# 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?

• $n=(n+1)+(n+2)-(n+3)$ ? May 26, 2014 at 17:32
• So painfully simple...!+1 May 26, 2014 at 17:39
• I presume the OP means consecutive positive integers strictly smaller than $n$. A fact that you can use: Every number that is not a power of $2$ is a sum of consecutive positive integers (all smaller than itself). For instance, $11 = 5 + 6$, $12 = 3 + 4 + 5$, $13 = 6 + 7$, $17 = 8 + 9$, $18 = 5 + 6 + 7$, $19 = 9 + 10$, etc. (For proof, see here or here.) So it's enough to ask the (suitably modified) question for powers of $2$. May 26, 2014 at 17:42
• Why do you presume such a thing, @ShreevatsaR ? Some of the examples of the OP don't even use integers...or even positive integers. May 26, 2014 at 17:48
• $(2-1)+(2-1)+(2-1)...$ $n$ times May 26, 2014 at 17:56

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.
• Rather, $(n+1)$, $(n+2)$ and $(n+3)$ are three consecutive integers. +1. May 27, 2014 at 2:30