Can I have a logical explanation for why this number is so ridiculously close to a whole number? $$e^ {\pi\sqrt{163}}=262537412640768743.9999999999992\cdots$$

Why does this number run so incredibly close to a whole number? Can I have a logical explanation for why this finding? I know how to calculate it, but I want an answer that would explain it to me by intuition only. Thank you very much for any help. It is not the same as the other similar problem because I am looking for a logical reason and the other is wondering whether or not it is concidence.
Here's an explanation of these numbers. Say you want $e^{\pi \cdot n}$ to be within a positive error called $\epsilon$ of a natural number $L$, but you want $n$ to be natural as well. In other words...

$$L-\epsilon \lt e^{\pi \cdot n} \lt L+\epsilon$$

You can "solve" for n. Take the natural log and divide by $\pi$ to get...

$${1 \over {\pi}} \cdot \ln{(L-\epsilon)} \lt n \lt {1 \over {\pi}} \cdot \ln{(L+\epsilon)}$$

Now it should be clear that if $L$ ranges through $1$ to $\infty$ that the boundary for $n$ will range $0$ to $\infty$. 

Since $L$ must be an natural number, there is little reason to think the values $n$ can take will be any more likely to not contain a natural number. 

So now you'd use the above and pick an error tolerance. Values that allowed for both a natural $n$ and $L$ will be within $\epsilon$ of $L$. I'm guessing these are the so called Heegner numbers, you'd just allow for $n$ which are the square root of other numbers.
Here's an explanation of these numbers. Say you want $e^{\pi \cdot n}$ to be within a positive error called $\epsilon$ of a natural number $L$, but you want $n$ to be natural as well. In other words...

$$L-\epsilon \lt e^{\pi \cdot n} \lt L+\epsilon$$

You can "solve" for n. Take the natural log and divide by $\pi$ to get...

$${1 \over {\pi}} \cdot \ln{(L-\epsilon)} \lt n \lt {1 \over {\pi}} \cdot \ln{(L+\epsilon)}$$

Now it should be clear that if $L$ ranges through $1$ to $\infty$ that the boundary for $n$ will range $0$ to $\infty$. 

Since $L$ must be an natural number, there is little reason to think the values $n$ can take will be any more likely to not contain a natural number. 

So now you'd use the above and pick an error tolerance. Values that allowed for both a natural $n$ and $L$ will be within $\epsilon$ of $L$. I'm guessing these are the so called Heegner numbers, you'd just allow for $n$ which are the square root of other numbers.