The product of $n$ positive integers is equal to their sum, which can be expressed by the following equation：

$\prod_{i=1}^na_i=\sum_{i=1}^na_i$

For example, when $n=3$, we noticed that $(1,2,3)$ is the only natural solution. When $n=5$, there are three natural solutions: $(1,1,1,2,5), (1,1,1,3,3), (1,1,2,2,2)$. But when $n$ takes any specific value, how many sets of natural solutions are there? I found the relevant sequence in OEIS (https://oeis.org/A033178).

The number of solutions for all natural $2≤n≤100$ are listed below:

$1, 1, 1, 3, 1, 2, 2, 2, 2, 3, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 1, 5, 4, 3, 3, 5, 2, 4, 3, 5, 2, 3, 2, 6, 3, 3, 4, 7, 2, 5, 2, 4, 4, 5, 2, 5, 4, 4, 3, 7, 2, 5, 4, 5, 4, 4, 2, 9, 3, 4, 4, 7, 2, 5, 5, 4, 3, 6, 3, 9, 4, 3, 3, 6, 3, 5, 2, 7, 4, 5, 2, 10, 5, 4, 5, 8, 2, 6, 3, 6, 3, 6, 5, 6, 5, 4, 5$ (From https://oeis.org/A033178/b033178.txt)

Let's call this sequence $A_n$. I noticed that when $n$ is a prime number, it seems that the value corresponding to $A_n$ will be larger than the terms near it. You can easily discover this by comparing these two sequences: Number of multisets of n positive integers with equal sum and product and Prime number.

However, this discovery only holds for most prime numbers, and a small portion of prime numbers do not satisfy this pattern. For example, $1487$ is a prime number, but it seems that $A_{1487}$ is not larger than the nearby terms ($A_{1486}=11$, $A_{1487}=7$， $A_{1488}=9$). A more common situation is that only the neighboring term on one side is larger, such as $A_{167}$. These are relatively few coincidences, overall, when $n$ is a prime number, $A_n$ appears larger, which I would like to call a peak.

My question: Why does $A_n$ generally have a peak when $n$ is a prime number?