# A funny number problem

Let each integer number from $1$ to $2n$ be ascribed to a vertex or a side of an $n$-gon (one number per vertex and per side). Let the weight of a side be the sum of three numbers "connected" to it (i.e. the number ascribed to the side and two numbers ascribed to its end points).

Find all possible combinations (disregarding reflections and rotations) of numbers (from 1 to 12) for hexagon such that the weights of all sides are equal. Please do not use computer.

I had a lot of fun solving the problem, and wonder if some simple recipes for finding the solutions for larger $n$ exist.

PS. The original problem was to find a single combination, but it was so funny that I decided to find them all.

• Do you count reflections and rotations of a solution as being different? – paw88789 Jan 25 '18 at 16:32
• I further notice that you can do a "half rotation" where a side label goes to the next vertex and the vertex label goes to the next side around the polygon. You get a solution. Is that type of solution to be counted as different? – paw88789 Jan 25 '18 at 16:53
• It should not be in general possible but if by chance it is, it should be regarded as a new solution. – user Jan 25 '18 at 17:00
• Actually, I think the 'half rotation' as described in my previous comment, does in general give a valid (new) solution. – paw88789 Jan 25 '18 at 17:54

Not an answer, but some numerical results. Denote the "first" vertex as $a_1$, the next side is $a_2$. There are 2 cases: either fix $a_1=1$ or fix $a_2=1$. So I do a brutal force search, and divide the result by 2 to exclude the mirror cases. Below I show the combined cases ("all") and the fix $a_1=1$ case separately.

n  all   a1=1
3  4     2
4  6     4
5  6     4
6  20    10
7  118   66
8  282   131
9  1540  732
10 7092  3399
11 36128 17298


A few findings.

1. The 2 cases are not one-to-one but their numbers are generally close.
2. The growth rate is close to and slightly higher than $e^x$.
• Your "all" numbers fit this OEIS sequence. – Jens Jan 25 '18 at 22:09
• @Jens, these are essentially the same problem, lol – lion Jan 25 '18 at 22:34
• Probably I have posed problem wrongly. I was mainly interested in rules which would allow to drastically reduce the application of brute force. I have found two such rules, and I am quite sure there are other ones. – user Jan 28 '18 at 10:20