There was an interesting discussion on the functional differential equation
$$f'(x) = f(f(x)) \tag{1a}$$
The essence of which was to find a solution of $(1a)$ by Taylor expansion of $f$ about a point $x=a$, i.e.
$$f(x,a) = \sum_{k\ge 0} \frac{(x-a)^k}{k!} f^{(k)}(a)\tag{2}$$
Matters simplify considerably if $a$ is assumed to be a fixed point of $f$, i.e.
$$f(a) = a\tag{1b}$$
This program was carried out to a large extent in the reference quoted.
Here I'd like to draw attention to the emerging sequence of integers which in itself is of interest, and could contribute the the estmation of the radius of convergence of $(2)$.
We can conveniently split our goal in two tasks:
task 1:
Calculate the $n$-th derivative of a function $f(x)$ which obeys $(1a)$ and $(1b)$
task 2:
Find the general term of the integer sequence found in task 1.
Here is what I did so far.
To start task 1 let us calculate the first few derivatives at a general $x$ as well as at the fixed point. We will be using the abbreviated notation $f^{(k)}(x)$ for the k-th derivative, and $f_k(x)$ for the k-fold iterated function $f(x)$.
Then we have
$$\begin{align*}\frac{d}{dx}f(x) &=f^{(1)}(x)\overset{\text{(1a)}}= f(f(x)) = f_2 (x) \end{align*}\tag{3a}$$ $$f^{(1)}(a) = f_2(a) = a\tag{3b}$$
$$\begin{align*}f^{(2)}(x) &= \frac{d}{dx}f^{(1)}(x) =\frac{d}{dx}f(f(x)) = f'(f(x))\cdot\frac{d}{dx}f(x)\\& \overset{\text{(1a)}}=f(f(f(x))) \cdot f(f(x))= f_3(x) \cdot f_2(x)\end{align*} \tag{4a}$$ $$f^{(2)}(a) =f_3(a) \cdot f_2(a) = a\cdot a = a^2\tag{4b}$$
Before we continue we note that the derivative of the k-fold iterated function is, for $k \ge 2$, given by
$$\frac{d}{dx} f_k(x) =f_{k+1}(x)f_{k}(x)f_{k-1}(x)\cdots f_{2}(x) \tag{5a}$$
and at the fixpoint we have (notice that since $f_{1}(a)=a$ and $f_{k+1}(a)=f(f_{k}(a))$ we find that $f_{k}(a)=a$ for all $k$) so that
$$\frac{d}{dx} f_k(x) |_{x \to a} =f_{k+1}(a)f_{k}(a)f_{k-1}(a)\cdots f_{2}(a)=a^k \tag{5b}$$
Now continuing
$$f^{(3)}(x)= \frac{d}{dx}f^{(2)}(x)=\frac{d}{dx}\left(f_3(x) \cdot f_2(x)\right)=\frac{d}{dx}\left(f_3(x)\right) \cdot f_2(x)+f_3(x) \frac{d}{dx}\left(f_2(x)\right)\\ = \left(f_4(x)f_3(x)f_2(x)\right)f_2(x) + f_3(x)\left(f_3(x)f_2(x)\right) =f_2(x)^2 f_3(x) f_4(x) + f_2(x) f_3(x)^2$$
$$f^{(3)}(x)|_{x \to a} =a^2 a a + a a^2= a^4+a^3$$
Now one step more
$$f^{(4)}(x)= \frac{d}{dx}f^{(3)}(x)= \frac{d}{dx}\left(f_2(x)^2 f_3(x) f_4(x) + f_2(x) f_3(x)^2\right)\\ =2 f_2 (f_3 f_2 ) f_3 f_4 + f_2^2 (f_4 f_3 f_2 ) f_4 +f_2 ^2 f_3 (f_5 f_4 f_3 f_2 ) + (f_3 f_2 ) f_3^2+ 2 f_2 f_3 (f_4 f_3 f_2 )\\ =2 f_2^2 f_3^2 f_4+f_2^3 f_3 f_4^2 +f_2 ^3 f_3^2 f_4 f_5+f_2 f_3^3 + 2 f_2^2 f_3^2 f_4 \\ =4 f_2^2 f_3^2 f_4+f_2^3 f_3 f_4^2 +f_2 ^3 f_3^2 f_4 f_5+f_2 f_3^3$$
$$f^{(4)}(x)|_{x \to a} =4 a^2 a^2 a + a^3 a a^2 +a^3a^3 a a + a a^3\\ =a^4+4a^5 + a^6+a^7$$
A litte Mathematica code yields more
$$ \begin{array}{l} \{\text{d1f(a)},a\} \\ \left\{\text{d2f(a)},a^2\right\} \\ \left\{\text{d3f(a)},a^4+a^3\right\} \\ \left\{\text{d4f(a)},a^7+a^6+4 a^5+a^4\right\} \\ \left\{\text{d5f(a)},a^{11}+a^{10}+4 a^9+8 a^8+11 a^7+11 a^6+a^5\right\} \\ \left\{\text{d6f(a)},a^{16}+a^{15}+4 a^{14}+8 a^{13}+22 a^{12}+22 a^{11}+60 a^{10}+58 a^9+66 a^8+26 a^7+a^6\right\} \\ \left\{\text{d7f(a)},a^{22}+a^{21}+4 a^{20}+8 a^{19}+22 a^{18}+38 a^{17}+76 a^{16}+122 a^{15}+220 a^{14}+319 a^{13}+387 a^{12}+553 a^{11}+424 a^{10}+302 a^9+57 a^8+a^7\right\} \\ \end{array} $$
Writing generally
$$f^{(n)}(x)|_{x \to a} =\sum_{k=0}^{\infty}c_{n,k}a^k$$
task 2 ist to find a formula for the coefficient fields $c_{n,k}$.
For small $n$ we have
$$c_{2} = (1)$$ $$c_{3} = (1,1)$$ $$c_{4} = (1,4,1,1)$$ $$c_{5} = (1,11,11,8,4,1,1)$$ $$c_{6} = (1,26,66,58,60,22,22,8,4,1,1)$$ $$c_{7} = (1,57,302,424,553,387,319,220,122,76,38,22,8,4,1,1)$$
I tried to identify sequences in the OEIS library but with only modest success:
If read horizontally, no match was found.
If read vertically,
(1,4,11,26,57) -> https://oeis.org/A000295, Eulerian numbers (Euler's triangle: column k=2 of A008292, column k=1 of A173018). (Formerly M3416 N1382)
(1,11,66,302) -> https://oeis.org/A000460 Eulerian numbers (Euler's triangle: column k=3 of A008292, column k=2 of A173018). (Formerly M4795 N2047)
(8,58,424) -> no match
Here I am stuck. Can you do better?
