Abstract
Here, we showthat if un = n2(n) +/- 1, then the largest prime factor of u(n) +/- m! for n >= 0, m >= 2 tends to infinity with max{m, n}. In particular, the largest n participating in the equation u(n) +/- m! = 2(a)3(b)5(c)7(d) with n >= 1, m >= 2 is n = 8 for which (8.2(8) + 1) - 4! = 3(4).5(2).