Two individuals Al and Bob go to a coffee shop every couple days. After going to the coffee shop, Al will go back to the shop in $k$ days with probability $p_k$, for $k \geq 1$. After going to the coffee shop, Bob will go back in $k$ days with probability $q_k$, for $k \geq 1$.
Suppose Al and Bob both came to the coffee shop today. What is the expected number of days until Al and Bob both come to the coffee shop on the same day again?
Context: I made this problem up. One notable case to consider is when $p_k = q_k = 0$ for $k > m$ for some $m$.
Attempt: I think the expected number of days $T$ is
$$\displaystyle\sum_{k \geq 1} \left(k \displaystyle\sum_{a_1,...,a_{j_1} \geq 1 \\ s.t. \ \ a_1 + ... + a_{j_1} = k} \ \ \ \ \ \ \displaystyle\sum_{b_1,...,b_{j_2} \geq 1 \\ s.t. \ \ b_1 + ... + b_{j_2} = k \ and \\ a_1+...+a_{l_1} \neq b_1 + ... + b_{l_2} \forall l_1 < j_1 \forall l_2 < j_2} \prod_{1 \leq i_1 \leq j_1} p_{a_{i_1}} \prod_{1 \leq i_2 \leq j_2} q_{b_{i_2}} \right).$$
Note that we need to rule out cases that they did not meet before. For example, if $k = 3, \ 1 + 1 + 1 = 2 + 1 = 3$ is not valid because then Al and Bob would have met at $k = 2,$ since $1 + 1 = 2$.
Is there any way to simplify this? If you assumed $p_k = q_k = 0$ for $k > m = 3$, then is there a nicer expression for $T$?
Maybe there is someway to write $T$ recursively, and then solve for $T$? Thank you!