Groups of numbers that are equal
1 - 0
2 - 0
3 - 1 - 1,2;3
4 - 1 - 1,4;2,3
5 - 0
6 - 0
Ect
How do I math these
wat chu talkin bout fgt :v
>>7797301
Ha! There's still no proof of this even though the ancient Egyptians were already working at this. It's not a famous theorem, but at the same time very important and difficult to work out even with the most advanced of mathematicians. Postulates have been made on how to math these but there's no verification..... yet....
We need a modern bright mind like Gauss to shed light on this.
[citation needed]
Op here
Got some sleep and worked on it a bit, here's what I've found
As I knew when I first posted
>to be split into two equal groups there must be an even number of odd numbers (duh)
>2,5;3,4;1,6 are equivalent (+1+(-1)) and can be swapped thus resulting in many similar solutions
And what I've since found
>7 has 3 groupings
>8 has 7 groupings
>groupings can be systematically found by keeping the small numbers left (as such that the groups 1,2,3... 1,3,4... Abd 1,2,4... Would be ordered 1,2,3...;1,2,4...;1,3,4...) then adding until you reach half the Nth triangle number
>for example: N=4 - 1 -> 1,2 -> 1,2,3 not valid -> 1,3 valid -> thus 1,3:2,4 -> 1,4 not valid -> return 1,3:2,4
>the difficulty of finding these increases very exponentially
I think that's as far as ill get but I might figure something out
Solving all the possibilities up to 8 served my applied math purposes anyways (I'm making symmetric tile patterns in minecraft lol) so I don't have much motivation to solve this now
