Problem 3

An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. The following is a four-row anti-Pascal triangle containing every integer from 1 through 10:

\begin{array}{ccccccccccc} &&&&4&&&&\\ &&&2&&6&&&\\ &5&&7&&1&&&\\ 8&&3&&10&&9 \end{array}

Does there exist an anti-Pascal triangle with 2018 rows which contains every integer from 1 to

1+2+\cdots+2018

Step 3 of 6: Force the apex path to be extremal

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A=p_1\to p_2\to\cdots\to p_n=B,\qquad B-A=\sum_{i=1}^{n-1}q_i,\qquad A+\sum_{i=1}^{n-1}q_i\ge 1+2+\cdots+n=N

Detailed analysis

Follow the arrows from the top entry A to the bottom entry B. Let $q_i$ be the sibling of $p_{i+1}$ at the i-th step. The n numbers consisting of A and all $q_i$ are distinct positive entries of the triangle, so their sum is at least the sum of the n smallest positive integers. The path identity says that this sum is exactly B. Since no entry exceeds N, equality is forced: B=N, and A together with the $q_i$ are precisely the numbers from 1 through n.