A sequence of numbers is platense if the first number is greater than $1$, and $a_{n+1}=\frac{a_n}{p_n}$ which $p_n$ is the least prime divisor of $a_n$, and the sequence ends if $a_n=1$. For instance, the sequences $864, 432,216,108,54,27,9,3,1$ and $2022,1011,337,1$ are both sequence platense. A sequence platense is cuboso if some term is a perfect cube greater than $1$. For instance, the sequence $864$ is cuboso, because $27=3^3$, and the sequence $2022$ is not cuboso, because there is no perfect cube. Determine the number of sequences cuboso which the initial term is less than $2022$.
Problem
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Tags: number theory
youthdoo
13.12.2022 17:41
Let $S$ be the set of all numbers $s$, for which the platense sequence beginning with $s$ is cuboso. For $x=\prod_{i=1}^kp_i^{\alpha_i}$, $p_1<p_2<\cdots<p_k$, we claim that $x\in S$ if and only if $3\le p_k$. The $\textit{if}$ is easy to prove. The term $a_{(\sum_{i=1}^k\alpha_i)-2}=p_k^3$ is a cube. For the $\textit{only if}$ part, assume that we arrive at a cube $c$ in the sequence. The prime factors $p_k$ are the last to be divided. If it's the only prime factor that $c$ has, then $p_k\ge3t\ge3$. Otherwise $3\mid p_k$, so $p_k\ge3$. Then enumerate, as shown below. $$\begin{array}{|c|c|} \hline p_k&x\\\hline 11&11^3\\\hline 7&7^3,2\times7^3,3\times7^3,2^2\times7^3,5\times7^3\\\hline 5&\begin{aligned}&5^4,2\times5^4,3\times5^4,\\&5^3,2\times5^3,3\times5^3,2^2\times5^3,\\&2\times3\times5^3,2^3\times5^3,3^2\times5^3,\\&2^2\times3\times5^3,2^4\times5^3\end{aligned}\\\hline 3&\begin{aligned}&3^6,2\times3^6,3^5,2\times3^5,2^2\times3^5,2^3\times3^5,\\&3^4,2\times3^4,2^2\times3^4,2^3\times3^4,2^4\times3^4,\\&2^i\times3^3(i=0,1,2,3,4,5,6)\end{aligned}\\\hline 2&2^i(i=3,4,5,6,7,8,9,10)\\\hline \end{array}$$