Existence of an entire function with algebraically independent derivatives

Solution 1:

If any $f^{(n)}$ is constant, then $f^{(n+1)}=f^{(n+2)}=\cdots=0$, and algebraic independence over $\Bbb Q$ (in the sense defined above) fails everywhere. Assume then that $f^{(1)}$ is not constant, so for some $K\in\{1,2\}$, $g_K:=Kf^{(1)}+f^{(2)}$ is not constant. Then by the Open Mapping Theorem, for any $z_0\in\Bbb C$ and $\epsilon>0$, $g_K(B_\epsilon(z_0))$ is open, so it contains algebraic numbers. However, if $g_K(z)$ is algebraic, then either neither of $f^{(1)}(z)$ and $f^{(2)}(z)$ is algebraic, or both are, and if neither are then they are algebraically dependent. Therefore, for any sequence, algebraic independence over $\Bbb Q$ (in the sense defined above) fails at a set which is dense in $\Bbb C$.

More generally, for any polynomial $P$ with rational coefficients, unless $P(f^{(i_1)}(z), \dots, f^{(i_k)}(z))$ is constant, the Open Mapping Theorem can likewise be applied to it, meaning that there will be a dense set on which $\{f^{(i_1)}(z),\dots,f^{(i_k)}(z)\}$ are algebraically dependent over $\Bbb Q$ (in the usual meaning of the phrase algebraically dependent, not the sense defined above.) So, I don't think that anything like what the questioner hopes for will be possible. The best you can do is to have a collection of dense sets of various forms of algebraic dependence, permeating the complex plane.