Is there a known solution to $f(x) = (1-x)f(x^2)$? The functional equation $f(x) = (1-x)f(x^2)$ (with $f(0)=1$) has a simple solution that can be expressed as a rapidly converging infinite product
$$f(x) = \prod_{n=0}^\infty (1 - x^{2^n}) = (1-x)(1-x^2)(1-x^4)(1-x^8)\cdots$$
and as a more slowly converging power series
$$f(x) = \sum_{n=0}^\infty (-1)^{H(n)}x^n = 1 - x - x^2 + x^3 -\cdots,$$
where $H(n)$ is the Hamming weight of $n$ (i.e., the number of 1's in the binary representation of $n$).
Is anything known about this function $f$?  Does it have a simple description in terms of more familiar functions?
 A: (I made this comment an answer so that the question can be "completed" by "accepting" this answer) 
There is an interesting discussion around the Thue-Morse-sequence/constant in the following two articles of J. Shallit/JP Allouche, where the function in question occurs in a wider context.
See links at Jeffrey Shallit's homepage: 


*

*The ubiquituous Thue-Morse-Sequence (this is a postscript file
and can be converted (for instance by Ghostscript) to pdf to be readable.) In the paper your
formula occurs in proposition 2.      

*A powerpoint-like overview is in this pdf-file (somehow a summary, possibly an extension) of the first paper. Again you find your function discussed (in short)
and referenced from a wider context (see page 36 "Another definition").                  


[update] I included the hint to Ghostscript due to the comments to this answer
A: 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) $ 
