Relaxed empty Pascal triangles
by Jan Kristian Haugland

Problem 3 of the 59th International Mathematical Olympiad defines an empty Pascal triangle with n rows having numbers such
that every number (except in the bottom row) is the absolute difference of the two numbers immediately below it. One can show
that when n is sufficiently large, then there is no empty Pascal triangle with n rows that contains all integers from 1 to n (n + 1) / 2.

Suppose we also allow such a number to be the sum of the two numbers below it. I.e., each upright triangle of unit side length contains
three numbers such that one of them is the sum of the other two. For which n can we now fill in the integers from 1 to n (n + 1) / 2?

A parity argument shows that there are no solutions when n = 2^k – 2 with k ≥ 3. Heuristically, it seems that solutions should exist
for all other n up to 13. However, with almost no known mathematical structure of the solutions, and an astronomical growth rate
of the number of permutations of (1, ..., n (n + 1) / 2), they may be exceedingly difficult to find for the highest feasible values of n.

The cases with n ≤ 8 have been completely solved. The lists of solutions give the number
in the first row of the Pascal triangle, then the two numbers in the second row, and so on.

n	Total number	Isometry classes
1	1	1
2	6	1
3	42	7
4	246	41
5	744	124
6	0	0
7	1158	193
8	2088	348

A solution for n = 9:

								8
							20		12
						2		22		34
					36		38		16		18
				33		3		41		25		43
			6		39		42		1		24		19
		23		29		10		32		31		7		26
	21		44		15		5		27		4		11		37
35		14		30		45		40		13		17		28		9