Need help for a long proof I am doing. I need to justify that if I examine the set of {Odd+Odd}, the unique pairs exist if I utilize {2n-1, {3,5...,2n-3,2n-1})
If we have:
- 3 5 7 9 11
3 (3,3) (3,5) (3,7) (3,9) (3,11)
5 (5,3) (5,5) (5,7) (5,9) (5,11)
7 (7,3) (7,5) (7,7) (7,9) (7,11)
9 (9,3) (9,5) (9,7) (9,9) (9,11)
11(11,3) (11,5) (11,7) (11,9) (11,11)
The unique sets are obvious as there's a line of reflection around (3,3); (5,5); (2n-1,2n-1). Thus unique sets are (3,3); (5,3), (5,5), (7,3), (7,5), (7,7), (9,3)... (9,9), (11,3)... (11,11)...
But I want the longer proof to be as complete as possible. What are my options here? Is there a stars and bars equation? Or a simple binomial expansion rule?
Thanks.
I apologize for weird formatting. I'm on my phone and this is urgent.
The only way in which an ordered pair of odd numbers could be non-unique is by inversion, i.e., $a\neq b$ and both $(a,b)$ and $(b,a)$ are counted. Thus, as you suggest correctly, then number of equivalence classes under the equivalence relation defined by inversion is equal to $k(k+1)/2$ where $k$ is the number of candidates that you started with. In your example $k=n-1,$ but is there any specific reason why you left the odd number $1$ out?