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Duro Kurepa conjectured that the function on the title is always nonzero in $\mathbb{Z}/{n \mathbb{Z}}$ provided $n>2.$ Daniel Barsky and B\'enali Benzahgou [MR2145571 (2006a:11025)] proved this. Thus, for all odd prime numbers $p$ $$ ku(p) = 0!+1!+ \cdots + (p-1)! \pmod{p} $$ is an inversible element of $\mathbb{Z}/{p\mathbb{Z}}.$

Question: There are infinite subsets $S$ of the set of odd composite numbers such that for each $s \in S$ one has that $ku(s) \pmod{s}$ is inversible in $\mathbb{Z}/{s \mathbb{Z}}.$ ?

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    $\begingroup$ By the way, the "proof" of the Kurepa conjecture turned out to be false. Here's an extract from the Erratum à l'article Nombres de Bell et somme de factorielles [MR2817943]: As pointed out to us by Farid Bencherif and Joseph Oesterlé, there are some irreparable calculation errors in the proof of Theorem 3 of our article. Theorem 3 and its proof (the Kurepa conjecture) are therefore withdrawn, and the Kurepa conjecture ($0!+1!+\cdots+(p−1)!\not\equiv0\pmod p$,for prime $p\ge3$) is not proved. $\endgroup$
    – Chandan Singh Dalawat
    Feb 23, 2013 at 2:08

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Yes, if $s$ is a power of an odd prime then your function gives a unit mod $s$.

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  • $\begingroup$ @Gerry: Nice. Do you think may be examples with all $\omega(s)>1$ ? $\endgroup$
    – Luis H Gallardo
    Jan 26, 2011 at 12:13
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    $\begingroup$ Thinking about it a bit more, it seems to me that it's true for all odd composite $s$. For let $p$ be a prime dividing $s$. Then $ku(s)\equiv ku(p)\pmod p$, so $ku(s)$ is invertible mod $p$, so it's invertible mod $p^r$ for all $r$. Thus $ku(s)$ is invertible modulo each prime power dividing $s$, so it's invertible modulo $s$. $\endgroup$
    – Gerry Myerson
    Jan 26, 2011 at 21:58
  • $\begingroup$ So, we have a complete answer !. I also wondered about the distribution of ordersof $ku(p)$ in $GF(p)$, but after some calculations this seems random... $\endgroup$
    – Luis H Gallardo
    Jan 26, 2011 at 22:59

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