One version of Ackermann’s function A(m,n) is defined recursively for m, n ∈ N by:
Bitmap
A(0, n)= n + 1, n ≥ 0;
A(m, 0) = A(m − 1, 1), m 0; and
A(m, n) = A(m − 1, A(m, n − 1)), m, n 0.

a. Calculate A(1, 3) and A(2, 3).
b. Prove that A(1, n) = n + 2 for all n ∈ N.
c. For all n ∈ N show that A(2, n) = 3 + 2n
d. Verify that A(3, n) = 2n+3 − 3 for all n ∈ N.

Question

One version of Ackermann’s function A(m,n) is defined recursively for m, n ∈ N by: 
Bitmap 
A(0, n)= n + 1, n ≥ 0; 
A(m, 0) = A(m − 1, 1), m > 0; and 
A(m, n) = A(m − 1, A(m, n − 1)), m, n > 0. 

a. Calculate A(1, 3) and A(2, 3). 
b. Prove that A(1, n) = n + 2 for all n ∈ N. 
c. For all n ∈ N show that A(2, n) = 3 + 2n 
d. Verify that A(3, n) = 2n+3 − 3 for all n ∈ N.

anonymous · Answer

While I appreciate the attachment, it is a bit difficult to read to as some of it appears to be faded.

anonymous · Answer

you don't understand it?

anonymous · Answer

A(1,3) is 3rd case, right? since m =1, n =3 both m, n>0, ok?

anonymous · Answer

from your Bitmap. confirm if you got me

anonymous · Answer

This sounds right, yet.

anonymous · Answer

so, A(1,3)= A(1-1, A(1, 3-1) = A(0, A(1,2) )get it?

anonymous · Answer

Not particularly. This is a subject that is very foreign to me. I am continuing to work on it.

anonymous · Answer

your bitmap, the third condition |dw:1384132124472:dw|