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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?

Some new observations suggest that this property exists when $n$ is simply odd. However, when we focus on the positions of record in the list, I find that a large part occurs at prime numbers (significantly greater than the proportion of prime numbers in odd numbers), indicating that there is a pattern in prime numbers that makes the term larger.

The positions of record in this list occur at: $n = 2, 5, 13, 25, 37, 41, 61, 85, 113, 181, 361, 421, 433, ...$ (Refer to https://arxiv.org/pdf/1908.03235)

Out of these $13$ numbers, $10$ are prime numbers. This indicates that $A_n$ generally tend to have a more significant peak when $n$ is a prime number. So, in my opinion, this still points to some particular prime specific pattern.

More rigorous question: Why does $A_n$ generally have a more significant peak when $n$ is a prime number?

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    $\begingroup$ The question here was cross-posted from this Math StackExchange question which was just asked yesterday. $\endgroup$ Commented yesterday
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    $\begingroup$ Looking at the values, to me it seems to be simply the fact that values for even $n$ are, usually, smaller than values for odd $n$. It's not always the case of course, but to me it suggests that the pattern is related to parity alone more than primality. $\endgroup$ Commented yesterday
  • $\begingroup$ I agree with your viewpoint. However, when we focus on the positions of record in the list, I find that a large part occurs at prime numbers (significantly greater than the proportion of prime numbers in odd numbers), indicating that there is a pattern in prime numbers that makes the term larger. Is that true? @Wojowu $\endgroup$ Commented yesterday
  • $\begingroup$ If $k$ is the number of elements different than $1$, for $k=2$ the solutions have this form: $(1,...,1,a,\frac{a+n-2}{a-1})$ which implies $n\equiv1 mod (a-1)$. So there is one solution for every distinct way $n-1$ can be expressed as a product of two numbers. In general, for any $k$, we have solutions where $n$ is not divisible by $P-1$, where $P$ is the product. $\endgroup$ Commented 19 hours ago

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