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Let $\omega(k)$ count how many distinct prime factors k has,

Then I can prove that for any coprime integers $a,b$ $\lim_{n\to\infty}\frac{\sum_{k=2}^n\omega(ak+b)}{\sum_{k=2}^n\omega(k)}=1$

Does anyone think this is a deep result?

Its essentially saying on average all linear polynomials with co prime coefficients

Have around the same number of prime factors

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    look at my lower sum, what you are trying to argue is incorrect2012-12-31

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This type of result, and the tools needed to prove it, would usually be discussed in a graduate level Analytic Number Theory text. My personal preference is Montgomery and Vaughn's "Multiplicative Number Theory I. Classical Theory."

Here is another way to see why this result is true. Lets examine the sum $\sum_{\begin{array}{c} n\leq x\\ n\equiv a\ (q) \end{array}}\omega(n).$ Since $\omega(n)=\sum_{p|n}1,$ the above becomes $\sum_{\begin{array}{c} n\leq x\\ n\equiv a\ (q) \end{array}}\sum_{p|n}1=\sum_{p\leq x}\sum_{\begin{array}{c} n\leq x,\ p|n\\ n\equiv a\ (q) \end{array}}1.$ This equals

$\sum_{\begin{array}{c} p\leq x\\ p\nmid q \end{array}}\sum_{\begin{array}{c} n\leq\frac{x}{p}\\ n\equiv p^{-1}a\ (q) \end{array}}1=\sum_{\begin{array}{c} p\leq x\\ p\nmid q \end{array}}\left(\frac{x}{pq}+O(1)\right) $

$=\frac{x}{q}\log\log x+\frac{B_{1}x}{q}+O\left(\frac{x}{\log x}\right),$ where $B_1$ is Mertens constant. From here, we can recover your asymptotic since $\log \log \left(\frac{x}{q}\right)\sim \log \log x$, and so $\sum_{n\leq x }\omega(nq+a)\sim\sum_{n\leq x }\omega(n)\sim x \log \log x.$

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    For Merten's Theorem in arithmetic progressions, there is a paper by Languasco and Zaccagnini. To derive what I said about $\omega_{q,a}(n)$, it suffices show that $\sum_{n\leq x}\omega_{q,a}(n)=\sum_{\begin{array}{c} p\leq x\\ p\equiv a\ (q) \end{array}}\left[\frac{x}{p}\right].$ From here, we find that $\sum_{n\leq x}\omega_{q,a}(n)=\frac{x}{\phi(q)}\log\log x+C(q,a)x+O\left(\frac{x}{\log x}\right)$ uniformely for $q\leq\left(\log x\right)^{A}.$2012-12-31