A sequence of numbers is [i]platense[/i] 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 [i]cuboso[/i] 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 $2022$.
To determine the number of sequences cuboso with an initial term less than \(2022\), we need to identify all initial terms \(s\) such that the platense sequence starting with \(s\) contains a perfect cube greater than \(1\).

1. **Understanding the Platense Sequence**:
   - A sequence is platense if it starts with a number greater than \(1\) and each subsequent term is obtained by dividing the previous term by its smallest prime divisor, until the sequence reaches \(1\).

2. **Identifying Cuboso Sequences**:
   - A sequence is cuboso if it contains a perfect cube greater than \(1\).

3. **Characterizing Numbers in Cuboso Sequences**:
   - Let \(x = \prod_{i=1}^k p_i^{\alpha_i}\) where \(p_1 < p_2 < \cdots < p_k\) are the prime factors of \(x\).
   - We claim that \(x\) is in a cuboso sequence if and only if \(p_k \geq 3\). This is because the smallest prime factor \(p_k\) must be at least \(3\) to ensure that a perfect cube can appear in the sequence.

4. **Verification**:
   - If \(p_k \geq 3\), then \(p_k^3\) is a perfect cube and will appear in the sequence.
   - If \(p_k < 3\), then the sequence will only contain powers of \(2\), which cannot form a perfect cube greater than \(1\).

5. **Enumerating Valid Initial Terms**:
   - We need to count all numbers less than \(2022\) that have a prime factor \(p_k \geq 3\).

6. **Counting Numbers with Prime Factors \(\geq 3\)**:
   - We will consider numbers with the largest prime factor being \(3, 5, 7, 11\), etc., up to the largest prime less than \(2022\).

7. **Detailed Enumeration**:
   - For \(p_k = 11\): \(11^3 = 1331\) (only one number less than \(2022\)).
   - For \(p_k = 7\): \(7^3 = 343\), \(2 \times 7^3 = 686\), \(3 \times 7^3 = 1029\), \(2^2 \times 7^3 = 1372\), \(5 \times 7^3 = 1715\) (5 numbers).
   - For \(p_k = 5\): \(5^3 = 125\), \(2 \times 5^3 = 250\), \(3 \times 5^3 = 375\), \(2^2 \times 5^3 = 500\), \(2 \times 3 \times 5^3 = 750\), \(2^3 \times 5^3 = 1000\), \(3^2 \times 5^3 = 1125\), \(2^2 \times 3 \times 5^3 = 1500\), \(2^4 \times 5^3 = 2000\) (9 numbers).
   - For \(p_k = 3\): \(3^3 = 27\), \(2 \times 3^3 = 54\), \(2^2 \times 3^3 = 108\), \(2^3 \times 3^3 = 216\), \(2^4 \times 3^3 = 432\), \(2^5 \times 3^3 = 864\), \(2^6 \times 3^3 = 1728\) (7 numbers).
   - For \(p_k = 2\): \(2^3 = 8\), \(2^4 = 16\), \(2^5 = 32\), \(2^6 = 64\), \(2^7 = 128\), \(2^8 = 256\), \(2^9 = 512\), \(2^{10} = 1024\) (8 numbers).

8. **Summing Up**:
   - Total number of valid initial terms: \(1 + 5 + 9 + 7 + 8 = 30\).

The final answer is \(\boxed{30}\).