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Shortened recurrence relations for Bernoulli numbers. (English) Zbl 1171.11010
The authors start with two little known results of {Saalschütz} giving recurrence relations for the Bernoulli numbers $B_{2n}$. The first of these results actually is contained in the second one which reads as $$\sum_{k=m}^n (-1)^{k+1} (2^{2k}-2) C(n-m,k-m) B_{2k}=\sum_{j=0}^m (-1)^j\frac{4^{n-j}(n-j)!}{2(n-j)+1} C^{(n)}(m,j)$$ with certain numbers $C(n,m)$ and $C^{(n)}(m,j)$ which may be given explicitly and where $0\leq j\leq m\leq n$. The case $m=0$ gives the first identity. Then the authors express these numbers $C(n,m)$, $C^{(n)}(m,j)$ in terms of Stirling numbers of both kinds. The also discuss results by {\it P. G. Todorov} [J. Math. Anal. Appl. 104, 309--350 (1984; Zbl 0552.10007)] of a similar taste involving Stirling numbers $s(k,l)$ and $S(m,j)$ of the first and second kind respectively. Finally, using generating functions for the numbers $S(n,k)$ they get for $1\leq m\leq n$ and $k\geq 0$ the formula $$\sum_{j=m}^n \binom{n+k}{j-m} S(n-j+k+m,k+m) B_j=\frac{(-1)^m}{k+m} \sum_{j=1}^{m+1} \frac{N(n,m,k,j)}{\binom{k+m-1}{j-1}},$$ where $N(n,m,k,j)=(n+k) S(m+1,j) S(n+k-1,k+m-j)-m S(m,j)S(n+k,k+m-j)$. From this they also derive some other formulas.

11B68Bernoulli and Euler numbers and polynomials
11B73Bell and Stirling numbers
05A19Combinatorial identities, bijective combinatorics
Full Text: DOI
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