## The complexity of an $n\times n$ binary matrix is the number of distinct rows and columns.

For example, consider the $3\times 3$ matrices

$$ \mathbf{A} = \begin{pmatrix} 1&0&1\\0&0&0\\1&0&1\end{pmatrix} \quad

\mathbf{B} = \begin{pmatrix} 0&0&0\\0&0&0\\1&1&1\end{pmatrix} $$

$\mathbf{A}$ has complexity $2$ because the set of rows and columns is $\{000,101\}$.

$\mathbf{B}$ has complexity $3$ because the set of rows and columns is $\{000,001,111\}$.

For $0 \le k \le n^2$, let $c(n, k)$ be the minimum complexity of an $n\times n$ binary matrix with exactly $k$ ones.

Let

$$C(n) = \sum_{k=0}^{n^2} c(n, k)$$

For example, $C(2) = c(2, 0) + c(2, 1) + c(2, 2) + c(2, 3) + c(2, 4) = 1 + 2 + 2 + 2 + 1 = 8$.

You are given $C(5) = 64$, $C(10) = 274$ and $C(20) = 1150$.

Find $C(10^4)$.

### The problem seems to hinge on the concept of a binary matrix and its complexity as defined in terms of unique rows and columns. However, the problem requires a deeper understanding and involves a degree of difficulty not easily digestible for some, since it deals with a high degree of combinatorial mathematics.

Let us use a generic solution for a binary matrix size $n \times n$. When creating a binary matrix, we are placing a specified number of ones (k) within an $n \times n$ matrix that is otherwise filled with zeros. In order to minimize complexity, one would want to place the ones such that they create the fewest number of unique rows and columns.

For $k = 0$ and $k=n^2$, it is simple to place the ones in a ways that the complexity is 1, because either all entries are zero for $k=0$, and all entries are one for $k=n^2$.

For $k$ in the range $1 \leq k \leq n^2 – 1$, one can form a $n \times n$ matrix such that it at least has one row or one column filled with all ones, and the rest are filled with zeros. In order to minimize the complexity, one needs to find a way that the ones can form the most common rows or columns.

However, when we try to find a pattern or formula to directly calculate $C(n)$, we may be facing a daunting challenge due to the variety of ways to fill the ones into an $n \times n$ matrix while satisfying the complexity definition. To solve this particular problem using direct computation, such as using a computer or a programming language, might be viable for smaller $n$, like $C(5)$, $C(10)$, even $C(20)$. But, given the problem asks for $C(10^4)$, this task seems to be computationally intensive if not impossible.

One may suspect a pattern with the given examples, as $C(2^n) = 2^{n^2}$. However, using this rule to calculate $C(10^4)$ does not directly apply as 10 is not a power of 2.

Given that the question is rather complex, it seems more research or methods would be needed to accurately determine $C(10^4)$.

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