Let $n$ a positive integer. We call a pair $(\pi ,C)$ composed by a permutation $\pi$$:$ {$1,2,...n$}$\rightarrow${$1,2,...,n$} and a binary function $C:$ {$1,2,...,n$}$\rightarrow${$0,1$} "revengeful" if it satisfies the two following conditions:

$1)$For every $i$ $\in$ {$1,2,...,n$}, there exist $j$ $\in$ $S_{i}=${$i, \pi(i),\pi(\pi(i)),...$} such that $C(j)=1$.

$2)$ If $C(k)=1$, then $k$ is one of  the $v_{2}(|S_{k}|)+1$ highest elements of $S_{k}$, where $v_{2}(t)$ is the highest nonnegative integer such that $2^{v_{2}(t)}$ divides $t$, for every positive integer $t$.

Let $V$ the number of revengeful pairs and $P$ the number of partitions of $n$ with all parts powers of $2$.
Determine $\frac{V}{P}$.
1. **Define the problem and notation:**
   - Let \( n \) be a positive integer.
   - A pair \((\pi, C)\) is called "revengeful" if it satisfies two conditions:
     1. For every \( i \in \{1, 2, \ldots, n\} \), there exists \( j \in S_i = \{i, \pi(i), \pi(\pi(i)), \ldots\} \) such that \( C(j) = 1 \).
     2. If \( C(k) = 1 \), then \( k \) is one of the \( v_2(|S_k|) + 1 \) highest elements of \( S_k \), where \( v_2(t) \) is the highest nonnegative integer such that \( 2^{v_2(t)} \) divides \( t \).

2. **Calculate the number of possible \( C \)'s for a given permutation \( \pi \):**
   - Suppose \( \pi \) has \( k \) cycles of lengths \( c_1, c_2, \ldots, c_k \).
   - For each cycle of length \( c_j \), there are \( 2^{v_2(c_j) + 1} - 1 \) possible \( C \)'s.
   - Therefore, the total number of possible \( C \)'s for this \( \pi \) is:
     \[
     \prod_{j=1}^k (2^{v_2(c_j) + 1} - 1)
     \]

3. **Count the number of permutations with specific cycle structures:**
   - Consider the contribution of permutations with \( n_i \) cycles of length \( i \) (where \( \sum i n_i = n \)).
   - The number of such permutations is given by:
     \[
     \frac{(\sum i n_i)!}{\prod n_i!} \prod \left(\frac{(i-1)!}{i!}\right)^{-n_i} = \frac{(\sum i n_i)!}{(\prod n_i!)(\prod i^{n_i})}
     \]

4. **Combine the contributions to find \( V \):**
   - The contribution of permutations of \( n \) elements to \( V \) is:
     \[
     n! \sum_{\sum i n_i = n} \prod (n_i!)^{-1} i^{-n_i} \prod_{i \ge 1} (2^{v_2(i) + 1} - 1)^{n_i} x^{i n_i}
     \]
   - We are computing \( F(n) = \sum V_n / n! x^n \).

5. **Generate function for \( n_1, n_2, \ldots \):**
   - Focus on the generating function:
     \[
     F(n) = \prod_{j \ge 1} \left(\sum_{n_j \ge 0} \frac{(2^{v_2(j) + 1} - 1)^{n_j} x^{j n_j} (1/j)^{n_j}}{n_j!}\right)
     \]
   - Simplify the inner sum:
     \[
     \sum_{n_j \ge 0} \frac{\left(\frac{x^j (2^{v_2(j) + 1} - 1)}{j}\right)^{n_j}}{n_j!} = \exp\left(\frac{x^j (2^{v_2(j) + 1} - 1)}{j}\right)
     \]
   - Therefore:
     \[
     F(n) = \prod_{j \ge 1} \exp\left(\frac{x^j (2^{v_2(j) + 1} - 1)}{j}\right) = \exp\left(\sum_{j \ge 1} \frac{x^j (2^{v_2(j) + 1} - 1)}{j}\right)
     \]

6. **Simplify the exponential expression:**
   - First, get rid of:
     \[
     \exp\left(\sum_{j \ge 1} \frac{x^j (-1)}{j}\right)
     \]
   - Notice:
     \[
     \int_0^t y^{j-1} dy = \frac{y^j}{j}
     \]
   - Thus:
     \[
     \exp\left(\sum_{j \ge 1} -\frac{x^j}{j}\right) = \exp\left(\sum_{j \ge 1} \int_0^x y^{j-1} dy\right) = \exp\left(\int_0^x \sum_{j \ge 1} y^{j-1} dy\right) = \exp\left(\int_0^x \frac{1}{1-y} dy\right) = \exp(-\log(1-x)) = \frac{1}{1-x}
     \]

7. **Deal with the remaining exponential term:**
   - Consider:
     \[
     \exp\left(\sum_{j \ge 1} \frac{x^j (2^{v_2(j) + 1})}{j}\right)
     \]
   - Since each \( j \) can be written as \( 2^m k \) where \( k \) is odd, rewrite the inner sum as:
     \[
     \sum_{m \ge 0} \sum_{k \text{ odd}} \frac{x^{2^m k} (2^{m+1})}{2^m k} = 2 \sum_{m \ge 0} \sum_{k \text{ odd}} \frac{x^{2^m k}}{k} = \sum_{m \ge 0} \sum_{k \ge 1} \frac{x^{2^m k} (1 - (-1)^k)}{k} = \sum_{m \ge 0} \sum_{k \ge 1} \frac{(x^{2^m})^k + (-x^{2^m})^k}{k}
     \]
   - Simplify further:
     \[
     = \sum_{m \ge 0} -\log(1 - x^{2^m}) + \log(1 + x^{2^m}) = -\log(1 - x) + \sum_{m \ge 0} \log(1 + x^{2^m}) - \log(1 - x^{2^{m+1}})
     \]
   - Therefore:
     \[
     = \exp\left(\sum_{j \ge 1} \frac{x^j (2^{v_2(j) + 1} - 1)}{j}\right) = \exp(\log(1 - x) - 2 \log(1 - x) - \sum_{m \ge 1} \log(1 - x^{2^m})) = \prod_{m \ge 0} (1 - x^{2^m})^{-1}
     \]

8. **Final simplification:**
   - We have:
     \[
     \prod_{m \ge 0} (1 - x^{2^m})^{-1} = \prod_{m \ge 0} (1 + x^{2^m} + \cdots) = \sum_{n \ge 0} P_n x^n
     \]
   - It follows that:
     \[
     \frac{V_n}{n!} = P_n
     \]

The final answer is \( \boxed{ n! } \)