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This is a bit of a follow up question to this question I asked a couple days ago. The main content of that post can be phrased as asking for a nontrivial lower bound on the sum $$ \sum_{n\leq x} \Lambda(\lfloor \alpha_1 n+\beta_1\rfloor) \Lambda(\lfloor \alpha_2n+\beta_2\rfloor), $$ where $\Lambda$ is the von Mangoldt function and $\alpha_1,\alpha_2,\beta_1,\beta_2\in\mathbb{R}$ with $\alpha_1,\alpha_2$ irrational and $\alpha_1\neq \alpha_2$.

A much more general question would be to ask for asymptotic formulas (or upper/lower bounds) for sums of the form $$ \sum_{n\leq x} f(a_1(n)) f(a_2(n)), $$ where $a_1(n)$ and $a_2(n)$ are two sequences of integers with the same/similar "structure" (a term which I will intentionally leave vaguely defined). Some examples might include $$ \begin{aligned} &\sum_{n\leq x} \Lambda(a_1n+b_1)\Lambda(a_2n+b_2),\\ &\sum_{n\leq x} d(\lfloor \alpha_1 n+\beta_1\rfloor) d(\lfloor \alpha_2n+\beta_2\rfloor),\\ &\sum_{n\leq x} \varphi(\lfloor n^{c_1}\rfloor)\varphi(\lfloor n^{c_2} \rfloor), \end{aligned} $$ for instance, where $d$ and $\varphi$ denote the usual divisor and Euler totient function. One could also study such sums on average over appropriate "families" of sequences. For instance, the sums $$ \sum_{n\leq x} d(n)d(n+h) \qquad \text{and} \qquad \sum_{h\leq H} \sum_{n\leq x} d(n)d(n+h) $$ have been studied and understood asymptotically by various authors. Thus my question:

For what functions and sequences/families of sequences has the above kind of problem been studied?

The first of the three sums above is related to Dickson's Conjecture. Are there conjectures/results of this form for other arithmetic functions/sequences?

I realize this is a fairly open ended question, but I'm curious as to what research has been done on problems like these. A natural approach would be to apply the circle/delta method, so perhaps there are general results related to this method that apply to this kind of problem.

Depending on the responses, I may collect answers given in the comments into a summary answer.

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