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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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    $\begingroup$ I doubt a formula exists, but you can compute it by computing differences between terms in this sequence. $\endgroup$
    – Wojowu
    Apr 21, 2020 at 11:58
  • $\begingroup$ These differences (excluding zeros) are given in this sequence. $\endgroup$
    – Wojowu
    Apr 21, 2020 at 12:01
  • $\begingroup$ Thanks @Wojowu for your answer, the sequence oeis.org/A292225 not clear ! the definition of the sequence is : "a(n) gives the total number of admissible tuples starting with 0 in the interval [0, 1, ..., n-1]. " but how a(1)=1, and a(3)=2 !! $\endgroup$
    – Lagrida Yassine
    Apr 21, 2020 at 12:14
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    $\begingroup$ $a(1)=1$ because there is exactly one admissible tuple $(0)$. $a(3)=2$ because you have tuples $(0)$ and $(0,2)$. $\endgroup$
    – Wojowu
    Apr 21, 2020 at 12:31
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    $\begingroup$ The answer would depend on how many primes there are smaller than $h_{k-1}$; this already shows that no simple formula is likely to exist. An asymptotic formula might be a reasonable thing to ask for, however $\endgroup$
    – Stanley Yao Xiao
    May 2, 2020 at 1:50

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