Question

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 $5/5=1$ at the 10-th prime. Here is a plot of this ratio up to $10^5$:

I would expect the 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 $\frac{1}{2}$:

Answer 1

Yes. This was proven in

C. Mauduit and J. Rivat, Sur un problème de Gelfond: la somme des chires 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.

Answer 2

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.