Problem 1879

In what follows, part of the generic configurations referred to may not exist for the first few values of $n$; in this case, simply consider the corresponding induced

We now show by induction on $n$ that a binary anti-Pascal $n$-triangle contains at least $\lfloor n(n+1)/6 \rfloor$ ones, of which at least $n-1$ lie on the bottom three rows (and since the transpose of a binary anti-Pascal triangle is again one such, the same holds for the leftmost three columns).
The cases $n=1, 2, 3$ are easily dealt with, so let $n \ge 4$, and let $A$ be a binary anti-Pascal $n$-triangle. The lower left $3 \times 3$ subarray $A'$ (induced if $n=4$) contains at least 3 ones, unless the lower left and central entries are both one and the other entries are all zero, in which case there are only 2 ones in $A'$.
In the former case, consider the binary anti-Pascal $(n-3)$-triangle consisting of the rightmost $n-3$ columns of $A$. By the induction hypothesis, the bottom three rows of this triangle contain at least $n-4$ ones, so the bottom three rows of $A$ contain at least $3 + (n-4) = n-1$ ones.
In the latter case, the statement clearly holds if $n=4$, so let $n \ge 5$ and notice that the bottom two entries of the column of $A$ adjoining $A'$ along the right flank are both one. Thus, the $3 \times 4$ subarray of $A$ extending $A'$ rightwards contains at least 4 ones. The rightmost $n-4$ columns of $A$ form a binary anti-Pascal $(n-4)$-triangle whose bottom three rows contain at least $n-5$ ones, by the induction hypothesis. Hence the bottom three rows of $A$ contain at least $4 + (n-5) = n-1$ ones.
In either case, the bottom three rows of $A$ contain at least $n-1$ ones. Finally, by the induction hypothesis, the binary anti-Pascal $(n-3)$-triangle atop the bottom three rows of $A$ contains at least $\lfloor(n-3)(n-2)/6\rfloor$ ones, so $A$ contains at least $n-1 + \lfloor(n-3)(n-2)/6\rfloor = \lfloor n(n+1)/6 \rfloor$ ones. This completes the induction and concludes the proof.