What is the positive integer number $n$ satisfying $(2n)!=n!\times(n+1)!$

Question

What is the positive integer number $n$ satisfying $(2n)!=n!\times(n+1)!$

I tried to expand the factorial but it didn't help much ?

Answer 1

If we use positive integers only for the factorial function:

Expanding $n!(n+1)!$ we get $n!(n!)(n+1) = (n!)^2(n+1)$. To compare how the two different functions grow, let us compute $\frac{(2n)!}{(n!)^2 * (n+1)}$. or $\frac{(2n)!}{(n!)^2}(n+1)$. We recognise $\frac{(2n)!}{(n!)^2}$ as the central binomial coefficient, which is always greater than 0 for all positive integers.

Since $n$ is also a positive integer, $n+1$ is a positive integer, so $\frac{(2n)!}{(n!)^2}(n+1)$ is a positive integer. Therefore, $(2n)!$ is always greater than $(n!)(n+1)!$, for $n > 1$ (as @Yves Daoutt mentioned). Indeed, calculating the cases $n=0$ and $n=1$ give $0! = (0!)(1!)$ and $2! = (1!)(2!)$, or $1 = 1$ and $2 =2$ respectively.

Answer 2

The simplest reason that you can't have any other solution (except $n=1$) is that $(2n)!$ is just too big.

The equation is equivalent to $(2n)!/(n!\times n!)=n+1$. This can be written (by expanding and cancelling) as

$$\frac{2n}{n}\times\frac{2n-1}{n-1}\times\cdots\times\frac{n+1}1=n+1.$$

Now it's awkward to calculate the LHS exactly, but much easier to show it's too big (if $n>1$). Each of the fractions is at least $2$ and there are $n$ of them, so we have $\text{LHS}\geq 2^n$. If $n>1$ then $2^n>n+1$, so the two sides aren't equal.

Answer 3

The first values yield

$$2=1\cdot2,\\ 24>2\cdot6,\\ 720>6\cdot24\\\cdots$$

and by induction

$$(2n)!>n!(n+1)!\\\implies\\(2(n+1))!=(2n)!(2n+1)(2n+2)>n!(n+1)(n+1)!(n+2)=(n+1)!(n+2)!$$

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Actually, the ratio LHS/RHS asymptotically grows like $4^n$.