Let $a$ be a nonnegative integer. For a given positive integer $k$, let $b_1,b_2,\ldots,b_k$ be odd integers greater than $1$. Using this result, it can be shown that, for each integer $n$, $$f_{a;b_1,b_2,\ldots,b_k}(n):=n^a\,\prod_{i=1}^k\,\left(n^{b_i}-n\right)$$ is divisible by $$\Gamma\left(a;b_1,b_2,\ldots,b_k\right):=2^{\min\left\{k+a\,,\,2k+\sum\limits_{i=1}^k\,v_2\left(b_i-1\right)\right\}}\,\prod_{\substack{{p\in\mathbb{Z}_{>2}}\\p\text{ prime}}}\,p^{t_p}\,,$$ where $$t_p:={\min\left\{k+a\,,\,\sum\limits_{\substack{{1\leq i\leq k}\\{p-1|b_i-1}}}\,\big(1+v_p\left(b_i-1\right)\big)\right\}} $$ for each prime $p>2$. Here, $v_q(m)$ denotes the largest exponent $\nu\in\mathbb{Z}_{\geq0}$ such that the $\nu$-th power of the given prime $q$ divides the integer $m$.
Question. Fix $a$ and $\left(b_1,b_2,\ldots,b_k\right)$. What is the greatest common divisor $G\left(a;b_1,b_2,\ldots,b_k\right)$ of all integers of the form $f_{a;b_1,b_2,\ldots,b_k}(n)$ (i.e., $n$ runs over all the integers)? Is it equal to $\Gamma\left(a;b_1,b_2,\ldots,b_k\right)$? If not, what is a counterexample?
At least, we know that $G\left(a;b_1,b_2,\ldots,b_k\right)=\Gamma\left(a;b_1,b_2,\ldots,b_k\right)$ in the following examples.
Case $k=1$: See this link.
Case $a=0$, $k=2$, and $\left(b_1,b_2\right)=\left(13,17\right)$: See this link.
Here is another related problem.