TITLE: The most number of points that realize only $k$ distinct distances
QUESTION [12 upvotes]: For $k \ge 1$, let $f_d(k)$ be the largest possible number of points $p_i$
in $\mathbb{R}^d$ that determine at most $k$ distinct (Euclidean) distances,
$\|p_i-p_j\|$.
Example. For points in the plane $\mathbb{R}^2$, 
$f_2(1)=3$ via an equilateral triangle, and
$f_2(2)=5$ via the regular pentagon.



It is clear that $f_d(k)$ is finite: it is not possible to "pack" an infinite
number of points into $\mathbb{R}^d$ while only determining a finite
number of point-to-point distances.

Q. What is the growth rate of a reasonable upper bound for $f_d(k)$?

I am particularly interested in $\mathbb{R}^3$. 
$f_3(1)=4$ via the regular simplex.
I am not even certain what is $f_3(2)$. Does anyone know?
Certainly $f_3(2) \ge 6$ just by placing one point immediately above the 
centroid of the pentagon.
But regardless of exact values, I would be interested in an upper bound
for $f_3(k)$.
As well as pointers to results in the literature.
This question has an Erdős-like flavor, and undoubtedly has been considered previously.
Thanks!

REPLY [8 votes]: Bannai, Bannai and Stanton proved that $f_d(k) \leq {d + k \choose k}$ in 1983. See: http://link.springer.com/article/10.1007%2FBF02579288
I don't think this bound has been improved in general. It is certainly not tight for  every value of the parameters. 
The first good bound for this function was for $k = 2$, as given by Larman Rogers and Seidel in 1977, "On Two-Distance Sets in Euclidean Space". They proved that $f_d(2) \leq (d+1)(d+4)/2$ using a nice dimension argument. This bound was later improved by Blokhuis in 1981 to $(d+1)(d+2)/2 = {d+2 \choose 2}$, which was later generalised to $f_d(k) = {d+k \choose k}$. 
This problem is also discussed in the manuscript "Linear Algebra Methods in Combinatorics" by Babai and Frankl, where you can find some of these proofs.