This is a minor twist on Legendre's conjecture, of course.
To restate: I submit that for all $n>1$, every interval $\left(n^2,(n+1)^2\right)$ contains at least one integer matching each form $p, 2p, 3p$ and $4p$, where $p$ is prime.
e.g. In the interval $(3^2,4^2)$ we have primes $11,13$; $2p$ semiprimes $10,14$; $3p$ semiprime $15$; and the $4p$ 3-almost prime $12$.
I've confirmed this through $n=500$ or so, to where it looked unlikely to fail.
To my initial surprise, the $4p$ is actually precisely equivalent to Oppermann's conjecture, which is stronger than Legendre's (and more elegant, for my money). Thus if my stated conjecture is always true for $4p$, it immediately implies it is always true for $p$; this means the $p$ is actually redundant and could be omitted from the conjecture.
I'll consider it an answer if someone provides a counterexample to my claim. Failing that, I'd like to know whether $2p$ and $3p$ are independent claims from $4p$, or if they somehow follow from it like $p$ did. I suspect this is not the case, but am not sure.
(Along the same lines, I'm curious if it follows that always having the form $ap$ is inherently a stronger statement than always having $bp$ where $a>b$. Is having $2p$ in the interval a stronger claim than Legendre and weaker than Oppermann?)
Update
Upon revisiting this, I noticed a generalization of the above which seems to hold.
Given some $a \geq 2$, for all $n\geq 2$ and for all $1 \leq k\leq \frac{3^a-1}{2}$, there is at least one prime $p$ such that
$$n^a<kp<(n+1)^a.$$
This allows us to easily extend from the squares ($a=2$), which gave us through $4p$, to:
- cubics ($a=3$) which cover through $13p$,
- quartics ($a=4$) covering through $40p$,
- quintics ($a=5$) through $121p$,
- ...and so forth. It looks very solid to me.