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 6 of 6: Exceed the maximum

Previous step Next step

D>C+\sum_{j=1}^{m}r_j\ge\sum_{j=1}^{1008}(2018+j)=2\,543\,688>N

Detailed analysis

Each step on the path from C to D adds its positive sibling increment. Therefore the value at D is larger than the sum of the displayed lower bounds for those increments. For n equal to 2018, that sum is $2,543,688$ , whereas N is $2,037,171$ . Thus D would be larger than N, impossible because N is already the largest entry in the triangle. The assumed anti-Pascal triangle does not exist.