How to find the optimal lines? Does anyone know anyway or any algorithm that can exactly and/or
numerically find lines $\left\{ l_{i}\right\} _{i=1}^{n_{k}+2}$ that maximizes $$\min_{1\le i<j\le n_{k}+2}\text{angle}\left(l_{i},l_{j}\right)$$ in $\mathbb{R}^{n_{k}}$ for some sequence of integers $1<n_{1}<n_{2}\cdots<n_{k}<\cdots?$ Thanks a lot.
 A: Bukh and Cox provide exact constructions for the infinite sequence $n_k\equiv 1\bmod 3$. Sloane maintains numerical solutions for $n_k\leq 16$. Computer-assisted proofs of optimality have been obtained for $n_k\leq 6$; see this paper and references therein.
Edit: The Bukh–Cox construction is contained in the proofs of their Lemma 16 and Theorem 3. In what follows, I isolate the construction for convenience. (Throughout, I use their notation whenever possible, so your $n_k$ is replaced by $d$, for example.) Put $k=2$ and $N=\binom{k+1}{2}=3$, and let $A$ denote any $k\times N$ equiangular tight frame. See this survey for general information on these objects, but in this case, we may take
$$ A = \left[\begin{array}{rrr}1&-\frac{1}{2}&-\frac{1}{2}\\0&\frac{\sqrt{3}}{2}&-\frac{\sqrt{3}}{2}\end{array}\right].$$
If we denote the columns of $A$ by $x_1,x_2,x_3$, then notice that each $x_i$ has unit norm and $|\langle x_i,x_j\rangle|=\frac{1}{2}=:\beta$ for every $1\leq i<j\leq 3$. Take
$$ C:=\frac{1}{\beta}(A^\top A-I)+I. $$
Then the diagonal entries of $C$ are $1$ and the off-diagonal entries of $C$ are $\pm1$. One may further show that $\lambda:=\lambda_\mathrm{max}(C)=\frac{1}{\beta}(\frac{N}{k}-1)+1$. Since $d\equiv -k\bmod N$ by assumption, we have $b:=\frac{d+k}{N}\in\mathbb{N}$. Put $\epsilon=\frac{1}{b\lambda-1}$ and consider the matrix $G:=(1+\epsilon)I-\epsilon(C\otimes J)$, where $I$ is the identity matrix of size $Nb=d+k$ and $J$ is the all-ones matrix of size $b$. One may show that $G$ is positive semidefinite with rank at most $d$, and so $G=XX^\top$ for some $X\in\mathbb{R}^{d\times (d+k)}$. The columns of $X$ are unit vectors that achieve equality in the Bukh–Cox bound (i.e., Theorem 2(a)). It follows that the lines they span maximize the minimum interior angle. 
