Max odds in a triangle Place $n(n+1)/2$ integers in an equilateral triangle such that each number is the sum of the two numbers below it. What is the maximum number of odds in the triangle?
My attempt: I suspect it's $\lceil n(n+1)/3\rceil$ and it's pretty easy to verify this for small $n$. The optimal pattern seems to be $110110\ldots $ in a spiral shape. By adding $1$ for each odd and $-2$ for each even, all you'd have to do is show the total tally is at most $2$. By supposing the tally is at least $3$ and considering certain positions of the first three $1$s (odds)in a row, we get a contradiction. But other cases are more annoying and I feel like there should be a more elegant solution given that the optimal pattern seems so simple.
Other potentially useful information is that rotating the triangle gives valid solutions, and you can swap the bottom row and keep the rest of the parities the same (so there might be a perturbation argument).
 A: Instead of odd and even, I will think of this as a triangle of zeroes and ones, where entry is the sum $\pmod 2$ of the entries below it. This is mostly a complete proof, with one bit at the end left to the reader.
We can prove this by induction on $n$. Assuming the result for $n-1$, and given a triangle of order $n$, the top $n-1$ rows of triangle will by induction have at most $\lceil n(n-1)/3\rceil$ ones. Therefore, you are immediately done in the case where the bottom row has at most
$$
\lceil n(n+1)/3\rceil -\lceil n(n-1)/3\rceil=\lceil n(n+1)/3-n(n-1)/3 \rceil =\lceil 2n/3\rceil
$$
ones. (We can combine the terms within the ceiling brackets because at least one of $n(n+1)/3$ and $n(n-1)/3$ is always an integer).
Assume, then, that there are more than $\lceil 2n/3\rceil$ ones in the bottom row. The key idea is that when there are lot of ones in the bottom row, there tend not to be too many ones in the second row from the bottom. Therefore, we can prove there are not too many ones in the bottom two rows combined, then leverage the inductive hypothesis for the triangle made of the first $n-2$ rows.
To this end, we know that there are at most $\lceil (n-1)(n-2)/3\rceil$ ones in the first $n-2$ rows, so we just need to prove there are at most
$$
\lceil n(n+1)/3\rceil -\lceil (n-1)(n-2)/3\rceil=\lceil n(n+1)/3-(n-1)(n-2)/3 \rceil =\lceil (4n-1)/3\rceil
$$
in the bottom two rows combined. Letting $X$ be the number of ones in the bottom row, and $Y$ be the number of ones in the row above that, then we know that $X>\lceil 2n/3\rceil$, and we want to prove that $X+Y\le \lceil (4n-1)/3\rceil$. Writing this as
$$
-X+(Y+2X)\stackrel{?}\le  \lceil (4n-1)/3\rceil,
$$
we know that $-X\le -\lceil 2n/3\rceil - 1$, so we would be done if we could show
$$
Y+2X\le \lceil (4n-1)/3\rceil+\lceil 2n/3\rceil + 1
$$
Since $2n\le \lceil (4n-1)/3\rceil+\lceil 2n/3\rceil + 1$, it suffices to prove
$$
Y+2X\le 2n$$or
$$
Y\le 2(n-X)\tag{$*$}
$$
That is, it suffices to prove the number of ones in the second row from the bottom is at most two times the number of zeroes in the bottom row. I leave this last bit to you. Up until now, we have not used the fact that every entry in the triangle is the sum of the two below it, so proving $(*)$ will use this property.
