# The number of admissible tuples with last element equal to $h_{k-1}$?

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.

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 : $$(0,8) ; \ (0, 2, 8) ; \ (0, 6, 8) ; \ (0, 2, 6, 8)$$

• I doubt a formula exists, but you can compute it by computing differences between terms in this sequence. Apr 21, 2020 at 11:58
• These differences (excluding zeros) are given in this sequence. Apr 21, 2020 at 12:01
• 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 !! Apr 21, 2020 at 12:14
• $a(1)=1$ because there is exactly one admissible tuple $(0)$. $a(3)=2$ because you have tuples $(0)$ and $(0,2)$. Apr 21, 2020 at 12:31
• 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 May 2, 2020 at 1:50