<s>[INST] When is $\prod_{n=2}^{k} n-\frac{1}{n}$ not an integer? 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?
 [/INST] 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:
$\prod_{n=2}^{2m+ 1} (n-\frac 1n) = \prod_{k=1}^m [(2k-\frac 1{2k})((2k+1)-\frac 1{2k+1})] = \prod_{k=1}^m (2k(2k+1) -2)=K\in \mathbb Z$.
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 $(2 - \frac 12)$.)
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