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?


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


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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