I know that in the literature there are interesting articles involving the sequence of Ramanujan primes, I refer the Ramanujan Prime from the online encyclopedia Wolfram MathWorld. This week I wondered what about experimental mathematics concerning this sequence of prime numbers (in the past I've known for example the article $1/f$ noise in the distribution of prime numbers by Marek Wolf, Physica A: Statistical Mechanics and its Applications Vol. 241, Issue 3–4, (1997), pp. 493-499). I have thought the following question inspired by the article Un nuevo patrón en los números primos that the author Bartolo Luque refers in his column Juegos matemáticos of a scientific journal, the Spanish edition of Scientific American, that is the journal Investigación y Ciencia, pages 91-93 (Julio 2019). He is author of [1].
Question. Do the first-digit frecencies of Ramanujan primes satisfy a Benford's law? Many thanks.
Thus I'm asking if you know how to deduce, or to disprove it, (showing your computational evidence or reasonings) if first-digit frecuencies of Ramanujan primes obey some Benford's law. I add that Wikipedia has an article dedicated to Benford's law.
I hope that my question is interesting, feel free to add comments about it. Also if my question is in the literature answer it as a reference request.
References:
[1] Bartolo Luque and Lucas Lacasa, The First-Digit Frequencies of Prime Numbers and Riemann Zeta Zeros, Proceedings: Mathematical, Physical and Engineering Sciences Vol. 465, No. 2107 (Jul. 8, 2009), pp. 2197-2216 (Royal Society).
Best Answer
If I am following what is being asked, the answer is no.
Set $R$ to be the set of Ramanujan primes. Let $R_d$ be the set of Ramanujan primes with lead digit $d$. For a set of positive integers integers $S$, we'll write $S(x)$ to be the number of elements in S which are at most $x$. Then you are asking whether for any $d=1,2, \cdots 9$ we have $$\lim_{x \rightarrow \infty}\frac{R_d(x)}{R(x)} =\log_{10} \left(\frac{d+1}{d}\right)$$
This statement is false. Let $Ram_n$ be the $n$th Ramanujan prime. Then by Sondow's theorem $Ram_n$ is asymptotic to the $2n$th prime, which is asymptotic to $2n \log 2n \sim 2n \log n$ by the prime number theorem. So $$R(x) \sim \frac{x}{2\log x}.$$
Now, set $x=(10^t)$ Then we have (neglecting small error terms) $$R_9(x) \geq R(x)) - R(\frac{9}{10}x) = \frac{x}{2 \log x} - \frac{\frac{9}{10}x}{2 \log x} = \frac{x}{10 (2)\log x} .$$
So for this set of value of $x$ we have $R_9(x)/R(x)$ is at least about $\frac{1}{10}$. But $\log_{10} \frac{10}{9}$ is much smaller, a little under $0.046.$
Note that this proof really doesn't use anything deep about the Ramanujan primes other than their asymptotic. In general, for any set of integers $S$ where $$S(x) \sim \frac{cx}{(\log x)^k}$$ for some positive constants $c$ and $k$, it will fail the base $10$ version of Benford's law. And this will apply to any other base $b>2$ by the same reasoning. Base $b$ will always have too many elements starting with $b-1$. (And of course in base $b=2$ Benford's law is trivial.) In order to have a Benford's law distribution one generally needs to be growing at least as slowly as $x^{\alpha}$ for some $\alpha <1$ or not have a good asymptotic at all. Edit: Actually see Will's comment below, one in fact needs a much stronger density requirement than this to have any hope of satisfying Benford's law.