Mhh maybe I have made some errors but the solution of the eqation =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=\beta\gt 1)
then
=\beta^{ln(0)}=\beta^{-\infty}=0)
because of the limit
=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=\gamma)
and continue with the recursion
=\gamma^{ln(2)})
=\gamma^{ln(2)ln(3)})
...
=\gamma^{{\Pi}_{i=2}^{2+n} ln(i)})
So actually the solution is of the form for some
=\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
)!
=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=\alpha \neq 0)
or 
=\exp_\alpha ({\Pi}_{i=0}^{n} ln(z_0+i))=\alpha^{{\Pi}_{i=0}^{n} ln(z_0+i)})
Let assume that
I'm not sure how to continue in this case... maybe we can start by fixing the value of
...
So actually the solution is of the form for some
------
Btw I see the relation with the Bennet's commutative Hyperoperations (denote it by
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