From a set of N elements how many reflexive relations are out of the anti symmetric relations As I know in a set of N elements there are $2^{n^2}$ (two to the power of $n$ squared) relations, in which there are $3^{\frac12(n^2-n)}$ anti-symmetric relations.
How can I find out of those anti-symmetric relations the amount that is also reflexive?

A binary relation on a set $S$ with $N$ elements can be represented by an $(N \times N)$-table in which the cell $(a, b)$ is marked if and only if $(a, b)$ belongs to the relation (i.e., $a$ is related to $b$).
For example, if $S = \{ 1, 2, 3, 4 \}$, a simple example of binary relation on $S$ is:
\begin{array}{c||c|c|c|c|}
& 1 & 2 & 3 & 4 \\
\hline
1 & & \times & \times & \\
\hline
2 & & & & \\
\hline
3 & \times & & & \\
\hline
4 & & & & \times \\
\hline
\end{array}
This table represents the relation $\{(1, 2), (1, 3), (3, 1), (4, 4) \}$.
Since the table has $N^2$ cells and each cell can be either marked or not ($2$ options), there are $2^{N^2}$ binary relations on $S$.

A binary relation on $S$ is reflexive if $(a, a)$ belongs to the relation for any $a \in S$. This means that all the cells on the diagonal are marked (but other cells can be marked as well). For example, we can turn the relation above into a reflexive one as follows (the diagonal is highlighted):
\begin{array}{c||c|c|c|c|}
& 1 & 2 & 3 & 4 \\
\hline
1 & \color{red}\times & \times & \times & \\
\hline
2 & & \color{red}\times & & \\
\hline
3 & \times & & \color{red}\times & \\
\hline
4 & & & & \color{red}\times \\
\hline
\end{array}
Again, the table has $N^2$ cells, of which the $N$ cells on the diagonal must be marked (only $1$ option), while the remaining $N^2 - N$ cells not on the diagonal can either be marked or not ($2$ options). Therefore there are $1^N \cdot 2^{N^2 - N} = 2^{N^2 - N}$ reflexive relations on $S$.

A reflexive relation on $S$ is also antisymmetric if at most one of the pairs $(a, b)$ and $(b, a)$ belong to the relation for any distinct $a, b \in S$. This means all the cells on the diagonal are marked as before, and for $a \neq b$ there are exactly $3$ options:


*

*Neither $(a, b)$ nor $(b, a)$ are marked;

*$(a, b)$ is marked and $(b, a)$ is not marked;

*$(b, a)$ is marked and $(a, b)$ is not marked.


The reflexive relation above isn't antisymmetric because both $(1, 3)$ and $(3, 1)$ are marked, but we can turn it into an antisymmetric one by removing e.g. the mark on the cell $(1, 3)$: 
\begin{array}{c||c|c|c|c|}
& 1 & 2 & 3 & 4 \\
\hline
1 & \times & \times & & \\
\hline
2 & & \times & & \\
\hline
3 & \times & & \times & \\
\hline
4 & & & & \times \\
\hline
\end{array}
There are $N$ elements on the diagonal which must be marked ($1$ option) and there are $\binom N 2 = \frac {N (N - 1)} 2$ ways of choosing two elements $a \neq b$ for which there are the $3$ options above. Therefore there are $1^N \cdot 3^{N(N-1)/2} = 3^{N(N-1)/2}$ reflexive antisymmetric relations on $S$.
