In order to obtain a valid and consistent extension all we need is to define how we calculate one fractional operation for example $H_{\frac{3}{2}}(x,y)$ which is something between addition and multiplication. So let us try to extend one work that says that we can use arithmetic-geometric mean and obtain such consistent extension. First thing first.

Notice that Ackermann function is directly related to hyperoperations.

$$A(m,n) = \begin{cases}
0[m]n & m=0 \\
2[m](n+3)-3 & m>0 \\
\end{cases}$$

So let us repeat a more or less standard notation
$$H_n(a,b)=a[n]b$$
$$H_0(a,b)=b+1$$
$$H_1(a,b)=a+b$$
$$H_2(a,b)=ab$$
$$H_3(a,b)=a^b$$

$$H_n(a,b)=a[n]b=a[n-1](a[n](b-1))$$

According to https://www.hindawi.com/journals/mpe/2016/4356371/ we can define $H_{\frac{3}{2}}$ using arithmetic-geometric mean, $M(x,y)$, and its inverse and one constant that we can calculate since for all hyperfunctions we know that

$$H_n(2,2)=4$$

because $2+2=4$ and $H_n(2,2)=2[n]2=2[n-1](2[n]1)=2[n-1]2$

(The author is arguing that the resulting $H_{\frac{3}{2}}$ coming from AG mean corresponds to physical process, where the analysis very likely came from, and polynomial interpolation as well.)

So it is $$H_{\frac{3}{2}}(x,y)=M^{(-1)}(M(x,y),\epsilon_{\frac{3}{2}})$$

$$\epsilon_{\frac{3}{2}}=M^{(-1)}(2,4)$$

Now that we have $H_{\frac{3}{2}}(x,y)$ we can obtain integer evaluations for $H_{\frac{5}{2}}(x,n)$ simply

$$H_{\frac{5}{2}}(x,n)=H_{\frac{3}{2}}(x,H_{\frac{3}{2}}(x,...\text{n times}))$$

The same as what we have with other operations $x \cdot n = x+x+...\text{n times}$

Now that we have integer $H_{\frac{5}{2}}(x,n)$ and $H_{\frac{5}{2}}(x,n+1)$ we can find the middle point. Notice for multiplication and exponentiation this is how we find value in the middle

$$\frac{xn+x(n+1)}{2}$$

$$\sqrt[2]{x^n x^{n+1}}$$

So $H_{\frac{5}{2}}(x,n+\frac{1}{2})$ is the solution of

$$H_{\frac{3}{2}}(x,2)=H_{\frac{3}{2}}(H_{\frac{5}{2}}(x,n),H_{\frac{5}{2}}(x,n+1))$$

And the same for whatever else middle integer we want.

Once we have $H_{n+\frac1{2}}$ we use the same division procedure to obtain $H_{n+\frac{1}{4}}$ and $H_{n+\frac{3}{4}}$ and so on and we are done. Once we have all rational fractional parts, we can extend it to real numbers even though we cannot express it in some closed form.

We can easily extend this down to $H_{\frac{1}{2}}$ using $x[1+r]y=x[r](x[1+r](y-1)), 0<r<1$ (Solve $x[1+r](z-1)=y$ and then $x[r]y=x[r+1]z$)

For the second argument $y$ in $H_n(x,y)$ (which is $n$ in $A(m,n)$) all operations: addition multiplication... naturally extend to real numbers, so there is nothing to add to that end.

Finally for $r,s \geq 0$ reals, we have

$$A(r,s) = \begin{cases}
0[r]s & r=0 \\
2[r](s+3)-3 & r>0 \\
\end{cases}$$

where $x[r]y=H_{r}(x,y)$ is an extension over reals obtained by the above procedure using the n/n+1 hyper-mean.

Essentially this procedure is the same if we decide to use some another average operation between addition and multiplication. All comes to define just that and the rest of the construction is all the same.