Suppose $f$ is a function in $z$ such that for all $z$, $$f\left(\frac{1}{z}\right)=-f(z).$$ (Examples are $\frac{1-z}{1+z}$ and $\log z$.) It turns out that for odd $k$, $f^{(k)}(1)$ is independent of $f^{(j)}(1)$ for all $j<k$, whereas for even $k$, $f^{(k)}(1)$ depends on all $f^{(j)}(1)$ for odd $j<k$.
Is there a way to see that without computing individual derivatives and successively solving for $f^{(k)}(1)$? Perhaps, combinatorially? This can be derived from the properties of the Lah triangle, but I wonder if there is a proof that avoids those heavy computations. To see why the Lah numbers are involved, see exercise 7 on page 157-158 of Comtet (the case of $\alpha=-1$).