I know that in the literature there are interesting articles involving the sequence of Ramanujan primes, I refer the [*Ramanujan Prime*](https://mathworld.wolfram.com/RamanujanPrime.html) 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.**
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> 
> 


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.*](https://en.wikipedia.org/wiki/Benford%27s_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:
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[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).



New experiments involving Ramanujan primes: Benford's law
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](https://arxiv.org/abs/0907.5232) $Ram\_n$ 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.


