Let $k \geq 2$ and $(h_1, h_2,\cdots,h_{k-1}) \in \mathbb{N}^{k-1}$.

Consider the $k$-tuple : $\mathcal{H}_k=(0,h_1,\cdots,h_{k-1})$ with $0<h_1<\cdots<h_{k-1}$.

The $k$-tuple $\mathcal{H}_k$ is **admissible** if and only if there is no **fixed prime number** $q$ dividing $p(p+h_1) \cdots (p+h_{k-1})$ for all $p\in\mathbb{P}, p \geq q$.

**Example 1:** $\mathcal{H}_3 = (0,2,4)$ is not admissible, because $p(p+2)(p+4)$ is always divisible by $3$ when $p\in\mathbb{P}, p \geq 3$.

**Example 2:** $\mathcal{H}_3=(0,2,6)$ is admissible.

Question:For a fixed $h_{k-1}$, can we know the exact number of admissible tuples with last element equal to $h_{k-1}$ ?

**Example 1:** Let $h_{k-1}=6$, then the admissible tuples are :
$$(0, 6); \ (0,2,6) ; \ (0, 4, 6)$$
**Example 2:** Let $h_{k-1}=8$, then the admissible tuples are :
$$(0,8) ; \ (0, 2, 8) ; \ (0, 6, 8) ; \ (0, 2, 6, 8)$$

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