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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