A note on Bernoulli numbers and polynomials (Q1844250)

Put \(S_k =S_k(n) = \sum_{n=0}^{n-1} a^k\). It is well known that \(S_1^2 = S_3\), \(2S_1^4 = S_5 + S_7\). Stern showed that [\textit{P. Bachmann}, Niedere Zahlentheorie. Tell II (Teubner, Leipzig, 1910, p. 20) (reprint Chelsea, Bronx, 1968; Zbl. 253.10001)) \[ 2^{m-1} S_1^m = \sum_{2j < m} \binom{m}{2j+1} S_{2m-2j-1}. \] In terms of the Bernoulli polynomial \(B_n(x)\) this gives the identity \[ (x(x - 1))^nm =2 \sum_{2j < m}\binom{m}{2j+1} \frac{B_{2m-2j}(x) - B_{2m-2j}}{2m - 2j}. \] A slightly simpler formula is \[ (m+1) (x - \tfrac12)^m = \sum_{2j \le m} \binom{m+1}{2j+1} 2^{-2j} B_{m-2j}(x). \tag{*} \] The inverse of (*) is given by \[ B_k(x) = \sum_{s=0}^k \binom{k}{s} D_s(x - \tfrac12)^{k - \varepsilon}, \tag{**} \] where \(D_k = 2^k B_k(\tfrac12) = 2 (1 - 2^{k -1}) B_k\). These results suggest the following two theorems. I. The set of equations \(\displaystyle(m + 1)x_m = \sum_{2j \le m} \binom{m+1}{2j+1} 2^{-2j} y_{m - 2j}\quad (m=0,1,2,\ldots)\) \quad\qquad is equivalent to the set \(\displaystyle y_m = \sum_{2j \le m} \binom{m}{2j} 2^{-2j} D_{2j} x_{m-2j}\quad (m=0,1,2,\ldots)\). II. The set of equations \(\displaystyle(2m + 1)x_m = \sum_{j=0}^m \binom{2m+1}{2j} 2^{2j - 2m} y_j \quad (m=0,1,2,\ldots)\) \quad\qquad is equivalent to the set \(\displaystyle y_m = \sum_{j=0}^m \binom{2m}{2j} 2^{2j - 2m} D_{2m - 2j} x_j \quad (m=0,1,2,\ldots)\). III. The set of equations \(\displaystyle(2m + 2)x_m = \sum_{j=0}^m \binom{2m+2}{2j+1} 2^{2j - 2m} y_j \quad (m=0,1,2,\ldots)\) \qquad is equivalent to the set \(\displaystyle y_m = \sum_{j=0}^m \binom{2m+1}{2j+1} 2^{2j - 2m} D_{2m - 2j} x_j \quad (m=0,1,2,\ldots)\).