Continuum between addition, multiplication and exponentiation? I noticed this old post which attempts to find the shades of grey between a linear and log scale where results are between zero and one.
However, I was looking for the more general case where we find the continuum between the operators themselves - add, multiply, and exponentiation or even higher.
The function could be a hyperoperation$(n,x,y)$, where $x$ and $y$ are the numbers to operate on, and $n$ is allowed to be a non-integer number for the type of operator. So $n=1$ would be 'addition', $n=2$ would be multiplication, and $n=3$ would be exponentiation. But one would also have $n=2.5$ to be the shade of grey BETWEEN multiplication and exponentiation, or $n=3.5$ or $4$ for beyond exponentiation and tetration).
I want the formula to work with real numbers, not just integers. Any thoughts?
These related posts may also be of interest:
Why are addition and multiplication commutative, but not exponentiation?
Algorithm for tetration to work with floating point numbers
 A: 
I was looking for the more general case where we find the continuum between the operators themselves - add, multiply, and exponentiation or even higher.

I think that this is stictly related to the extension of the Ackermann Function to non-integer values: I guess we can call the problem "non-integer ranks problem" because it has to do with Hyperoperations with non integer rank/index $s$
$$H_s(x,y)=x[s]y=x\uparrow^{s-2}y=G(s,x,y)$$
where

$H_1(x,y)=x[1]y=x\uparrow^{-1}y=G(1,x,y)=x+y$
$H_2(x,y)=x[2]y=x\uparrow^{0}y=G(2,x,y)=xy$
$H_3(x,y)=x[3]y=x\uparrow^{1}y=G(3,x,y)=x^y$
$H_4(x,y)=x[4]y=x\uparrow^{2}y=G(4,x,y)={}^{y}x$ (Tetration)

Few weeks ago I wrote a detailed answer about this on MSE and I think that it can be interesting to your question
Example $x$, $y$ and $z$ values for $x\uparrow^\alpha y=z$ where $\alpha\in \Bbb R-\Bbb N$
Anyways in short words what I say there is that there is not, as far as I know, a known and accepted method to find non-integer ranks hyperoperations.
A: My research doesn't focus on tetration as much as it does iterated functions, which give me a line of mathematical attack against tetration and the higher hyperoperators. 
First identify a fixed point for an iterated function which can be a hyperoperator, 
$f^n(x)$. Now $f$ is a "horizontal or intra-hyperoperator function." And $g$ is a "vertical or inter-hyperoperator function".
Consider that $g(1\uparrow^x1)=(1\uparrow^{x+1}1)$ with $(1\uparrow^\infty1) = 1$ as a fixed point. 

Mathematica 12 Code
$g[1]=1\uparrow1, g[2]=1\uparrow^21, g[3]=1\uparrow^31$
g[x_] := Interpolation[{2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}, 
     InterpolationOrder -> 12][x]
Plot[g[x], {x, 1, 10}, AxesOrigin -> {1, 1}]


Note the period two nature of the function $g$ and the convergence to the fixed point $(1\uparrow^\infty1) = 1$. So $(1\uparrow^x1)$ extends the Ackermann function to real values. 

Let $1\le a<2$. I conjecture that $a\rightarrow\infty\rightarrow\infty=a$
