How to find Ackermann(3,n)?

Question

             { y+1               if x=0; 
     A(x,y)= { A(x-1,1)          if y=0;
             { A(x-1,A(x,y-1))   otherwise

So I was trying to prove A(3,y) = $2^{(y+3)} -3 $

Though I was able to figure out

A(1,y) = y + 2
A(2,y) = 2y + 3 

But for A(3,y) = A(2,A(3,y-1)) 
               = 2A(3,y-1) + 3
               = 2(A(2,A(3,y-2))) + 3
               = 2(2(A(3,y-2) + 3) + 3
               = 2(2(A(3,y-2)) + 6 + 3
               = 2(2(2....(y times)..(A(3,0) + 3))))) + 3.2^y + 3.2^(y-1) ... + 3
               = 2^y(A(3,0) + 3 ) + 3(1-2^y)/(1-2)

and as A(3,0) = 5 $$=2^y.8 + 3(1-2^y)/(1-2)$$ $$=2^{(y+3)} - 3(1-2^y)$$

However it seems like I am doing something wrong as A(3,y) = $2^{(y+3)} -3 $and not $2^{(y+3)} - 3(1-2^y) $

Answer 1

There is a mistake in your decomposition. The result looks like this: $$2(2(...(2\cdot A(3,0) + 3)+3)+3)...+3$$

So, in fact you have $5\cdot2^y$ and then there is a geometric sequence with overall sum $3(2^y-1)$, add it together and you get $8\cdot 2^y-3$.