I am looking for all the possible ways of gluing tetrahedra to form simply connected manifolds with the topology of a sphere.
I am interested in a function $f(n) = N$ that tells me, that $n$ number of tetrahedra can form $N$ unique "Triangulations". It is important that the configurations cannot be closed so at least $1$ free triangle has to be at the end.
As an example, let a vertex denote a tetrahedron, then the whole picture discusses a $4$-valent graph. For $n=1$ we have $1$ configuration with boundary (number of sticking out indices) $4.$, for $n=2$ we have $1$ configuration with boundary $6$, for $n = 3$ there are $2$ of them : chain with boundary $8$ and triangle with boundary $6$... And so on.
Is there any closed formula for that? If there is, i would be also interested in some distribution of the boundaries if there is any.
In dimension 3, the best known (to my knowledge) result is
Eran Nevo and Stedman Wilson, How many n-vertex triangulations does the 3 -sphere have?, 2013.
and the follow-up result in all odd dimensions:
Nevo, Eran; Santos, Francisco; Wilson, Stedman, Many triangulated odd-dimensional spheres, Math. Ann. 364, No. 3-4, 737-762 (2016). ZBL1351.52013.
In dimension 3, they constructed $\ge 2^{An^2}$ pairwise non-isomorphic triangulations of the 3-dimensional sphere $S^3$ where $n$ is the number of vertices. It is not hard to get a similar lower bound in terms of the number of tetrahedra. It is also easy to see that there is an upper bound of the form $2^{Bn^2}$.
For the treatment of even-dimensional spheres, see
Kalai, Gil, Many triangulated spheres, Discrete Comput. Geom. 3, No. 1-2, 1-14 (1988). ZBL0631.52009.