It is known that for triangle-free graphs, if they are $d$-regular, then $2d\leq n$, where $n$ is the number of vertices. In words, the degree is less than or equal to half the number of vertices (complete bipartite for $2d = n$).



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> My question is, for all $d, n$ with $2d\leq n$, can we always find a
> $d$-regular triangle-free graph on $n$ vertices? Do you know any related results in the
> literature?
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> 
> 



Existence of triangle-free graphs for regular graphs of degree at most n/2
**Yes**, it is always possible to find regular triangle-free graphs of any degree up to half the number of vertices (as long as the number of vertices is even). To see this, by Hall's Theorem the edges of $K\_{n,n}$ can be partitioned into $n$ disjoint perfect matchings. The union of $d$ of these perfect matchings is a $d$-regular bipartite graph (and hence triangle-free).


It is obviously not true if the number of vertices is odd. If $n$ and $d$ are both odd, then there are no $d$-regular graphs on $n$ vertices and hence no triangle-free ones either.


