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Fix a finite collection of distinct prime numbers $(p_1, p_2, \dots, p_s)$, denote their product by $N$. For a natural number $n$ let $\beta(n)$ be the number of $k$, $k\leq n$, for which $k$ and $N$ are relatively prime. Consider the quantity $$ K(p_1, p_2, \dots, p_s)=\sum_{n\geq 1}\left(\frac{\beta(n)}{n}-\theta\right)^2, $$ where $$ \theta=\frac{\beta(N)}{N}=\prod_{i=1,\dots, s}\left(1-\frac{1}{p_i}\right)=\lim_{n\to\infty}\frac{\beta(n)}{n}. $$

Now consider the natural partial order (by inclusion) on the set of all finite collections of prime numbers. It gives us a net. My question is: does $K(p_1, p_2, \dots, p_s)$ tend to infinity along this net?

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interesting. But it seems it can be stated more simply! –  user59671 Feb 12 '13 at 2:55

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Well, this is a 'possibly open problem'. What you are asking basically gives a reformulation of Riemann Hypothesis. Note that, $\frac{\beta (n)}{n} = \sum_{d|N}\mu(d)[\frac{n}{d}]$. If you are familiar with Legendre Sieve you may note that $K(p_1, p_2, \dots, p_s)=\sum_{n\geq 1}\left(\frac{\beta(n)}{n}-\theta\right)^2 = \sum_{n\geq 1}\frac{\left(\sum_{d|N}\mu(d)\left\{\frac{n}{d}\right\}\right)^2}{n^2}$. If you use formal sieve method you may be able to get a 'good bound' of the function $K(p_1,...,p_s)$. See the notes. I would like to suggest a different procedure. Introduce a Weighted $l^2$ space where $||x||=\sum_{n=1}^{\infty}\frac{x(n)^2}{n^2}$ whenever convergent. Consider the Hilbert space generated by $\left\{\gamma_n|\gamma_n(k)=\left\{\frac{k}{n}\right\},k=1,2,...\right\}$(See the paper Bagchi's Reformulation which describes the methodology in detail). Consider the vector $x_N=\sum_{d|N}\mu(d)\gamma_d$, and therefore $K(p_1,...,p_s)$ is nothing but $||x_N||^2$. From this one can have a conjecture that RH is true iff $K(p_1,...,p_s)\to 1$ as $N\to\infty$. Hope this helps.

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The OP knew that if the Riemann Hypothesis fails, then $K(p_1, \dots, p_s)$ tends to infinity. Could you please give more details concerning your assertion about the equivalence of RH and the fact that $K(p_1, \dots, p_s)\to 1$. –  limanac Jan 22 at 14:47
    
Moreover, here you ask this as a question. –  limanac Jan 22 at 14:56
    
By deduce I wanted to mean, One 'may try to prove'. Sorry for bad word choice. –  Kunnysan Jan 22 at 17:11
    
Thanks. To prove the RH, if suffices to show that $K(p_1, \dots, p_s)$ fails to tend to infinity. Actually, this was the original idea of the question. –  limanac Jan 24 at 22:01

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