First 10 terms of a sequence My problem:

Give the first 10 terms of $f(5,b)$ in the recursive sequence defined by
  \begin{equation}
f(a,b) = \begin{cases}
    -2 & \text{if } b = 0\\
    3 & \text{if } b = 1\\
    1 & \text{if } b = 2\\
    f(f(b+2, b-3), b-1)-a & \text{if } b \geq 3\\
\end{cases}
\end{equation}

I think the answer may be  -4, -9, -14, -19, -24, -29, -34, -39, -44, -49. Can anyone confirm this?

We have $f(a,3)=f(f(5,0),2)-a=-a+1$
$f(a,4)=f(f(6,1),3)-a=f(3,3)-a=-a-2$
$f(a,5)=f(f(7,2),4)-a=f(1,4)-a=-a-3$
$f(a,6)=f(f(8,3),5)-a=f(-7,5)-a=-a+4$
$f(a,7)=f(f(9,4),6)-a=f(-11,6)-a=-a+15$
$f(a,8)=f(f(10,5),7)-a=f(-13,7)-a=-a+28$
$f(a,9)=f(f(11,6),8)-a=f(-7,8)-a=-a+35$
$f(a,10)=f(f(12,7),9)-a=f(3,9)-a=-a+32$
$f(a,11)=f(f(13,8),10)-a=f(15,10)-a=-a+17$
$f(a,12)=f(f(14,9),11)-a=f(21,11)-a=-a-4$
So substituting $a=5$ we get: 
$f(5,0)=-2; f(5,1)=3; f(5,2)=1; f(5,3)=-4$
$f(5,4)=-7,f(5,5)=-8,f(5,6)=-1,f(5,7)=10$
$f(5,8)=23,f(5,9)=30,f(5,10)=27,f(5,11)=12,f(5,12)=-9$
