TITLE: Odd-bit primes ratio
QUESTION [18 upvotes]: Say that a number is an odd-bit number if
the count of 1-bits in its binary representation is odd.
Define an even-bit number analogously.
Thus $541 = 1000011101_2$ is an odd-bit number,
and $523 = 1000001011_2$ is an even-bit number.

Are there, asymptotically, as many odd-bit primes as even-bit primes?

For the first ten primes, we have
$$
\lbrace 10, 11, 101, 111, 1011, 1101, 10001, 10011, 10111, 11101
\rbrace
$$
with 1-bits
$$
\lbrace 1, 2, 2, 3, 3, 3, 2, 3, 4, 4 \rbrace
$$
and so ratio of #odd to $n$ is $5/10=0.5$ at the 10-th prime.
Here is a plot of this ratio up to $10^5$:




(Vertical axis is mislabeled: It is #odd/$n$.)


I would expect the #odd/$n$ ratio to approach $\frac{1}{2}$, except perhaps the fact that primes ($>2$) are
odd might bias the ratio.  The above plot does not suggest convergence
by the 100,000-th prime (1,299,709).
Pardon the naïveness of my question.
Addendum: Extended the computation to the $10^6$-th prime (15,485,863), where it still
remains 1.5% above $\frac{1}{2}$:

REPLY [24 votes]: Yes. This was proven in
C. Mauduit and J. Rivat, Sur un problème de Gelfond: la somme des chiffres des nombres premiers, Ann. Math.  
I found this by searching for "evil prime" and "odious prime" in the OEIS. More precisely, they prove the Gelfond conjecture: 

Let $s_q(p)$ denote the sum of the digits of $p$ in base $q$.  For $m, q$ with $\gcd(m, q-1) = 1$ there exists $\sigma_{q,m} > 0$ such that for every $a \in \mathbb{Z}$ we have
$$| \{ p \le x : s_q(p) \equiv a \bmod m \} | = \frac{1}{m} \pi(x) + O_{q,m}(x^{1 - \sigma_{q,m}})$$
where $p$ is prime and $\pi(x)$ the usual prime counting function.

REPLY [21 votes]: Yes, the ratio approaches 1/2.  This was proven in
C. Mauduit et J. Rivat, Sur un probléme de Gelfond: la somme des chiffres des nombres premiers.
See Three topics in additive prime number theory for exposition.  Also, the poorly-named sequences in Sloane: A027697 and A027699.