Summing $\sum\frac{1}{i}$ with two constraints on $i$

Question

Let's fix a positive integer $n$ and consider two positive composites $a,b \le n$. We can consider the prime factorizations $a=2^{a_2}3^{a_3}\cdots p^{a_p}$ and $b=2^{b_2}3^{b_3}\cdots p^{b_p}$, where $p$ is the largest prime $\le n.$

I am trying to sum $\displaystyle\sum\frac{1}{i}$, over positive integers $i$ such that:

1. $i$ consists of no prime factor larger than $n$ (but $i$ may be larger than $n$); that is, $i$ is of the form $i=2^{i_2}\cdots p^{i_p}$.
2. The $i_k$'s must satisfy both of the following conditions: $$(a_2-i_2)1_{\lbrace a_2-i_2>0\rbrace}+\cdots+(a_p-i_p)1_{\lbrace a_p-i_p>0\rbrace}\ge 2,\\(b_2-i_2)1_{\lbrace b_2-i_2>0\rbrace}+\cdots+(b_p-i_p)1_{\lbrace b_p-i_p>0\rbrace}\ge 2.$$ ($1_{\lbrace>0\rbrace}$ is a characteristic function which is $1$ when the difference is positive, $0$ otherwise.)

Is it possible to determine which $i$'s satisfy (i) and (ii)? If not, is there any chance of finding naive bounds for $\displaystyle\sum\frac{1}{i}$ over such $i$?

Answer 1

$\def\peq{\mathrel{\phantom{=}}{}}$Suppose $p_1 < \cdots < p_s$ are all prime numbers no greater than $n$ and$$ a = \prod_{k = 1}^s p_k^{a_k}, \quad b = \prod_{k = 1}^s p_k^{b_k}. $$ Define\begin{align*} A_0 &= \{(j_1, \cdots,j_s) \in \mathbb{N}^s \mid j_k \geqslant a_k\ (\forall 1 \leqslant k \leqslant s)\},\\ A_1 &= \{(j_1, \cdots,j_s) \in \mathbb{N}^s \mid \exists 1 \leqslant k_0 \leqslant s,\ j_{k_0} = a_{k_0} - 1,\ j_k \geqslant a_k\ (\forall k ≠ k_0)\},\\ B_0 &= \{(j_1, \cdots,j_s) \in \mathbb{N}^s \mid j_k \geqslant b_k\ (\forall 1 \leqslant k \leqslant s)\},\\ B_1 &= \{(j_1, \cdots,j_s) \in \mathbb{N}^s \mid \exists 1 \leqslant k_0 \leqslant s,\ j_{k_0} = b_{k_0} - 1,\ j_k \geqslant b_k\ (\forall k ≠ k_0)\}, \end{align*} and $A = \mathbb{N}^s \setminus (A_0 \cup A_1)$, $B = \mathbb{N}^s \setminus (B_0 \cup B_1)$. For $j = (j_1, \cdots, j_s) \in \mathbb{N}^s$, define $\displaystyle p^j = \prod_{k = 1}^s p_k^{j_k}$, then the sum to be found is\begin{align*} \sum_{j \in A \cap B} \frac{1}{p^j} &= \sum_{j \in \mathbb{N}^s} \frac{1}{p^j} - \sum_{j \in A^c \cup B^c} \frac{1}{p^j} = \prod_{k = 1}^s \frac{p_k}{p_k - 1} - \sum_{j \in A^c \cup B^c} \frac{1}{p^j}. \end{align*}

Note that $A^c = A_0 \cup A_1$, $B^c = B_0 \cup B_1$, $A_0 \cap A_1 = \varnothing$, $B_0 \cap B_1 = \varnothing$, thus\begin{align*} \sum_{j \in A^c \cup B^c} \frac{1}{p^j} &= \sum_{j \in A^c} \frac{1}{p^j} + \sum_{j \in B^c} \frac{1}{p^j} - \sum_{j \in A^c \cap B^c} \frac{1}{p^j}\\ &= \sum_{j \in A_0} \frac{1}{p^j} + \sum_{j \in A_1} \frac{1}{p^j} + \sum_{j \in B_0} \frac{1}{p^j} + \sum_{j \in B_1} \frac{1}{p^j} - \sum_{j \in A^c \cap B^c} \frac{1}{p^j}. \end{align*} For the first four sums,$$ \sum_{j \in A_0} \frac{1}{p^j} = \prod_{k = 1}^s \sum_{m = a_k}^∞ \frac{1}{p_k^m} = \prod_{k = 1}^s \frac{1}{(p_k - 1) p_k^{a_k - 1}} = \frac{1}{a} \prod_{k = 1}^s \frac{p_k}{p_k - 1}, $$\begin{align*} \sum_{j \in A_1} \frac{1}{p^j} &= \sum_{a_k \geqslant 1} \frac{1}{p_k^{a_k - 1}} \prod_{l ≠ k} \sum_{m = a_l}^∞ \frac{1}{p_l^m} = \sum_{a_k \geqslant 1} \frac{1}{p_k^{a_k - 1}} \prod_{l ≠ k} \frac{1}{(p_l - 1) p_l^{a_l - 1}}\\ &= \sum_{a_k \geqslant 1} (p_k - 1) \prod_{l = 1}^s \frac{1}{(p_l - 1) p_l^{a_l - 1}} = \frac{1}{a} \left( \sum_{a_k \geqslant 1} (p_k - 1) \right) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right), \end{align*} and analogously,$$ \sum_{j \in B_0} \frac{1}{p^j} = \frac{1}{b} \prod_{k = 1}^s \frac{p_k}{p_k - 1},\quad \sum_{j \in B_1} \frac{1}{p^j} = \frac{1}{b} \left( \sum_{b_k \geqslant 1} (p_k - 1) \right) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right). $$

Next, note that$$ \sum_{j \in A^c \cap B^c} \frac{1}{p^j} = \sum_{j \in A_0 \cap B_0} \frac{1}{p^j} + \sum_{j \in A_1 \cap B_0} \frac{1}{p^j} + \sum_{j \in A_0 \cap B_1} \frac{1}{p^j} + \sum_{j \in A_1 \cap B_1} \frac{1}{p^j}. $$ Define $c_k = \max(a_k, b_k)$ for $1 \leqslant k \leqslant s$, then$$ \sum_{j \in A_0 \cap B_0} \frac{1}{p^j} = \prod_{k = 1}^s \sum_{m = c_k}^∞ \frac{1}{p_k^m} = \prod_{k = 1}^s \frac{1}{(p_k - 1) p_k^{c_k - 1}} - \frac{1}{[a, b]} \prod_{k = 1}^s \frac{p_k}{p_k - 1}, $$\begin{align*} \sum_{j \in A_1 \cap B_0} \frac{1}{p^j} &= \sum_{a_k - b_k \geqslant 1} \frac{1}{p_k^{a_k - 1}} \prod_{l ≠ k} \sum_{m = c_l}^∞ \frac{1}{p_l^m} = \sum_{a_k - b_k \geqslant 1} \frac{1}{p_k^{a_k - 1}} \prod_{l ≠ k} \frac{1}{(p_l - 1) p_l^{c_l - 1}}\\ &= \left( \sum_{a_k - b_k \geqslant 1} (p_k - 1) p_k^{c_k - a_k} \right) \left( \prod_{k = 1}^s \frac{1}{(p_k - 1) p_k^{c_k - 1}} \right)\\ &= \frac{1}{[a, b]} \left( \sum_{a_k - b_k \geqslant 1} (p_k - 1) p_k^{c_k - a_k} \right) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right)\\ &= \frac{1}{[a, b]} \left( \sum_{a_k - b_k \geqslant 1} (p_k - 1) \right) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right), \end{align*} $$ \sum_{j \in A_0 \cap B_1} \frac{1}{p^j} = \frac{1}{[a, b]} \left( \sum_{b_k - a_k \geqslant 1} (p_k - 1) \right) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right). $$ Since there exists no $k$ such that $a_k - b_k \geqslant 1$ and $b_k - a_k \geqslant 1$, then\begin{align*} \sum_{j \in A_1 \cap B_1} \frac{1}{p^j} &= \sum_{\substack{a_{k_1} - b_{k_1} \geqslant 1\\b_{k_2} - a_{k_2} \geqslant 1}} \frac{1}{p_{k_1}^{a_{k_1} - 1} p_{k_2}^{b_{k_2} - 1}} \prod_{l ≠ k_1, k_2} \sum_{m = c_l}^∞ \frac{1}{p_l^m}\\ &= \sum_{\substack{a_{k_1} - b_{k_1} \geqslant 1\\b_{k_2} - a_{k_2} \geqslant 1}} \frac{1}{p_{k_1}^{a_{k_1} - 1} p_{k_2}^{b_{k_2} - 1}} \prod_{l ≠ k_1, k_2} \frac{1}{(p_l - 1) p_l^{c_l - 1}}\\ &= \Biggl( \sum_{\substack{a_{k_1} - b_{k_1} \geqslant 1\\b_{k_2} - a_{k_2} \geqslant 1}} (p_{k_1} - 1)(p_{k_2} - 1) p_{k_1}^{c_{k_1} - a_{k_1}} p_{k_2}^{c_{k_2} - b_{k_2}} \Biggr) \left( \prod_{k = 1}^s \frac{1}{(p_k - 1) p_k^{c_k - 1}} \right)\\ &= \frac{1}{[a, b]} \Biggl( \sum_{\substack{a_{k_1} - b_{k_1} \geqslant 1\\b_{k_2} - a_{k_2} \geqslant 1}} (p_{k_1} - 1)(p_{k_2} - 1) \Biggr) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right)\\ &= \frac{1}{[a, b]} \left( \sum_{a_k - b_k \geqslant 1} (p_k - 1) \right) \left( \sum_{b_k - a_k \geqslant 1} (p_k - 1) \right) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right). \end{align*} Therefore, after combining terms,\begin{align*} \sum_{j \in A \cap B} \frac{1}{p^j} &= \prod_{k = 1}^s \frac{p_k}{p_k - 1}\\ &\peq -\left( \frac{1}{a} \left( \sum_{a_k \geqslant 1} (p_k - 1) + 1 \right) + \frac{1}{b} \left( \sum_{b_k \geqslant 1} (p_k - 1) + 1 \right) \right) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right)\\ &\peq + \frac{1}{[a, b]} \left( \sum_{a_k - b_k \geqslant 1} (p_k - 1) + 1 \right) \left( \sum_{b_k - a_k \geqslant 1} (p_k - 1) + 1 \right) \left( \prod_{k = 1}^s \frac{p_k}{p_k - 1} \right). \end{align*}