Interesting commutative hyperoperators ?

[MphLee]

\[b\bullet _{n+2} c = n +1+ b + c +  ((b-1)\bullet _{n+2} c) \bullet _{n+1} (b\bullet _{n+2}(c-1)) +((b-1)\bullet _{n+2} c) \bullet _{n} (b\bullet _{n+2}(c-1)) \]

\[b \bullet _{1} c=1\]
\[b \bullet _{0} c=0\]

Eg.
\(b\bullet _{2} c =1+ b + c +  ((b-1)\bullet _{2} c) \bullet _{1} (b\bullet _{2}(c-1)) +((b-1)\bullet _{2} c) \bullet _{0} (b\bullet _{2}(c-1)) \)
\(b\bullet _{2} c =b + c +  2 \)

\(b\bullet _{3} c = 2+ b + c +  ((b-1)\bullet _{3} c) \bullet _{2} (b\bullet _{3}(c-1)) +((b-1)\bullet _{3} c) \bullet _{1} (b\bullet _{3}(c-1)) \)
\(b\bullet _{3} c = 2+ b + c +  ((b-1)\bullet _{3} c) + (b\bullet _{3}(c-1)) +2 +1 \)
\(b\bullet _{3} c = 5+ b + c +  ((b-1)\bullet _{3} c) + (b\bullet _{3}(c-1)) \)

fix \(b=1\) define the function \(f(\,c)=1\bullet_3 c \) and \(g(\,c)=0\bullet_3 c+1\)

\(1\bullet _{3} (c+1)= 6 + c +  (0\bullet _{3} c+1) + 1\bullet _{3}c \)
\(f (c+1)= 6 + c +  g(\,c) + f(\,c) \) i.e. \( fS=6+I+g+f\)...

Why this should matter, is it completely defined?

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If I had to define a Fibonacci-like recursion I'd do it like that: let \(F_n:\mathbb R^2\to \mathbb R\) be the "definendum" family of binary functions, and \(A(x,y),B(x,y)\) two fixed binary functions.

Additive Fibonacci operations
\(F_0(x,y)=A(x,y),\,\,\,F_1(x,y)=B(x,y)\)
\(F_{n+2}(x,y)=F_{n}(x,y)+F_{n+1}(x,y)\)

If we let \(A\) be addition and \(B\) be multiplication we get

\(F_2(x,y)=x+y+xy\) an important operation in the field of formal groups.
\(F_3(x,y)=x+y+2xy\), \(F_3(x,y)=2x+2y+3xy\), \(F_4(x,y)=3x+3y+5xy\) ...

Outer Compositional Fibonacci operations be composition of binary function defined as diagonal followed by composition
\(F_0=A,\,\,\,F_1=B\)
\(F_{n+2}=F_{n}\circ F_{n+1}\)

I.e. \(F_{n+2}(x,y)=F_{n}(F_{n+1}(x,y),F_{n+1}(x,y))\).

The computations shows that for first steps addition and multiplication \(F_{2}(x,y)=2xy\), \(F_3(x,y)=2x^2y^2\),...

Inner Compositional Fibonacci operations be composition of binary function defined as diagonal followed by composition
\(F_0=A,\,\,\,F_1=B\)
\(F_{n+2}=F_{n+1}\circ F_{n}\)

I.e. \(F_{n+2}(x,y)=F_{n+1}(F_{n}(x,y),F_{n}(x,y))\).

For \(A=+,\,B=\cdot\) we get 
\(F_{2}(x,y)=(x+y)^2\),
\(F_3(x,y)= 4(xy)^2\),
\(F_3(x,y)= 4(x+y)^4=4 x^4 + 16 x^3 y + 24 x^2 y^2 + 16 x y^3 + 4 y^4\),
\(F_4(x,y)= 64(xy)^8\)...