Let $n=2^{2018}$ and let $S=\{1,2,\ldots,n\}$. For subsets $S_1,S_2,\ldots,S_n\subseteq S$, we call an ordered pair $(i,j)$ [i]murine[/i] if and only if $\{i,j\}$ is a subset of at least one of $S_i, S_j$. Then, a sequence of subsets $(S_1,\ldots, S_n)$ of $S$ is called [i]tasty[/i] if and only if:

1) For all $i$, $i\in S_i$. 
2) For all $i$, $\displaystyle\bigcup_{j\in S_i} S_j=S_i$. 
3) There do not exist pairwise distinct integers $a_1,a_2,\ldots,a_k$ with $k\ge 3$ such that for each $i$, $(a_i, a_{i+1})$ is murine, where indices are taken modulo $k$. 
4) $n$ divides $1+|S_1|+|S_2|+\ldots+|S_n|$.

Find the largest integer $x$ such that $2^x$ divides the number of tasty sequences $(S_1,\ldots, S_n)$.

[i]Proposed by Vincent Huang and Brandon Wang
1. **Define the problem in terms of graph theory:**
   Consider a directed graph where you draw an edge from \(i\) to \(j\) if \(j \in S_i\). The given conditions imply that if \(i \to j\) and \(j \to k\), then \(i \to k\) as well. This is because the union of the sets \(S_j\) for \(j \in S_i\) must be equal to \(S_i\).

2. **Analyze the implications of the conditions:**
   - Condition 1: For all \(i\), \(i \in S_i\). This means each vertex has a self-loop.
   - Condition 2: For all \(i\), \(\bigcup_{j \in S_i} S_j = S_i\). This implies that the sets \(S_i\) are closed under the union operation.
   - Condition 3: There do not exist pairwise distinct integers \(a_1, a_2, \ldots, a_k\) with \(k \geq 3\) such that for each \(i\), \((a_i, a_{i+1})\) is murine, where indices are taken modulo \(k\). This means there are no directed cycles of length greater than 2.
   - Condition 4: \(n\) divides \(1 + |S_1| + |S_2| + \ldots + |S_n|\).

3. **Identify the possible structures of the graph:**
   - A single vertex.
   - An oriented tree (with \(n > 1\) vertices).
   - A cycle of length 2 (i.e., \(i \to j \to i\)).

4. **Calculate the number of edges:**
   The sum of the sizes of the sets is equal to \(n + \#(\text{edges})\), so we have \(\#(\text{edges}) = n - 1\). This means that our answer is a single oriented tree plus an arbitrary number of cycles.

5. **Count the number of tasty sequences:**
   - For a single oriented tree, the number of ways to form such a tree is given by \(2n^{n-2}\) (Cayley's formula for counting labeled trees, multiplied by 2 for orientation).
   - For cycles of length 2, we need to consider the combinations and permutations of vertices forming these cycles.

6. **Combine the results:**
   The total number of tasty sequences can be expressed as a sum of terms involving binomial coefficients and factorials:
   \[
   \binom{n}{n} \cdot 2n^{n-2} + \binom{n}{n-2} \cdot 2(n-2)^{n-4} \cdot 1!! + \binom{n}{n-4} \cdot 2(n-4)^{n-6} \cdot 3!! + \ldots + \binom{n}{4} \cdot 2(4)^2 \cdot (n-5)!! + \binom{n}{2} \cdot 2(2)^0 \cdot (n-3)!!
   \]

7. **Determine the largest power of 2 dividing the number of tasty sequences:**
   Since \(n = 2^{2018}\), we need to find the largest integer \(x\) such that \(2^x\) divides the number of tasty sequences. It is true that \(2^{2019}\) divides everything except the last expression, so our answer is \(2018\).

The final answer is \(\boxed{2018}\).