I don't know, whether the following counts as "is anything known", but I found it interesting.

Consider the separation according to the Hamming weight:
$$ \begin{array}{}
a_0(x) &= 1 &\\
a_1(x) &= & x^1+x^2 + x^4 + x^8 + ... \\
a_2(x) &= & x^3+x^5 + x^6 + x^9 + ... \\
a_3(x) &= & x^7+x^{11} + x^{13} + x^{14} + ... \\
... & = ...
\end{array}$$ which are all convergent for $|x| \lt 1$, then
$$ s_0(x) = a_0(x) +a_1(x)+a_2(x)+ ... \\
= \sum_{k=0}^\infty a_k(x) \\
= \sum_{k=0}^\infty x^k \\
= { 1 \over 1-x } \tag2$$
and
$$ s_1(x) = a_0(x) -a_1(x)+a_2(x)- ... \\
= \sum_{k=0}^\infty (-1)^k a_k(x) \\
= \sum_{k=0}^\infty (-1)^{H(k)} x^k \\
= f(x) \tag3$$
I found it then interesting, that at $x=1/2$ the evaluation $a_1(1/2)$ is known to be a transcendental number. But what about the other $a_k(1/2)$ ? If I recall it correctly they are all transcendental numbers, but don't have it at hand how this has been shown; I think they might be rational multiples of each other): just (another) infinite set of transcendental numbers adding up to a rational one.

*[update]* If I recall correctly the $a_k(1/2)$ can be generated as rational compositions of $a_1(1/2),a_1(1/2^2),a_1(1/2^3),...$ (or very similar) applying the Newton-method of converting a sequence of powersums into symmetric polynomials. For instance,
$$a_1(1/2) \approx 0.816421509022 \\
a_2(1/2) \approx 0.175061285686 \\$$
and also
$$ [(a_1(1/2)^2-(a_1(1/2) -1/2) )/2,a_2(1/2)] \\
[0.175061285686, 0.175061285686]
$$
using Pari/GP, reflecting the second power $(x^1 + x^2 + x^4+x^8+...)^2= 2(x^3 + x^5+x^6 + ...) + (x^2+x^4+x^8+...) \\ = 2 a_2(x)+ a_1(x^2) = 2 a_2(x)+ (a_1(x)-x) $