Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? For example:
$10 = 1 + 2 + 3 + 4$
$11 = 1 - 2 + 3 \times 4$
$12 = 3 \times 4$
$13 = -(1 - 2) + 3 \times 4$
$14 = 2 + 3 + 4 + 5$
$15 = 1 + 2 + 3 + 4 + 5$
$16 = (2/3)(4/5)(6 + 7 + 8 + 9)$
$17 = 1 + (2/3)(4/5)(6 + 7 + 8 + 9)$
$18 = (1 + 2)3!$
$19 = 4! - 5$
Obviously any triangular number has such an expression, as well as numbers one less than a triangular number. But how to prove it for other numbers? Or is there a counterexample?

Every integer $n$ can be written as
$$n= (n+1)+(n+2)-(n+3)$$
where $(n+1)$ and $(n+2)$ are of course two consecutive integers.
