Difficult question 3

Using method of induction

f(0,n) = n+1
f(1,0) = f(0, 1) = 2
f(1,n) = f(0, f(1,n-1))
......... = f(1,n-1) + 1
..........= f(1, n-2) + 2
..........= ...
..........= f(1,0) +n
..........= n+2

f(2,0) = f(1,1) = 1+2 =3
f(2,n) = f(1, f(2, n-1))
..........= f(2,n-1) + 2
..........= f(2, n-2) + 4
..........= ...
..........= f(2,0) + 2n
..........= 2n+3

f(3,0) = f(2,1) = 2+3 = 5
f(3,n) = f(2, f(3,n-1))
..........= 2f(3, n-1)) + 3
..........= (2^2)*f(3, n-2)) +2*3 + 3 = (2^2)*f(3, n-2) + (2^1+1)*3
..........= (2^3)*f(3, n-3))+2*(2^1+1)*3 + 2*3 + 3 = (2^3)*f(3, n-3))+(2^2+2^1+1)*3
..........= ...
..........= (2^n)* f(3, 0) + (2^(n-1)+...+2+1)*3
..........= 5*2^n + 3*(2^n - 1)
..........= 2^(n+3) - 3

f(4,0) = f(3,1) = 2^4 - 3 = 13
f(4,n) = f(3, f(4, n-1))
..........= 2^(f(4,n-1)+3) - 3
therefore, we have f(4,n)+3 = 2^(f(4,n-1)+3)
if we let g(n) = f(4,n) + 3 and g(0) = f(4,0)+3 = 16
then the relation becomes g(n) = 2^g(n-1) = 2^2^g(n-2) = 2^2^2^g(n-3) = ...
= 2^2^2^...^2^g(0) = 2^2^2^...^2^16 (here we have n numbers of 2)
therefore
f(4,1981) = 2^2^2^...^2^16 - 3 (here we have 1981 numbers of 2)