Counting even and odd numbers in the columns of a triangular arrangement of the integers I write the positive numbers starting at $1$ in a triangle:$$\mathbb{N}_\triangle = \begin{matrix}
    &&&&&21&\ldots         \\
    &&&&15&20&\ldots       \\
    &&&10&14&19&\ldots     \\
    &&6&9&13&18&\ldots   \\
    &3&5&8&12&17&\ldots \\
    1&2&4&7&11&16&\ldots 
    \end{matrix}$$
I would like some help proving the following claim

I denote the count of even numbers in column $n$ of
  $\mathbb{N}_\triangle$ by $e(n)$. Then for $n>1$ $$e(n) = \Bigg\lfloor{n+2
 \above 1.5pt 4}\Bigg\rfloor+\Bigg\lfloor{n+1
 \above 1.5pt 4}\Bigg\rfloor$$ Note  the $n^{th}-column$ of $\mathbb{N}_\triangle$ has $n$ integers so I can set  $f(n)$ to count the number of odd integers in
  $\mathbb{N}_\triangle$ and  $$f(n) = n-e(n)$$ It appears that 
  $e(n)\in$ A004524 

This is being driven by hunches and computer checks. $e(n)$ can be simplified but I think it is easier to remember it the way it is written. 

Note that each column forms an interval of integers, and so the numbers in the column alternate between being odd and even.
Hence if $n$ is even, so that there are an even number of integers in the $n$th column, exactly half of them will be even, so $e(2k) = k$ for every $k \in \mathbb{N}$.
If $n$ is odd, there will either be $\frac{n-1}{2}$ or $\frac{n+1}{2}$ even numbers, depending on whether the column starts with an odd or even number respectively.  Notice that the difference between the initial numbers of the $2k-1$st and $2k+1$st columns is $4k-1$, since the $2k-1$st column has $2k-1$ numbers, and the $2k$th column in between has a further $2k$ numbers.  As $4k-1$ is odd, that means that the initial numbers in the $2k-1$st and $2k+1$st odd columns have different parities.
Hence the first odd column starts with an odd number (namely $1$), the next odd column starts with an even number, the next odd column after that starts with an odd number again, and so on.  This means we have the formula
$$\begin{array}{c|cccccccccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 &  7 & 8 & ... & 4k + 1 & 4k + 2 & 4k + 3 & 4k + 4 & ... \\ \hline
e(n) & \frac{n-1}{2} & \frac{n}{2} & \frac{n+1}{2} & \frac{n}{2} & \frac{n-1}{2} & \frac{n}{2} & \frac{n+1}{2} & \frac{n}{2} & ... & 2k & 2k + 1 & 2k + 2 & 2k + 2 & ... \\
e(n) - \frac{n}{2} & - \frac12 & 0 & \frac12 & 0 & - \frac12 & 0 & \frac12 & 0 & ... & - \frac12 & 0 & \frac12 & 0 & ...
\end{array}. $$
As you can see, $e(n) - \frac{n}{2}$ is periodic with period $4$, as explained by the dependence on the parity of $n$, and then the alternating behaviour in the odd case.  Your formula captures this periodicity, and is correct.  Other ways to express it, which I find delightfully ridiculous, include
$$ e(n) = \frac12 \operatorname{Re} \left( n + i^{n+1} \right) = \frac12 \left( n - \sin \left(\frac12 n \pi\right)\right). $$
