Solutions for $Re(z) > 2$ ,$f(z) = \sum_{1<i}^{\infty} f(i)^{\dfrac{z-i+1}{i}}$?

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Consider the functions defined by the equations and conditions

$$Re(z) > 2$$

$$f(i+1) > f(i)-2$$

$$ f(z) = \sum_{1<i}^{\infty} f(i)^{\dfrac{z-i+1}{i}}$$

Does that make sense ? Does that converge ?

Suppose it does not make sense for some divergent solutions, does there exist a kind of continuation ?

What solutions exist ? How many solutions exist ? What are the asymptotics ?

Is it always analytic for all $2 < Re(z)$ ?

In particular for $f(i)$ being real and positive.