Mhh maybe I have made some errors but the solution of the eqation \( f(x+1)=f(x)^{ln(x)} \) seems to have some problems in the first values... that should make it not determinate for natural values
Let assume that \( f(0)=\beta\gt 1 \) then
\( f(1)=\beta^{ln(0)}=\beta^{-\infty}=0 \) because of the limit
\( f(2)=0^{ln(1)}=0^0 \) is not determinate (multivalued?)
I'm not sure how to continue in this case... maybe we can start by fixing the value of \( f(2)=\gamma \) and continue with the recursion
\( f(3)=\gamma^{ln(2)} \)
\( f(4)=\gamma^{ln(2)ln(3)} \)
...
\( f(2+n+1)=\gamma^{{\Pi}_{i=2}^{2+n} ln(i)} \)
So actually the solution is of the form for some \( \gamma \)
\( f(z+1)=\gamma^{{\Pi}_{i=2}^{z} ln(i)} \)?
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Btw I see the relation with the Bennet's commutative Hyperoperations (denote it by \( a\odot^e_i b=a\odot_i b \))!
\( f_i(z+1)=f_i(z)\odot_i z \)
This is like a kind of "Hyper-factorial" that uses the Bennet's Hyperops..
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Edit: I think that the correct forumla is something of the form
Given \( f(z_0)=\alpha \neq 0 \) or \( \neq 1 \)
\( f(z_0+n+1)=\exp_\alpha ({\Pi}_{i=0}^{n} ln(z_0+i))=\alpha^{{\Pi}_{i=0}^{n} ln(z_0+i)} \)
Let assume that \( f(0)=\beta\gt 1 \) then
\( f(1)=\beta^{ln(0)}=\beta^{-\infty}=0 \) because of the limit
\( f(2)=0^{ln(1)}=0^0 \) is not determinate (multivalued?)
I'm not sure how to continue in this case... maybe we can start by fixing the value of \( f(2)=\gamma \) and continue with the recursion
\( f(3)=\gamma^{ln(2)} \)
\( f(4)=\gamma^{ln(2)ln(3)} \)
...
\( f(2+n+1)=\gamma^{{\Pi}_{i=2}^{2+n} ln(i)} \)
So actually the solution is of the form for some \( \gamma \)
\( f(z+1)=\gamma^{{\Pi}_{i=2}^{z} ln(i)} \)?
------
Btw I see the relation with the Bennet's commutative Hyperoperations (denote it by \( a\odot^e_i b=a\odot_i b \))!
\( f_i(z+1)=f_i(z)\odot_i z \)
This is like a kind of "Hyper-factorial" that uses the Bennet's Hyperops..
--------------
Edit: I think that the correct forumla is something of the form
Given \( f(z_0)=\alpha \neq 0 \) or \( \neq 1 \)
\( f(z_0+n+1)=\exp_\alpha ({\Pi}_{i=0}^{n} ln(z_0+i))=\alpha^{{\Pi}_{i=0}^{n} ln(z_0+i)} \)
Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)
\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)