Solutions to $f(x+1) = 1 + f(x) + f(x/2)^2 + f(x/3)^3 + f(x/4)^4+...$ ? $f(x) = O(\frac{\exp(x)}{x+1})$?

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Let $x$ be real and let $n$ be a natural number.

Consider the following functional equation :

$$f(0) = 0$$ $$f(1)=1$$ $$f(x+1) = 1 + f(x) + f(x/2)^2 + f(x/3)^3 + f(x/4)^4+...$$

What are the solutions and what are the asymptotics ? How many free parameters do we have ? Are all solutions analytic ?

obviously related

$$g(0) = 0$$ $$g(1)=1$$ $$g'(x) = 1 + g(x/2)^2 + g(x/3)^3 + g(x/4)^4+...$$

And

$$h(0) = 0$$ $$h(1)=1$$ $$h(n+1) = 1 + h(n) + h(n/2)^2 + h(n/3)^3 + h(n/4)^4+...$$

Where the division is rounded below to the integer part.

Notice the connection to number theory.

If $n$ is close to an interval with many primes, it will relatively grow slower as usual. ( A similar thing probably happens for $f$ and $g$ )

My brutal estimate for $f,g,h$ and $x>0$ is

$$f(x) = O(g(x)) = O(h(x)) = O(\frac{\exp(x)}{x+1})$$

Using big-O notation.

Maybe asymptotics relate to number theory functions ? Or to differential equation ideas ?

I noticed

$$f(x) = O(\frac{\exp(x)}{x+1}) = O(\pi(\exp(x)))$$

maybe ??

Notice the higher derivatives of $f(x)$ can be given by the defining equation and taylors theorem. Not sure if that helps.

I wonder if and how $f(-x)$ and $f(x)$ relate and similar with $g$.

The focus here is on the real imput $x$ or natural integer $n$, although comments about complex $z$ ($f(z),g(z)$) are welcome ofcourse.