Given a unit radius $n$-sphere, and a constant $c = cos(\theta)$, $0 \le \theta \le \pi$, what is the size of the largest possible set of unit vectors $U = \{u_1, u_2, ..., u_n\}$ such that $u_i \cdot u_j \le c$ for all $1 \le i < j \le n$ (i.e. the angle between every distinct pair of unit vectors is $\ge \theta$).
Here's what I've got so far. Denote the answer $s(n), n \ge 1$.
$s(1) = 2$ since there are only two unit vectors, and the angle between them is always $\pi \ge \theta$.
$s(2) = \frac{2\pi}{\theta}$ by distributing them uniformly around the unit circle.
$s(3) = (s(2) - s(1))\frac{s(2)}{s(1)} + s(1) = \frac{s(2)^2}{s(1)} - s(2) + s(1)$, because we can rotate the unit circle $\frac{s(2)}{s(1)}$ times to tile the sphere, after accounting for the duplicates at the poles introduced by $s(1)$.
Then I'm going to guess that when $n \ge 3$, $$s(n) = \frac{s(n-1)^2}{s(n-2)} - s(n-1) + s(n-2)$$
But this last step is just a "guess". And it would be nice to have a closed-form expression.
Edit: I think that the $s(3)$ case might even be completely wrong, because you could get away with rotating the unit circle a lesser amount if you also rotated it along the other axis to alternate where the points are tiled. But I have no clue how to express this in math.
When $c \le 0$, $s(n) = 2n$, by taking an orthonormal basis plus its negation.
When $0 < c < 1$, http://en.wikipedia.org/wiki/Johnson%E2%80%93Lindenstrauss_lemma gives $$s(n) \ge \left \lfloor e^{\frac{1}{8}nc^2} \right \rfloor$$
But this bound seems to only give useful answers for large $n$ and $c$.
Source: https://mathoverflow.net/questions/24864/almost-orthogonal-vectors