Let be the largest subset of the tetrahedral grid:
which contains no tetrahedrons (a + r,b,c,d),(a,b + r,c,d),(a,b,c + r,d),(a,b,c,d + r) with r > 0; call such sets tetrahedron-free.
These are the currently known values of the sequence:
There are no tetrahedrons, so no removals are needed.
Removing any one point on the grid will leave the set tetrahedron-free.
Suppose the set can be tetrahedron-free in two removals. One of (2,0,0,0), (0,2,0,0), (0,0,2,0), and (0,0,0,2) must be removed. Removing any one of the four leaves three tetrahedrons to remove. However, no point coincides with all three tetrahedrons, therefore there must be more than two removals.
Three removals (for example (0,0,0,2), (1,1,0,0) and (0,0,2,0)) leaves the set tetrahedron-free with a set size of 7.
A lower bound of 2(n-1)(n-2) can be obtained by keeping all points with exactly one coordinate equal to zero.
You get a non-constructive quadratic lower bound for the quadruple problem by taking a random subset of size cn2. If c is not too large the linearity of expectation shows that the expected number of tetrahedrons in such a set is less than one, and so there must be a set of that size with no tetrahedrons. However, c = (241 / 4) / 6 + o(1 / n), which is lower than the previous lower bound.
With coordinates (a,b,c,d), consider the value a+2b+3c. This forms an arithmetic progression of length 4 for any of the tetrahedrons we are looking for. So we can take subsets of the form a+2b+3c=k, where k comes from a set with no such arithmetic progressions. [This paper, Corollary 1] gives this formula for a lower bound on the proportion of retained points: , for some absolute constant C.
An upper bound can be found by counting tetrahedrons. For a given n the tetrahedral grid has n(n+1)(n+2)(n+3)/24 tetrahedrons. Each point on the grid is part of n tetrahedrons, so (n+1)(n+2)(n+3)/24 points must be removed to remove all tetrahedrons. This gives an upper bound of (n+1)(n+2)(n+3)/8 remaining points.