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Given $n > 0$, consider the equation $$ \prod_{i = 1}^{k}(x_{i} + 1) - \sum_{i = 1}^{k}x_{i} - 1 = (x_1 + 1)(x_2 + 1)\cdots (x_k + 1) - (x_1 + x_2 + \cdots + x_k) - 1 = n$$ where $0 < x_{1} \leq x_{2} \leq \cdots \leq x_{k}$.

Let $S(n)$ be the number of positive integer solutions of this equation.

Here are some properties of $S(\cdot)$

  1. $S(n) \geq 1$, since $(1, n)$ ia always a solution when $k = 2$.

  2. $S(n) \geq \lceil d(n)/2 \rceil$, where $d(n)$ is the number of the factors of $n$. It is because $n = x_{1}x_{2}$ when $k = 2$. And it leads that $$\varlimsup_{n \to +\infty}S(n) = +\infty$$

  3. S(n) = 1 if $n =1, 2, 3, 5, 23$ and $S(n) = 3$ if $n = 4$.

My questions are as follow.

Q1: Determine the set $\{ n \mid S(n) = 1\}$, is it finite?

Q2: Compute $$\varliminf_{n \to +\infty}S(n) $$

Q3: If $$\lim_{n \to +\infty}S(n) = \varliminf_{n \to +\infty}S(n) = \varlimsup_{n \to +\infty}S(n) = +\infty$$ find $l_{m} = \max \{ n \mid S(n) < m\}$

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  • $\begingroup$ For the specific values in your point 3: $S(1) = \infty$ because of solutions $k=1$, $a_1$ arbitrary. But the description starts with $n>1$, so 1 should just be left out. And I get $S(5)=3$: $(1,4), (2,2), (1,1,1)$. $\endgroup$ – aschepler Apr 15 '20 at 16:57
  • $\begingroup$ @aschepler Yes, you are right. Let me correct it. $\endgroup$ – TeamBright Apr 16 '20 at 4:05
  • $\begingroup$ Something to think about $$P_k(x)=(x+x_1)(x+x_2)...(x+x_k)=\\ (x-(-x_1))(x-(-x_2))...(x-(-x_k))=\\ x^k+a_{k-1}x^{k-1}+...+a_1x+a_0,\space a_i>0$$ where $$(-x_1)+(-x_2)+...+(-x_k)=-a_{k-1}$$ as a result $$ (x_1 + 1)(x_2 + 1)\cdots (x_k + 1) - (x_1 + x_2 + \cdots + x_k) - 1 = n \iff \\ P_k(1)-a_{k-1}=n+1 \iff \\ 1+a_{k-1}+...+a_1+a_0-a_{k-1}=n+1 \iff \\ a_{k-2}+...+a_1+a_0=n$$ $\endgroup$ – rtybase Apr 17 '20 at 21:08

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