Are there known results about $n,k,r$ such that $n$-vertex $k$-uniform $r$-regular hypergraphs exist? If this is too large a class of hypergraphs, what if $k=\tilde{\theta}(n)$? What if an extra condition holds, requiring that for any two vertices, there is at most one hyperedge containing them?

_{Here a hypergraph consists of a vertex set $V$ and a hyperedge set $\mathcal{E}$, which is a family of subsets of $V$, and '$n$-vertex' means $|V|=n$, and '$k$-unform' means each set in $\mathcal{E}$ is a $k$-subset of $V$, and '$r$-regular' means for every vertex, there are $r$ hyperedges containing it.}

This question is motivated by projective planes and affine planes over finite fields. The former are $(q^2+q+1)$-vertex $(q+1)$-uniform $(q+1)$-regular hypergraphs, for a given prime powers $q$, i.e., there, $n=q^2+q+1$, $k=q+1$, and $r=q+1$.

Are there known results about $n,k,r$ such that $n$-vertex $k$-uniform $r$-regular hypergraphs exist? If this is too large a class of hypergraphs, what if $k=\tilde{\theta}(\sqrt{n})$? What if an extra condition holds, requiring that for any two vertices, there is at most one hyperedge containing them?

_{Here a hypergraph consists of a vertex set $V$ and a hyperedge set $\mathcal{E}$, which is a family of subsets of $V$, and '$n$-vertex' means $|V|=n$, and '$k$-unform' means each set in $\mathcal{E}$ is a $k$-subset of $V$, and '$r$-regular' means for every vertex, there are $r$ hyperedges containing it.}

This question is motivated by projective planes and affine planes over finite fields. The former are $(q^2+q+1)$-vertex $(q+1)$-uniform $(q+1)$-regular hypergraphs, for a given prime powers $q$, i.e., there, $n=q^2+q+1$, $k=q+1$, and $r=q+1$.

Are there known results about $n,k,r$ such that $n$-vertex $k$-uniform $r$-regular hypergraph exist? If this is too large a class of hypergraphs, what if $k=\tilde{\theta}(n)$? What if an extra condition holds, requiring that for any two vertices, there is at most one hyperedge containing them?

_{Here a hypergraph consists of a vertex set $V$ and a hyperedge set $\mathcal{E}$, which is a family of subsets of $V$, and '$n$-vertex' means $|V|=n$, and '$k$-unform' means each set in $\mathcal{E}$ is a $k$-subset of $V$, and '$r$-regular' means for every vertex, there are $r$ hyperedges containing it.}

This question is motivated by projective planes and affine planes over finite fields. The former are $(q^2+q+1)$-vertex $(q+1)$-uniform $(q+1)$-regular hypergraphs, for a given prime powers $q$, i.e., there, $n=q^2+q+1$, $k=q+1$, and $r=q+1$.

Are there known results about $n,k,r$ such that $n$-vertex $k$-uniform $r$-regular hypergraphs exist? If this is too large a class of hypergraphs, what if $k=\tilde{\theta}(n)$? What if an extra condition holds, requiring that for any two vertices, there is at most one hyperedge containing them?

_{Here a hypergraph consists of a vertex set $V$ and a hyperedge set $\mathcal{E}$, which is a family of subsets of $V$, and '$n$-vertex' means $|V|=n$, and '$k$-unform' means each set in $\mathcal{E}$ is a $k$-subset of $V$, and '$r$-regular' means for every vertex, there are $r$ hyperedges containing it.}

This question is motivated by projective planes and affine planes over finite fields. The former are $(q^2+q+1)$-vertex $(q+1)$-uniform $(q+1)$-regular hypergraphs, for a given prime powers $q$, i.e., there, $n=q^2+q+1$, $k=q+1$, and $r=q+1$.

Are there known results about $n,k,r$ such that $n$-vertex $k$-uniform $r$-regular graphs exist? If it involves a too large class of graphs, what if $k= \tilde{\theta}(n)$? What if an extra constant holds that for any two vertices, there is at most one hyperedge containing them?

Here a hypergraph consists of a vertex set $V$ and a hyperedge set $\mathcal{E}$ which is a family of subsets of $V$. $n$-vertex means $|V|=n$. $k$-unform means each set in $\mathcal{E}$ is a $k$-subset of $V$. $r$-regular means for every vertex, there are $r$ hyperedges contain it.

And this question is motivated by projective planes and affine planes over finite fields, for the former one, it is $(q^2+q+1)$-vertex $(q+1)$-uniform $(q+1)$-regular hypergraphs for all prime powers $q$.

Are there known results about $n,k,r$ such that $n$-vertex $k$-uniform $r$-regular hypergraph exist? If this is too large a class of hypergraphs, what if $k=\tilde{\theta}(n)$? What if an extra condition holds, requiring that for any two vertices, there is at most one hyperedge containing them?

_{Here a hypergraph consists of a vertex set $V$ and a hyperedge set $\mathcal{E}$, which is a family of subsets of $V$, and '$n$-vertex' means $|V|=n$, and '$k$-unform' means each set in $\mathcal{E}$ is a $k$-subset of $V$, and '$r$-regular' means for every vertex, there are $r$ hyperedges containing it.}

This question is motivated by projective planes and affine planes over finite fields. The former are $(q^2+q+1)$-vertex $(q+1)$-uniform $(q+1)$-regular hypergraphs, for a given prime powers $q$, i.e., there, $n=q^2+q+1$, $k=q+1$, and $r=q+1$.