P. Hall gave a formula for the number of generators of $G^n$ for any finite simple group $G$. One famous example is the fact that $A\_5^{19}$ is 2-generated, but $A\_5^{20}$ is not. The question of computing $d(G^n)$ has extensively been studied, starting with the work of Wiegold. However, what we need is a bound for the direct product of different factors. More precisely, for $k\_1<k\_2<\dots<k\_r$, how large can we take the exponents $e\_i$, such that $A\_{k\_1}^{e\_1}\times\dots\times A\_{k\_r}^{e\_r}$ is still 2-generated?


The closest related result I could find is due to Kravchenko and Petrenko (Some formulas for the smallest number of
generators for finite direct sums of matrix
algebras, Theorem 2.6), where the corresponding problem for full matrix algebra is dealt with. I would assume that the answer for simple groups is similar to the one for matrix algebras. The formula $d(A\_{k\_1}^{e\_1}\times\dots\times A\_{k\_r}^{e\_r})=\max\_i d(A\_{k\_i}^{e\_i})$ might be too optimistic, but I guess that it is not far from the truth.



How many generators does a direct product of alternating groups need?
To summarize what was indicated in the comments:


* as @DerekHolt points out, $d(A\_{k\_1}^{e\_1}\times\dots\times A\_{k\_r}^{e\_r})=\max\_i d(A\_{k\_i}^{e\_i})$. This is a simple
observation: if one has $d$ generators for $A\_{k\_i}^{e\_i}$
for all $i$, this is a surjective homomorphism $\phi\_i:F\_d\to A\_{k\_i}^{e\_i}$,
where $F\_d$ denotes the free group on $d$ generators. Then the diagonal $\phi\_1\times \cdots\times \phi\_r$ will have image generating $A\_{k\_1}^{e\_1}\times\dots\times A\_{k\_r}^{e\_r}$. To see this, note
the subgroup it generates will surject each factor, and thus will
have $A\_{k\_i}^{e\_i}$ in its composition series. By uniqueness of the
composition series ([Jordan-Hölder Theorem](https://en.wikipedia.org/wiki/Composition_series#Uniqueness:_Jordan.E2.80.93H.C3.B6lder_theorem)), the group must therefore be a product of these factors.
* As you note, Philip Hall proved that the maximal number $n$ such that 
$G^n$ is $k$-generated is $h\_k(G)$, the Eulerian function, which counts the number of $k$-generating sets for $G$, divided by $|Aut(G)|$. Thus,
the problem is reduced to determining the number of $k$-generating sets for $G$ for all $k$ (assuming one knows $Aut(G)$). In the literature,
this is usually expressed as the probability that $k$ elements generate,
so is divided by $|G|^k$ to get a number $< 1$, denoted by $p\_k(G)$.
* For $A\_n$, we have $Aut(A\_n)=S\_n$ for $n\geq 7$. The [best
estimates on $p\_2(A\_n)$ appear to be](http://www.ams.org/mathscinet-getitem?mr=3391478)


$$1−\frac{1}{n}−\frac{8.8}{n^2} ≤ p\_2(A\_n)<1−\frac{1}{n}−\frac{0.93}{n^2}.$$


Putting these facts together should give the maximal known possible
value of $e\_r$ that you desire. 


