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On congruences involving Bernoulli numbers and irregular primes. II. (English) Zbl 0856.11009
Denote by $S(n, k)$ the Stirling numbers of the second kind ($n$, $k$ positive integers), which are defined by the following recurrence formula $$S(n+ 1,k+ 1)= S(n, k)+ (k+ 1) S(n, k+ 1), \qquad S(n, 1)= S(n, n)=1.$$ In this paper the author proves the following two congruences concerning the Stirling numbers and the Bernoulli numbers $B_{2k}$: $$\align S(p+ 2k, p- 1) &\equiv \bigl( {\textstyle {{p-1} \over 2}}+ k\bigr) pS (p- 1+ 2k, p- 1) \pmod {p^4},\\ S(p+ 2k, p- 1) &\equiv \bigl( (2k- 1)/ 4k) p^2 B_{2k} \pmod {p^3}, \endalign$$ where $p$ is a prime and $k$ an integer, $1\leq k\leq {{p-3} \over 2}$. The proof uses the congruence $(*)$ $S(p- 1+ r,p- 1)\equiv pB_r/ r\pmod {p^2}$, $1\leq r\leq p-2$, from the preceding paper of the author [ibid. 30, 1-9 (1990)] and some equalities among the numbers $R_m (n, s)= \sum^{n- 1}_{j= 0} (-1)^j S(m+s, m-n- j) S(n, n-j)$. Using again the congruence $(*)$ and polynomials $\varphi_n (x)= \sum^n_{j=1} (-1 )^{j-1} (j-1)! S(n, j)x^j$ $(n\geq 1)$ the author gives the interesting equality $$\sum^r_{j=1} {{2r} \choose {2j-2}} B_{2j}/ 2j= {1\over {(2r+ 1)(2r+ 2)}}.$$
Reviewer: L.Skula (Brno)

11B68Bernoulli and Euler numbers and polynomials
11B73Bell and Stirling numbers