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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?

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Perhaps the thoughts which I followed recently give a satisfactory answer. I've looked at the "ZETA"-matrix and fiddled out the connection to the integral-representation in the Euler-MacLaurin-Formula. The article is not yet ready, needs some brushing and completing references and such. But as it might be helpful here: here is the link the integral in Euler-MacLaurin in connection with the Pascal-matrix

[update]: added the re-translation into bernoulli-numbers in the final formulae in the text

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Suppose there are $v_1,v_2,\cdots,v_n$ and an $x$ with

$$\begin{align} xv_1&=-\frac{1}{2}v_1-\frac{1}{6}v_2\cdots-\frac{1}{n(n-1)}v_n\\ xv_2&=+\frac{1}{1}v_1\\ xv_3&=+\frac{1}{2}v_2\\ \vdots\\ xv_{n}&=+\frac{1}{n-1}v_{n-1}\\ \end{align}$$ If $v_1=1$ and $N:=n+2$ then $$p_N(x):=\sum_{k=1}^{N-1}\frac{N!}{(N-k)!}x^{k-1}=0$$ solves the system. These polynomials also appear here.

Then using $$q_N(x) = \frac{x^{N-1}}{N!} p_N(1/x)$$ where $q'_N(x)$ is a partial sum of the power series for $e^x$, and $q_N(x)$ is a partial sum of the power series for $e^x-1$ (see here).
Apply the General Leibniz rule $$(fg)^{(n)}(x)=\sum_{k=0}^n {n \choose k} f^{(n-k)}(x) g^{(k)}(x)$$ to $$-xq'_N(x)f(x)=q_N(x)$$ to find out the series expansion at $0$ of $$-x \frac{q'_N(x)}{q_N(x)}=\rightarrow \frac{-x}{1-e^{-x}}.$$ This leads to a linear system as stated in the question.

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