Ytukyg

For which values of $k$ does the following product not evaluate to an integer: $$prod_{n=2}^{k} n-frac{1}{n}$$

I find it somewhat surprising that it not only can evaluate to an integer, but also that it does so most of the time.

The only values of $k$ that I could find so far, such that it does not evaluate to an integer are $2$ and $18$. Do these numbers have any special characteristics which would explain this?

$$prod_{n=2}^k frac{(n+1)(n-1)}{n} = frac{frac{(k+1)!}{2}cdot(k-1)!}{k!} = frac{(k+1)(k-1)!}{2} $$
The only number for which it can't be an integer is $k=2$, and that's indeed the case.

It is surprising that this is often an integer. But because it is surprising there is probably a simple reason that will be become clear when we try a few examples.

$(2 - frac 12)(3 - frac 13) = 2*3 - frac 32 - frac 23 + frac 1{2*3}$.

Okay, one thing that becomes clear is that if we have $(n - frac 1n)(m - frac 1n)$ our terms are going to be $nm$ (which is clearly an integer) $-frac nm, -frac nm$ and $frac 1{mmn}$. When we add $-frac mn -frac nm$ we get $-(frac {m^2}{mn} + frac {n^2}{mn})=-frac {m^2 + n^2}{nm}$ and we get $mn - frac {m^2 + n^2-1}{mn}$.

Well, no reason that $mn|m^2 + n^2 -1$ but if $m$ and $n$ are consecutive, i..e. $(n = m+1)$ we get $$m(m+1) - frac {m^2 + (m+1)^2 - 1}{m(m+1)} = m(m+1) - frac {2m^2 + 2m}{m(m+1)} = m(m+1) - frac {2m(m+1)}{m(m+1)} = m(m+1) -2$$ and that.... is an integer.

So $(m - frac 1m)((m+1) - frac 1{m+1}$ must be an integer and as integers multiply together to integers we have for a product of even number of terms:

$prod_{n=2}^{2m+ 1} (n-frac 1n) = prod_{k=1}^m [(2k-frac 1{2k})((2k+1)-frac 1{2k+1})] = prod_{k=1}^m (2k(2k+1) -2)=Kin mathbb Z$.

But if we have a product of odd terms:

$prod_{n=2}^{2m+ 2}(n-frac 1n) =(prod_{n=2}^{2m+ 1}(n-frac 1n))((2m+2)-frac 1{2m+2})=K( 2m+2 - frac 1{2m+2})$ and as $( 2m+2 - frac 1{2m+2})$ is not an integer $K( 2m+2 - frac 1{2m+2})$ won't be an integer.....

.... unless $2m+2|K$.

Okay.... we are halfway done.

When will $2m+2|K$?

I'm embarrassed to admit how long the following took me to figure out but:

$K=prod_{k=1}^m(2k(2k+1) -2)$ and for $2m(2m+1) -2=2(m+1)(2m -1)$ so $2m+2$ will always divide $K$ and this will always be an integer. (Unless we only multiply one term $(2 - frac 12)$.)