How many pairs of $(x,y,z)$ for $x+y+2z=n$ 
How many pairs of $(x,y,z)$ for $x+y+2z=n$ when $n$ is an odd integer $\geq 5$

For $n=5$ there are 2 pairs. $n=7$ there are 6 pairs. $n=9$ there are 12 pairs. $n=11$ there are 20 pairs.
$an^2 + bn + c = f(n)$
$25a + 5b + c = 2$
$49a + 7b + c = 6$
$81a + 8b + c = 12$
$f(n) = (n^2)/4 - n + 3/4 = (n-3)(n-1)/4$
Is there a better way to solve it?

put z=2 then x+y=n-2 has n-3 ways (x can be anything from 1 to n-3).
then put z=4 then x+y=n-4 has n-5 ways (x can be anything from 1 to n-5).
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put z =n-3  then x+y =3 have 2 ways.
total number of cases = 2+4+ 6.......n-3 = 2(1+2+3....$\frac{n-3}{2}$) = 2 $\frac{(\frac{n-3}{2})(\frac{n-1}{2})}{2}$ = $\frac{(n-3)(n-1)}{4}$ 

