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A relaxed $\zeta$-extensions of the Recursive Hyperoperations

I want to show you an easy extension for hyperoperations.
I don't want it to be the most natural, but I want to ask if someone already used this extension and if it can be usefull for something.

Since is a bit different I want to use the plus-notation ($+_{\sigma}$) for the hyperoperators.

I start with these basic definitions over the naturals $b,n,\sigma \in \mathbb{N}$:

$o)\,\,\,S(n)=n+1$

$o')\,\,\,B_b( \sigma+1):=
\begin{cases}
b, & \text{if} \sigma=0 \\
0, & \text{if} \sigma=1 \\
1, & \text{if} \sigma\gt 1 \\
\end{cases}$

Then the recursive definitions of the operators $b,n,\sigma \in \mathbb{N}$

$i)\,\,\,b+_0 n=S(n)$

$ii)\,\,\,b+_{\sigma+1}0=B_b(\sigma+1)$

$iii)\,\,\,b+_{\sigma+1}S(n)=b+_{\sigma}(b+_{\sigma+1}n)$

Observation before the extension's definitons

$b+_0 n=1+n$

$b+_1 n=b+n$

we can see that from rank zero to rank one we can define infinite functions $b+_\sigma n=\zeta_\sigma+n$ with $1\lt\zeta_\sigma\lt b$

$b+_0 n=\zeta_{0}+n=1+n$

$b+_{0.5} n=\zeta_{0.5}+n$

$b+_1 n=\zeta_{1}+n=b+n$

Generalizing, now we can define $\zeta_b$ as a continous functions from the interval $0$ to $1$, to the interval $1$ to $b$:

$Eiv)\,\,\,\zeta_b:[0,1]\rightarrow [1,b]$

$Ev)\,\,\,\zeta_b(\varepsilon)=\begin{cases}
1, & \text{if \varepsilon=0} \\
b, & \text{if \varepsilon=1 } \\ \end{cases}$

And we can define the operations with fractional rank starting from the interval $[0,1]$

$Evi)\,\,\, b +_{\varepsilon}n=\zeta _b(\varepsilon)+_{1}n \,\, \text{ and} \,\, \varepsilon \in [0,1]$

Other operations are these ( $\varepsilon \in ]0,1]$ and $k \in \mathbb{N}$ ):

$b +_{k+\varepsilon}n=\zeta _b(\varepsilon)+_{k+1} n$

Example of $\zeta _b$ functions and the generated $\zeta _b$-hyperoperations:

$\zeta _b(\varepsilon)=b^\varepsilon$ and $k \in \mathbb{N}$

for $\varepsilon \in ]0,1]$ and $k \in \mathbb{N}$
$b +_{k+\varepsilon}n=b^\varepsilon+_{k+1} n$