Boring sequence

Since very few people solved this easy problem, I decided to reformulate it in simpler language.

Take a sequence of positive integers

$a_1, a_2, \ldots, a_i, a_{i+1}, \ldots$

with this only one rule:

$a_{i+1}$ is the number of (positive) divisors of $a_i$ .

PROBLEM IS: Determine $a_{15}$ , knowing that $a_{75} = 1$ .

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The following is the PREVIOUS formulation (that in fact ask the same thing). If $x \in \mathbb N \setminus \{0\}$ ( $x$ is a positive integer) define $D_x$ as the set of the positive divisors of $x$ . Also denote with $\vert D_x \vert$ the number of its elements (the cardinality of the set $D_x$ ).

For istance $D_6 = \{1,2,3,6\}$ , and $\vert D_6\vert = 4$ .

Define a sequence of natural positive numbers in this way. $a_0 \in \mathbb N \setminus \{0\}$

$a_{n+1} = \vert D_{a_n} \vert \quad \quad \forall \ n \geq 0$

We know that $a_{75} = 1$ .

Find $a_{15}$ .