# Why will this function always be an integer?

1. Nov 10, 2014

### Avichal

f(n) is defined as 11x22x33....xnn

Then it seems as if f(n)/(f(r).f(n-r)) is always an integer for 0 < r < n.
I tried a few cases. Its true for them. Is it always true? I cannot seem to figure out any ways to prove it.

2. Nov 10, 2014

### mikeph

I presume r and n are integers?

3. Nov 10, 2014

### Staff: Mentor

x means multiplication? * would be clearer.

You can prove it with the fact that ${n \choose c} = \frac{n!}{c! (n-c)!}$ is always an integer.

4. Nov 12, 2014

### glappkaeft

It is trivially true since the set of all integers is closed under the operation of multiplication (exponentiation being a special case of multiplication).

5. Nov 12, 2014

### Staff: Mentor

There is a division...

6. Nov 12, 2014

### glappkaeft

I think I might have missread the f(n)/(f(r).f(n-r)) statement in the OP by reading to quickly. When I look at it now it makes no sence at all unless someone defines the "."-operator.

7. Nov 12, 2014

### Staff: Mentor

I assumed it to be a multiplication, too.

8. Nov 25, 2014

### AMenendez

This is the "hyperfactorial", $H(n)$ (assuming the x is multiplication). It is defined as
$H(n) = \prod_{i=1}^{n} i^i$.
Just by looking at it, I don't see any reason to believe it wouldn't be an integer at any point. I don't know if there's an explicit proof of this out there, but generally, an integer multiplied any number of times by itself is an integer, and when added to another integer that has undergone the same process, it should still be an integer. Of course, $n$ can be really large, so I can't say with any certainty what happens as the product has more and more (perhaps infinite) terms.
Interesting question--I'm curious now!
I'd love for other members to contribute to this.

9. Nov 25, 2014

### Staff: Mentor

There is a division! It makes the question not trivial. Here it is in red:
$$\frac{f(n)}{f(r)f(n-r)}$$

Still always an integer, but you have to prove it.