The linear recurrence is $\begin{align} v_1^{(k+1)}&:=-\frac{1}{2}v_1^{(k)}-\frac{1}{6}v_2^{(k)}\cdots-\frac{1}{n(n-1)}v_n^{(k)}\\ v_2^{(k+1)}&:=+\frac{1}{1}v_1^{(k)}\\ v_3^{(k+1)}&:=+\frac{1}{2}v_2^{(k)}\\ \vdots \\v_{n}^{(k+1)}&:=+\frac{1}{n-1}v_{n-1}^{(k)}\\ \end{align}$
Consider the following sequence $\begin{array} &&v_1^{(0)}=1,&v_2^{(0)}=0,&v_3^{(0)}=0,&v_4^{(0)}=0,&v_5^{(0)}=0,&\cdots,&v_n^{(0)}=0\\ &v_1^{(1)}=-\frac{1}{2},&v_2^{(1)}=1,&v_3^{(1)}=0,&v_4^{(1)}=0,&v_5^{(1)}=0,&\cdots,&v_n^{(1)}=0\\ &v_1^{(2)}=\frac{1}{12},&v_2^{(2)}=-\frac{1}{2},&v_3^{(2)}=\frac{1}{2},&v_4^{(2)}=0,&v_5^{(2)}=0,&\cdots,&v_n^{(2)}=0\\ &v_1^{(3)}=0,&v_2^{(3)}=\frac{1}{12},&v_3^{(3)}=-\frac{1}{4},&v_4^{(3)}=\frac{1}{6},&v_5^{(3)}=0,&\cdots,&v_n^{(3)}=0\\ &v_1^{(4)}=-\frac{1}{720},&v_2^{(4)}=0,&v_3^{(4)}=\frac{1}{24},&v_4^{(4)}=-\frac{1}{12},&v_5^{(4)}=\frac{1}{24},&\cdots,&v_n^{(4)}=0\\ \end{array}$
We see that $k!v_i^{(k)}$ are the coefficients of the bernoulli polynomials. These coefficients are realated to the Bernoulli Numbers by $B_n(x) = \sum_{k=0}^n{n\choose k}B_kx^{n-k}$ see wikipedia.
The system of linear equations is $\begin{array}~ {2\choose 0}B_1&+&{2\choose 1}B_2&&&&&&&=0\\ {3\choose 0}B_1&+&{3\choose 1}B_2&+&{3\choose 2}B_3&&&&&=0\\ {4\choose 0}B_1&+&{4\choose 1}B_2&+&{4\choose 2}B_3&+&{4\choose 3}B_4&&&=0\\ {5\choose 0}B_1&+&{5\choose 1}B_2&+&{5\choose 2}B_3&+&{5\choose 3}B_4&+&{5\choose 4}B_5&=0\\ \end{array}$ If $B_1:=1,B_2:=-\frac{1}{2},B_3:=\frac{1}{6},B_4:=0$ and $B_5:=-\frac{1}{30}$ (notice eq. $(34)$ on this page) the system is solved.
Is there some specific realation between the recurrence and the linear system, other than that they both produce this number sequence?