**Edit:** The outline of the proof which I posted previously was hard to follow, and omitted an enormous amount of details. This has now been updated and much has been added, however some details are still omitted.

**Question 2:**

Since we know $$E_n\sim \frac{1}{4}\frac{x}{\log x},$$ we need only find an asymptotic for $\sum_{pq^2\leq x} 1$, the number of integers of the form $pq^2$ less then $x$.

**Theorem:** We have that $$\sum_{\begin{array}{c}
n\leq x\\
n=pq^{2}
\end{array}}1=\sum_{pq^2\leq x} 1 \sim \frac{x}{\log x}\sum_{p} \frac{1}{p^2}. $$

**Heuristic:** If we looked at the case where one of the primes was $2$, we got the asymptotic $\frac {1}{4} \frac{x}{\log x}$. In the same way, if one of the primes must be $3$ we get $\frac {1}{9} \frac{x}{\log x}$ and if one of the primes must be $5$ we get $\frac {1}{25} \frac{x}{\log x}$. Summing over everything then gives $\sum_p \frac{1}{p^2}$ as the constant.

*Proof:*

**Step 1:** First, we show that $$\log x\sum_{pq^2\leq x} 1\sim \sum_{pq^2\leq x} \log (pq^2).$$ Clearly $\log x\sum_{pq^2\leq x}1\geq \sum_{pq^2\leq x} \log (pq^2)$. For the other direction let $2\leq f(x)\leq x$ be some soon to be chosen function. Then $$\sum_{pq^{2}\leq x}\log(pq^{2})\geq\sum_{f(x)<pq^{2}\leq x}\log(pq^{2})\geq\log\left(f(x)\right)\sum_{f(x)<pq^{2}\leq x}1.$$ Taking $f(x)=\frac{x}{\log^{2}x}$ we see that $$\log\left(f(x)\right)\sum_{f(x)<pq^{2}\leq x}1\sim\log x\sum_{pq^{2}\leq x}1.$$ Since $\sum_{pq^2\leq x} \log (pq^2)$ is bounded above and below by something asymptotic to $\log x\sum_{pq^2\leq x} 1$ we conclude the desired asymptotic.

**Step 2:**
We prove that $$\sum_{pq^2\leq x} \log (q^2) =o(x).$$ Rearranging we have
$$\sum_{pq^{2}\leq x}\log\left(q^{2}\right)=2\sum_{p\leq x}\sum_{q\leq\sqrt{\frac{x}{p}}}\log\left(q\right).$$ Chebyshevs estimate gives $\sum_{q\leq u}\log\left(q\right)\ll u $ so that $$\sum_{pq^{2}\leq x}\log\left(q^{2}\right)\ll\sum_{p\leq x}\sqrt{\frac{x}{p}}=\sqrt{x}\sum_{p\leq x}\frac{1}{\sqrt{p}}.$$ Let $2\leq f(x)\leq x$ be some soon to be chosen function. Then splitting we see that $$\sqrt{x}\sum_{p\leq x}\frac{1}{\sqrt{p}}\leq\sqrt{x}\sum_{n\leq f(x)}\frac{1}{\sqrt{n}}+\sqrt{x}\sum_{f(x)<p\leq x}\frac{1}{\sqrt{p}}.$$ By the prime number theorem and partial summation $$\sqrt{x}\sum_{f(x)<p\leq x}\frac{1}{\sqrt{p}}=\sqrt{x}\int_{f(x)}^{x}\frac{1}{\sqrt{t}\log t}dt+O\left(xe^{-c\sqrt{\log f(x)}}\right)$$ so that if we take $f(x)=\frac{x}{\log^{2}x}$ with some computation we find $$\sqrt{x}\sum_{p\leq x}\frac{1}{\sqrt{p}}\ll\frac{x}{\log x}=o(x).$$

**Step 3:** We use the hyperbola method to show that $$\sum_{pq^2\leq x} \log (p)\sim x \sum_{p} \frac{1}{p^2}.$$ Although it is straightforward why we should have this asymptotic, the hyperbola method and splitting the sum must be used to control the error term. We have that $$\sum_{pq^{2}\leq x}\log(p)=\sum_{q^{2}\leq x}\sum_{p\leq\frac{x}{q^{2}}}\log p=\sum_{q^{2}\leq f(x)}\sum_{p\leq\frac{x}{q^{2}}}\log p+\sum_{f(x)<q^{2}\leq x}\sum_{p\leq\frac{x}{q^{2}}}\log p $$

$$=\sum_{q^{2}\leq f(x)}\sum_{p\leq\frac{x}{q^{2}}}\log p+\sum_{p\leq\frac{x}{f(x)}}\log p\sum_{f(x)<q^{2}\leq\frac{x}{p}}1.$$

By the prime number theorem $$\sum_{p\leq\frac{x}{q^{2}}}\log p=\frac{x}{q^{2}}+O\left(\frac{x}{q^{2}\log\left(\frac{x}{q^{2}}\right)}\right)=\frac{x}{q^{2}}+O\left(\frac{x}{q^{2}\log\left(\frac{x}{f(x)}\right)}\right)$$ where the last term comes from the fact that $q^{2}\leq f(x)$. Then

$$\sum_{q^{2}\leq f(x)}\sum_{p\leq\frac{x}{q^{2}}}\log p=x\sum_{q^{2}\leq f(x)}\frac{1}{q^{2}}+O\left(\frac{x}{\log\left(\frac{x}{f(x)}\right)}\sum_{q\leq f(x)}\frac{1}{q^{2}}\right)=x\sum_{q}\frac{1}{q^{2}}+O\left(\frac{x}{\log\left(\frac{x}{f(x)}\right)}+\frac{x}{\sqrt{f(x)}}\right).$$

For the other sum, by chebyshevs estimate $$\sum_{p\leq\frac{x}{f(x)}}\log p\sum_{f(x)<q^{2}\leq\frac{x}{p}}1\ll\frac{1}{\log\left(\sqrt{f(x)}\right)}\sum_{p\leq\frac{x}{f(x)}}\log p\left(\sqrt{\frac{x}{p}}\right).$$ Taking $f(x)=\frac{x}{\log x}$ we obtain $$\sum_{pq^{2}\leq x}\log(p)=x\sum_{q}\frac{1}{q^{2}}+o(x)$$ as desired.

**Combining Steps 1-3:**

By step 1 we have $$\sum_{pq^2\leq x} 1\sim \frac{1}{\log x}\sum_{pq^2\leq x} \log(pq^2) =\frac{1}{\log x}\sum_{pq^2\leq x} \log(p)+\frac{1}{\log x}\sum_{pq^2\leq x} \log(q^2).$$ Applying steps 2 and 3 to the right hand side we see that $$\sum_{pq^2\leq x} 1 \sim \frac{x}{\log x} \sum_p \frac{1}{p^2}$$ proving our asymptotic.

**Consequences:** We then have that $$O_N\sim \frac{N}{\log N} \sum_{p>2} \frac{1}{p^2}.$$ Notice this means that $E_N>O_N$, and the ratio is $$\frac{O_N}{E_N}\sim 4\sum_{p>2} \frac{1}{p^2}.$$

To keep in line with your graph above, $$\frac{E_N}{O_N}\sim \frac{1}{4\sum_{p>2} \frac{1}{p^2}}$$

**Remark about the proof:** The proof of this fact is loosely based on some ideas in E. M. Wright's 1951 paper "A Simple Proof of a Theorem of Landau." That paper looks at $\sum_{pq\leq x} 1$ and the higher products.