The Jacobi identity in terms of the structure constants $c_{p,q}^r$ of an $N$-dimensional Lie algebra with $p,q,r=1,\ldots,N$ reads $$ J_{i,j,k}^l \equiv c_{i,j}^m \, c_{k,m}^l + c_{j,k}^m \, c_{i,m}^l + c_{k,i}^m \, c_{j,m}^l = 0 $$ (Einstein summation convention is implied.) Since the structure constants are anti-symmetric, $c_{p,q}^r = -c_{q,p}^r$, we only need to consider the $N^2 \cdot (N-1) \cdot (N-2) \; / \; 6$ equations $J_{i,j,k}^l =0$ with $i = 1,\ldots,N \quad j=i+1,\ldots,N \quad k=j+1,\ldots,N \quad l=1,\ldots,N$.
Recently, Paiva and Teixeira (arxiv:1108.4396) argued that, if the $N\cdot(N-1)$ structure constants $c_{p=1,q}^r$ are given, the system of $N \cdot (N-1) \cdot (N-2)\;/\;2$ equations $$ J_{i=1,j,k}^l = 0 $$ with $j=2,\ldots,N \quad k=j+1,\ldots,N \quad l=1,\ldots,N$ is linear in the unknown structure constants and (provided the system $J_{i=1,j,k}^l = 0$ is not degenerate) can be solved to determine all remaining unknowns $c_{p>1,q}^r$.
Interestingly, the $N \cdot (N-1) \cdot (N-2) \cdot (N-3) \; / \; 6$ equations $$ J_{i>1,j,k}^l = 0 $$ with $i=2,\ldots,N \quad j=i+1,\ldots,N \quad k=j+1,\ldots,N \quad l=1,\ldots,N$ (which are quadratic in $c_{p,q}^r$) appear to be redundant.
Is there a simple explanation for this redundancy?