The number of the positive integer solutions of $n = \prod_{i = 1}^{k}(x_{i} + 1) - \sum_{i = 1}^{k}x_{i} - 1$ given $n$.

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 $$l_{m} = \max \{ n \mid S(n) < m\}$$

• 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 $S(5)=3$: $(1,4), (2,2), (1,1,1)$. – aschepler Apr 15 '20 at 16:57
• @aschepler Yes, you are right. Let me correct it. – TeamBright Apr 16 '20 at 4:05
• 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 $$(x_1 + 1)(x_2 + 1)\cdots (x_k + 1) - (x_1 + x_2 + \cdots + x_k) - 1 = n \iff \\ P_k(1)-a_{k-1}=n+1 \iff \\ 1+a_{k-1}+...+a_1+a_0-a_{k-1}=n+1 \iff \\ a_{k-2}+...+a_1+a_0=n$$ – rtybase Apr 17 '20 at 21:08