smc

We know that $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ $-$ $\left(\genfrac{}{}{0ex}{}{n}{1}\right)$ $=$ $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ $-$ $\left(\genfrac{}{}{0ex}{}{n}{2,}\right)$ because they form an arithmetic sequence, and expanding, we have by the definitions in the problem: $$\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}-n=\frac{n(n-1)(n-2)(n-3)(n-4)...}{6((n-3)(n-4)...)}-\frac{n(n-1)(n-2)(n-3)...}{2(n-2)(n-3)...}.$$ Canceling out the $(n-2)!$ and the $(n-3)!$ from each side of the equals sign, we have $\frac{n(n-1)}{2}-n=\frac{n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}.$ Getting rid of the fractions by cross multiplication, and getting n on one side, we have $n^3-9n^2+14n=0,$ and we can factor out the n, so n(n^2-9n+14)=0, and we are looking for two integers x and y such that $x+y=-9$ and $xy=14.$ By guess and check, our integers are -7 and -2, so $n(n-7)(n-2)=0!!!$ According to the problem, $n>3,$ so we have n=7 or 2, thus $\boxed{7}$