Question 27
Ana writes lists of numbers, according to the following rules. The first number of the list must be an integer greater than 1. To compute the next number, Ana finds the least prime number that divides the last written number, and divides this number by that prime. Ana repeats this procedure many times until she writes the number 1.
For example, if the first number on the list is $864$, since the least prime that divides $864$ is $2$, the next number on the list is $864 \div 2 = 432$. The complete list will be: $864$, $432$, $216$, $108$, $54$, $27$, $9$, $3$, $1$.
Instead, if the first number of the list is $2022$, the list will be $2022$, $1011$, $337$, $1$. If on the list that begins with number $N$ there is at least one perfect cube greater than $1$, we say that $N$ is *cuboso*. For example, $864$ is *cuboso*, because the number $216 = 6^3$ is on the list. Instead, $2022$ is not *cuboso*, because none of the numbers $2022$, $1011$, $337$, $1$ is a perfect cube greater than $1$.
Determine how many positive integers less than $2022$ are *cuboso*.
For example, if the first number on the list is $864$, since the least prime that divides $864$ is $2$, the next number on the list is $864 \div 2 = 432$. The complete list will be: $864$, $432$, $216$, $108$, $54$, $27$, $9$, $3$, $1$.
Instead, if the first number of the list is $2022$, the list will be $2022$, $1011$, $337$, $1$. If on the list that begins with number $N$ there is at least one perfect cube greater than $1$, we say that $N$ is *cuboso*. For example, $864$ is *cuboso*, because the number $216 = 6^3$ is on the list. Instead, $2022$ is not *cuboso*, because none of the numbers $2022$, $1011$, $337$, $1$ is a perfect cube greater than $1$.
Determine how many positive integers less than $2022$ are *cuboso*.