# Boring sequence

Algebra Level pending

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 $$a_{75} = 1$$.

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 $$a_{15}$$.

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