Can this notation be computed in Mathematica?

Question

I defined a notation as follows:

(a, b) = $a^b$,

(a, b(c)) = (a, b-1(c)) * (a+1, b(c-1)),

(a, b(0)) = (a, b), and (a, 1(c)) = (a, a(c-1)).

For c = 4, (a, b(c)) is already tedious to evaluate. I next defined (a, b(0,1)) to be equal to (a, b($a^b$)) and (a, b(c, 1)) as (a, b-1(c,1)) * (a+1, b(c-1,1). (2, 2(1,1)) is greater than $10^{10000}$, and has so far taken me over a week to evaluate as I only have formulas for (a, b(2)) and (a, b(3)). Can (a, b(c)) be programmed in Mathematica? Any help would be appreciated.

Answer 1

Roman's solution gets very slow as the value b or c increases. I use Memoization below to prevent recomputing values. We should also ensure b and c are positive integers to avoid infinite recursion. My improved version that does all that is here.

Clear[f];
f[a_, b_] := a^b;
f[a_, b_, 0] := a^b;
f[a_, 1, c_Integer?Positive] := f[a,1,c] = f[a,a,c-1]
f[a_, b_Integer?Positive, c_Integer?Positive] := f[a,b,c] = f[a,b-1,c]*f[a+1,b,c-1]

Answer 2

You can write $(a,b(c))$ as f[a,b,c]:

f[a_, b_] = a^b;
f[a_, b_, 0] = f[a, b];
f[a_, 1, c_] := f[a, a, c - 1]
f[a_, b_, c_] := f[a, b - 1, c]*f[a + 1, b, c - 1]

examples:

f[1, 2, 3]
(*    15552    *)

f[3, 2, 1]
(*    432    *)

f[4, 4, 4]
(*    19393983827718432382942097517695828647998395142202754928834510298705433141104807237509816216475619585550730636347520400973355250128278033815331515175151864224862276548634782796395405263354886224119286448795750466960120269181334154615594100146469053340917944759980663872462182165749651283641595899665986474356606642225885307385859438341031898136637724087018412941823577813683669698799074676100519019046540622228028082879315049653248249344010970369039835332134345637888000000000000000000000000000000000000000000000000000000    *)