TITLE: Sequence of numbers with prime factorization $pq^2$
QUESTION [34 upvotes]: I've been considering the sequence of natural numbers with prime factorization $pq^2$, $p\neq q$; it begins 12, 18, 20, 28, 44, 45, ... and is A054753 in OEIS. I have two questions:

What is the longest run (a subsequence that consists of consecutive integers)? 
The first run of length 3 is 603, 604, 605. There are another 256 runs of length 3, and no longer runs, less than $10^7$. Can it be shown that runs of length 4 and 5 are not possible?
All I've managed to work out is that runs of length 6 are impossible (because one of the numbers would have to have 6 as a factor), and that any run of length 5 needs to consist of the numbers $48k+i$ for some $k$ and $i=1 \dots 5$.
Are there more even or odd numbers in the sequence? More specifically, what is the asymptotic ratio, $\rho=\lim\limits_{N\to\infty}E_N/O_N$, of the number of even elements to the number of odd elements less than $N$.
The following two graphs suggest that $E_N$ exceeds $O_N$ for large $N$. The first plots $E_N-O_N$, and the second $E_N/O_N$, against $N$. It appears that even numbers finally gain the ascendancy for good at $N=222436$.



All I've managed to determine is that $E_N \sim\frac{N}{4 \log \left(\frac{N}{4}\right)}+\frac{\sqrt{2N}}{\log \left(\frac{N}{2}\right)}$. I'm also pretty sure than $O_N$ is also $\Theta(\frac{N}{\log N})$ and hence $\rho$ is a constant.
I'm not a number theory expert (more an amateur combinatorialist), so I've no idea how hard these questions are. Any insight would be gratefully received.

REPLY [9 votes]: Question 1:
The following approach can be used to find runs of length 4 or 5:
Any run of length 4 or more contains an element $a$ such that $a \bmod 4 \equiv 2$, so $a=2p^2$ for some prime $p$.
Since, for all prime $p>5$, $p^2\bmod120\in\{1,49\}$, we have that $a=240k+2$ or $240k+98$ for some $k$.
We also have that either $a-2$ or $a+2$ is in the run and equals $4q$ for some prime $q$. Since none of $240k/4$, $(240k+96)/4$ and $(240k+100)/4$ is prime, we know that $a=240k+2$ and that the run contains $a$, $a+1$ and $a+2$, with $a+1=9r$ or $3r^2$ for some prime $r$. Also, if $a+3$ is in the run, it equals $25s$ or $5s^2$ for some prime $s$.
For each prime $p$, with $a=2p^2$, we check the following:

Is $a\equiv2\pmod{240}$?
Is $(a+2)/4$ prime?
Is $(a+1)/9$ or $\sqrt{(a+1)/3}$ a prime?
Is $(a+3)/25$ or $\sqrt{(a+3)/5}$ a prime?
Does $a-1$ have prime decomposition of the form $uv^2$?

If all 5 tests succeed, we have a run of length 5; if tests 1-4, or tests 1-3 and 5, succeed, we have a run of length 4.
The 16 runs of length five less than $7 \times 10^{18}$ begin with the numbers:
$$
\begin{align*}

10093613546512321&=7^2\times 205992113194129 \\
14414905793929921&=7^2\times 294181750896529 \\
266667848769941521&=7^2\times 5442200995304929 \\
562672865058083521&=7^2\times 11483119695062929 \\
1579571757660876721&=7^2\times 32236158319609729 \\
1841337567664174321&=7^2\times 37578317707432129 \\
2737837351207392721&=7^2\times 55874231657293729 \\
4456162869973433521&=7^2\times 90942099387212929 \\
4683238426747860721&=7^2\times 95576294423425729 \\
4993613853242910721&=7^2\times 101910486800875729 \\
5037980611623036721&=7^2\times 102815930849449729 \\
5174116847290255921&=7^2\times 105594221373270529 \\
5344962129269790721&=23^2\times 10103898164971249 \\
5415192610051711921&=7^2\times 110514134899014529 \\
6478494344271550321&=7^2\times 132214170291256129 \\
6644601589030969921&=7^2\times 135604114061856529

\end{align*}
$$