For simple, connected graph $G$ with minimum degree $\geq k$, if $k\geq 3$, does $G$ always have a cycle of length exactly $k+1$?

Question

Let $G$ be a simple, connected graph such that $\delta(G)\geq k$ (where $\delta(G)$ is the minimum degree). If $k$ is at least $3$, does $G$ always have a cycle of length exactly $k+1$?

P.S: I feel this is somwthat an extension to this question below:

Let $G$ be a graph of minimum degree $k>1$. Show that $G$ has a cycle of length at least $k+1$

I can't construct a graph with minimum degree 3 but not having cycles of length 4. Thanks a lot if you can show one!

Answer 1

Vertices and edges of truncated tetrahedron.

Similarly for the truncated dodecahedron. Every vertex has degree 3. There are many cycles length greater than 4, but none length 4.

Answer 2

The Petersen graph is another example. It is 3-regular and has no cycle of length less than 5.

An easy construction for even $k$ is $K_{k,k}$. Every vertex has degree $k$ but since the graph is bipartite, there are no odd cycles in the graph (i.e. no cycle with length $k+1$).

For odd $n$, any $2$-connected graph satisfies your constraint.