New commutative hyperoperator?

Question

After reading about Ackermann functions , tetration and similar, I considered the commutative following hyperoperator ?

$$ F(0,a,b) = a + b $$ $$ F(n,c,0) = F(n,0,c) = c $$ $$ F(n,a,b) = F(n-1,F(n,a-1,b),F(n,a,b-1)) $$

I have not seen this one before in any official papers. Why is this not considered ? Does it grow to slow ? Or to fast ?

It seems faster than Ackermann or am I wrong ?

Even faster is The similar

$$ T(0,a,b) = a + b $$ $$ T(n,c,0) = T(n,0,c) = n + c $$ $$ T(n,a,b) = T(n-1,T(n,a-1,b),T(n,a,b-1)) $$

which I got from a friend.

Notice if $nab = 0 $ then $T(n,a,b) = n + a + b $.

One possible idea to extend these 2 functions to real values , is to extend those “ zero rules “ to negative ones.

So for instance for the case $F$ :

$$ F(- n,a,b) = a + b $$ $$ F(n,-a,b) = -a + b $$ $$ F(n,a,-b) = a - b $$

The downside is this is not analytic in $n$.

Any references or suggestions ??

Answer 1

Let $F_n^{(a)}(b)=F_n(a,b)=F(n,a,b)$ and likewise $A_n(m)=A(n,m)$ for the Ackermann function.

Let $F_n(a)=F_n(a,a)$ as the symmetric case.

It is trivial to see this is increasing in all arguments. Thus we can make the following bounds:

$$F_n(a,b)\le F_{n-1}(\max\{F_n(a-1,b),F_n(a,b-1)\})$$

By induction, we can show that the optimal choice of $m$ will only subtract from one side, leading to

$$m=F_n(\min\{a,b\}-1,\max\{a,b\})$$

WLOG assume $a>b$. This thus gives us:

$$F_n^{(a)}(b)\le F_{n-1}(F_n^{(a)}(b-1)),~F_n^{(a)}(0)=a$$

In the end, this makes $F_n(a,b)$ behave like the usual Ackermann function with the base case modified from $A_n(0)=A_{n-1}(1)$ to $F_n^{(a)}(0)=a$.

So it's easy to see that this grows similarly to the Ackermann function and does not grow much faster or slower. The same analysis holds for $T$.