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I have removed my bold claims and the naive question so I can link to this post from another.

Edit

With each prime, we construct square-free numbers that have that prime as the greatest prime factor and merge those with the square-free number that have the previous primes as factors.

Example: after we have processed $p_{2}$ our square-free numbers are:

$\{1,2,3,6\}$ where you can see that 2 is a factor of half the numbers, as is 3. Since we double the count of square-free number for each prime, the proportionality is absolute (thank-you, Euclid).

We next apply $p_{3}$ to the previous square-frees and get:

$\{1,2,3,6,5,10,15,30\}$ where you can see that the primes $2,3,5$ are each factors of $\frac{1}{2}$ of the square-free numbers. This proportionality holds to infinity.

Edit 2

The reason I asked about the critical line is because you can consider the $p_{n} \propto \textit{all-square-free-numbers} = \frac{1}{2}$ as the state of the data. It remains the same whether you are calculating the first non-trivial zero or the last. It is a constant.

Edit 3

Square-free numbers with even number of factors in the numerators and those with odd number of factors in the denominators:

$\left\{\frac{1}{2}\times\frac{6}{3}\times\frac{10}{5}\times\frac{15}{30}\right\}=1$

Because the proportionality of the primes to the square-free numbers is always $\frac{1}{2}$, this holds to $\infty$.

Coda
The series identified in Edit 3 is the infinite product equivalent to the infinite Merten's sum $= 0$. However, when we restructure the product series into the real-world sequence, we can show that neither sum nor product can converge.

$\left\{1\times\frac{1}{2}\times\frac{1}{3}\times\frac{1}{5}\times6\times\frac{1}{7}\times10\times\frac{1}{11}\times\frac{1}{13}\times14\right\}$

We can see that there will always be one or more uncancelled primes in the denominator at all times, thus no convergence.

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5  
What exactly do you mean, "1/2 of the square-free numbers"? There are infinitely many square-free numbers. Do you mean natural density? –  Alex Becker Feb 25 '13 at 23:30
    
@AlexBecker, no. I mean that this proportionality is fixed as soon as the prime is encoutered. –  Fred Kline Feb 25 '13 at 23:33
4  
I have no idea what "this proportionality is fixed as soon as the prime is encoutered" means. Perhaps you could give an example? –  Alex Becker Feb 25 '13 at 23:35
    
I do not understand the comment above, so do not understand the meaning of "$1/2$ of the natural numbers." –  André Nicolas Feb 25 '13 at 23:37
6  
Let's try it for $p=7$. The square free-numbers from $p$ upwards are 7 8 10 11 13 14 15 17 19 21 22 23 26 29 30 31 33 34 35 37 38 39 41 42 43 46 47 51 ..., where the bold ones are multiples of $7$. Doesn't look like half of them "as soon as 7 is encountered" -- so if you don't mean a limiting denstity, then what do you mean? –  Henning Makholm Feb 25 '13 at 23:42
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closed as not a real question by Jacob Black, anon, Micah, Asaf Karagila, Andres Caicedo Feb 26 '13 at 0:35

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2 Answers

up vote 5 down vote accepted

I see. So what you mean, rigorously, is for any $n\in\mathbb{N}$, let $S_n$ be the set of squarefree products of primes $p_1,\ldots,p_n$. Then the fraction of $a\in S_n$ for which $p_n$ divides $a$ is $1/2$.

Let's view this in a different way. Let's just take a set of objects $a_1,\ldots, a_n$ and choose a subset of them randomly. For each $a_i$, we have that $a_i$ can be included or not included in the set. Since these are independent, we see that the number of subsets of $S$ is $2^n$. (Note, by the way, that this is the proof for $\sum_{k=0}^n {n \choose k}=2^n$.) Out of those, $2^{n-1}$ (or exactly half) do not include $a_i$, for whichever given $a_i$ you choose.

Do you see how this relates to your problem?

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Yes, finally someone understands where I'm going with this. I break it down into even number of factors and odd number of factors to show that at $\infty$ (supposing we get there) that these two sets are equal and thus the sums net to zero. –  Fred Kline Feb 26 '13 at 0:25
    
It's actually a pretty neat observation. I don't know anything about the Riemann hypothesis though, if there is a connection somebody else will have to explain it. –  Alexander Gruber Feb 26 '13 at 0:32
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"If proved true, does anyone think this could explain why the critical line is $1/2$?"

It is true, and trivial. Perhaps someone, somewhere, thinks this could explain why the critical line is $1/2$, but it doesn't, any more than $\cos^2x=(1/2)(\cos2x+1)$ explains why the critical line is $1/2$, or $x=(1/2){-b\pm\sqrt{b^2-4ac}\over a}$ explains why the critical line is $1/2$. Read up a little on the zeta function, and you will see why the critical line is at $1/2$.

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