Yes, some years ago I thought about a similar hierarchy, though in a different context of converting exponential binary trees into analytic functions.
To give an impression of this operator I compute the first three functions starting with \( f(x)=x+1 \). To make the notation more extensible I use \( I \) for the operator, that maps \( f \) to \( x\mapsto f^{\circ f(x)}(x) \).
\( I^0 f = f = x \mapsto x+1 \), \( (I^0f)^{\circ n}(x) = x + n \)
\( I^1 f = x \mapsto 2x+1 \), \( (I^1f)^{\circ n}(x) = 2^n x + 2^n - 1 \)
\( I^2 f = x\mapsto 2^x(x+1) - 1 \), \( (I^2 f)^{\circ n} = ? \)
And of course next we would put \( f(x) \) into the exponent of the operator and so on, getting nice hierarchy of just too quickly growing functions.
This hierarchy can already be built on positive integer functions, so no analytic something needed.
I guess the growth of \( I^n (x\mapsto x+1) \) is equal to the growth of \( x\mapsto b [n] x \).
To give an impression of this operator I compute the first three functions starting with \( f(x)=x+1 \). To make the notation more extensible I use \( I \) for the operator, that maps \( f \) to \( x\mapsto f^{\circ f(x)}(x) \).
\( I^0 f = f = x \mapsto x+1 \), \( (I^0f)^{\circ n}(x) = x + n \)
\( I^1 f = x \mapsto 2x+1 \), \( (I^1f)^{\circ n}(x) = 2^n x + 2^n - 1 \)
\( I^2 f = x\mapsto 2^x(x+1) - 1 \), \( (I^2 f)^{\circ n} = ? \)
And of course next we would put \( f(x) \) into the exponent of the operator and so on, getting nice hierarchy of just too quickly growing functions.
This hierarchy can already be built on positive integer functions, so no analytic something needed.
I guess the growth of \( I^n (x\mapsto x+1) \) is equal to the growth of \( x\mapsto b [n] x \).
