for considering my question. It is a very simple question. But it is one I need to find the answer to for something I am working on.

The question is this:

Is it conceivably, morally, imaginably POSSIBLE for there ever to be more primes from n to 2n than from 0 to n. Like, for example, could there be more primes from 18 to 36 than from 0 to 18 (of course, we know it's not true in this particular case, but I want to generalize it.)

Yes, it seems counter-intuitive to imagine that this would be the case. Yes, it smacks in the face of our immediate reason. But can it be PROVEN? HAS it been proven? And proven by whom?

I couldn't for the life of me try to prove it myself.

It might involve assuming a zero derivative for pi(n), which might be logarithmic. But we know only that it's close to some logarithmic function n/(ln n). Not that it's in fact a logarithmic function itself (or DO we?).

This would be the last step in a very important theorem I am working on. Please keep in mind that I have no degree in anything and have no formal training in mathematics. So any gaping holes in my knowledge should be made understandable to you.

Thank you.

I don't think this is known. Your statement is roughly equivalent to asking whether . The Second Hardy Littlewood Conjecture states that (so let x = y = n), but it's not known whether the conjecture is true.

[0, 2) and [2, 4), is one corner case. Or, do you guys consider 1 to be a prime? (I heard that analysis folks do that).

Salahuddin
http://maths-on-line.blogspot.in/

Thanks, guys. Salahuddin559: even if we consider 1 a prime, it wouldn't present a problem as far as the particular thesis I am working on. So let us say we discount 1 as a prime in my question.

Originally Posted by Petek

I don't think this is known. Your statement is roughly equivalent to asking whether . The Second Hardy Littlewood Conjecture states that (so let x = y = n), but it's not known whether the conjecture is true.

Petek, is it known whether or not is itself a simple logarithmic function (i.e. one not involving trigonometric functions)? What about n/(ln n) - ? Would that difference be itself a logarithmic function?

If there ever were a case where , then if we define continuously, then there must be some point where there are exactly as many primes from 0 to x as from x to 2x, even if both that number of primes and that number x are transcendental numbers. In that case, we could easily re-solve for n/(ln n) and say to ourselves, "Okay, in this case is n times some transcendental number t divided by the logarithm of n times that transcendental number t". That is (nt)/(ln (nt)). But if we use that to solve for some number higher than x, we are never going to find that is higher from (>x) to (2 (>x)) than from 0 to (>x), unless x happens to be a certain very rarely occurring type of number.

I'll have to work this out in more detail. This will be just one more obstacle to overcome.

Thank you.

Please define what you mean by a "simple logarithmic function" and a "logarithmic function" in this context. Thanks.

Can't we use the sieve of eratosthanese here? It also says about prime number density as, on average, Product of (1 - 1/Pi). Basically, for first n numbers, the number of primes used here will be additional, since those are prime for 0 .. n, but not for n .. to 2n. Basically we have to prove if this ratio can vary more than the number of terms in it (for n and for 2n)?

Salahuddin
Maths online

Originally Posted by Petek

Please define what you mean by a "simple logarithmic function" and a "logarithmic function" in this context. Thanks.

By that I mean a function that can be expressed as any formula that you wish, as long as A) it contains the value "ln (n)" or "ln (nr)" [where "r" represents any positive real number], and (b) there is no zero derivative anywhere in that formula.

Edit: Or, rather, a zero cannot be DERIVED anywhere in that formula upon a first derivative.

Salahuddin559: I just realized you were absolutely right about the case of 4, if we define pi(n) continuously. This is because, if we define pi(n) continuously, then pi(4) wouldn't be 2, but 2 and some change. Perhaps 2.5 (after all, 4 is halfway between two primes). And since pi(2) is exactly one - i.e. even where we define pi(n) continously - then, pi(4) - pi(2) would be 1 and some change. Say, 1.5. Which is certainly larger than 1. But I guess the next question would be: is that the last time that happens?

Sorry, I don't have any other suggestions or comments.