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About the period of Bell numbers modulo a prime. (English) Zbl 1195.11034
Let $p$ be a prime number and let $g(p)=(p^p-1)/(p-1)$. It is well-known that the order $o(r)$ of a root $r$ of the irreducible polynomial $x^r-x-1$ over ${\Bbb F}_p$ divides $g(p)$. {\it S. Wagstaff} in [Math. Comput. 65, No. 213, 383--391 (1996; Zbl 0852.11008)] has conjectured that $o(r)=g(p)$ holds for all primes $p$. In this paper, the authors point out some subsets $S\subset \{1,\ldots,g(p)\}$ that do not contain $o(r)$. Here are a couple of examples of the authors results. First it is not hard to see that if we write the positive integer $d$ in base $p$ as $$ d=d_0+d_1p+\cdots+d_{p-1} p^{p-1}, $$ with $0\le d_i\le p-1$ for all $i=0,\ldots,p-1$, and we put $$ P(x)=x^{d_0}(x+1)^{d_1}\cdots (x+p-1)^{d_{p-1}}-1, $$ then $d$ is a multiple of $o(r)$ if and only if $P(r)=0$. In particular, $d=o(r)$ if and only if the only solution of the exponential equation $P(r)=0$ is obtained when $d_0=d_1=\cdots=d_{p-1}$. Armed with the above fact, the authors prove that under the assumption that $d<o(r)$ the following hold: 1. $$ 2p-1\le d_0+\cdots+d_{p-1}\le p^2-3p+1. $$ 2. At least five of the $d_i$’s are positive. The proofs involve clever manipulations of algebraic equations in ${\Bbb F}_p$. We point out that $o(r)$ is also the period of the sequence of Bell numbers modulo $p$, whence, the title.

11B73Bell and Stirling numbers
11T06Polynomials over finite fields or rings
11T55Arithmetic theory of polynomial rings over finite fields
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