For how many integers n where

These will occur in pairs......

The first will occur at  C ( 4n -2, 2)   and the second will occur at C(4n - 1, 2)

So......there are 25 pairs of these from C(2, 2) to C(100, 2)  =  50 integers

Proof of the first :

[ 4n - 2] ! /   ( [4n - 2 - 2]! * 2! ]  =

[4n -2] ! / [(4n-4)! *2! ]  =

[4n -2] * [4n - 3] / 2 =

[4n -2]/2 * [4n -3]  =

[2n -1] * [4n -3]  ..........this is the product of two odds which always  = an  odd

Proof of the second

[4n - 1]!/[ (4n -1 -2)! * 2!] =

[ 4n -1]! / [(4n -3)! 2!]  =

[4n -1][4n-2] / 2  =

[4n -2] /  2  * [4n -1]  =

[2n -1] [4n -1]   =   odd