Factorials and Arithmetic Progression.

Question

Are there sets of factorials $(a_1!,a_2!,a_3!,\dots,a_n!)$, such that they exist in Arithmetic progression.

$n$ is a natural number

I don't see any such examples(Except for $n=2$). And I don't see it for any $n\ge2$.

Answer 1

There is no AP of length $3$. For suppose that $a! \lt b! \lt c!$ are in AP. Then $2b!=a!+c!$. Dividing through by $a!$, we find that $$2(b)(b-1)\cdots(b-a+1)a!=\left(1+(c)(c-1)\cdots (c-a+1)\right)a!.$$ Divide through by $a!$. We get $$2(b)(b-1)\cdots(b-a+1)=(1+(c)(c-1)\cdots (c-a+1)).$$

One side is even and the other is odd.

Answer 2

Let $a!<b!<c!$ be an arithmetic progression. Then we have $$ \frac{2\cdot b!}{a!}=1+\frac{c!}{a!} $$ Hence $\frac{c!}{a!}$ is odd. Therefore $c$ is odd and $a=c-1$. Contradiction.