On the primality of \(n! \pm 1\) and \(2 \times 3 \times 5 \times \dotsm \times p \pm 1\) (Q2759112)

For every prime number \(p\) write \(p\#\) to be the product of all the prime numbers less than or equal to \(p\). In the paper under review, the authors explain the computation they had performed in order to locate new primes of the form \(n!\pm 1\) and \(p\#\pm 1\). The previous range in which such computations were performed was \(n\leq 4580\) and \(p\leq 35000\) in \textit{C. Caldwell} [Math. Comput. 64, 889-890 (1995; Zbl 0823.11072)]. NEWLINENEWLINENEWLINEThe paper under review reports the results found by extending this search to \(n\leq 10000\) and \(p\leq 120000\). Three new primes were found, namely \(6380!+1\), \(6917!-1\) and \(42209\#+1\). The search took over one year of CPU time. The authors also give heuristics as to how many numbers of the form \(n!\pm 1\) or \(p\#\pm 1\) are expected to exist for \(n\leq N\) and \(p\leq N\), respectively. In both cases, their heuristics suggest that the number of such primes should be about \(e^\gamma\log N\), and the computations (albeit in a limited range) seem to support these conjectures. NEWLINENEWLINENEWLINEPart (iv) of Theorem 2.4 contains a typographical error that is corrected in an ``Added after Posting'' section at the end of the paper.