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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).

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    $\begingroup$ I think the answer depends on what distribution over Ramanujan primes you choose. $\endgroup$
    – Will Sawin
    Commented Aug 31, 2021 at 15:09
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    $\begingroup$ Well, here is a question for you: Is there a Benford's law for the first digits of natural numbers? Can you state it? How would you test it empirically? $\endgroup$
    – Will Sawin
    Commented Aug 31, 2021 at 21:22
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    $\begingroup$ @Will, I don't see the problem here. You look at the first $n$ terms of a sequence; you count how many begin with the digit one (say); you divide that by $n$; you see whether the quotient has a limit as $n\to\infty$. If it does, and if that limit is $\log2$, then the sequence satisfies Benford (for the digit "one"). For the sequence of natural numbers, the limit doesn't exist. $\endgroup$
    – Gerry Myerson
    Commented Aug 31, 2021 at 23:48
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    $\begingroup$ It wouldn't be too hard for OP to include the definitions of "Ramanujan prime" and "Benford's Law" instead of making others users chase them offsite, would it? $\endgroup$
    – Gerry Myerson
    Commented Aug 31, 2021 at 23:52
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    $\begingroup$ Benford's law is essentially nothing but an expression for the invariant (Haar) measure on the multiplicative group of positive reals. Therefore, asking whether a certain distribution satisfies Benford's law is the same as asking whether this distribution is asymptotically invariant in the multiplicative scale. $\endgroup$
    – R W
    Commented Sep 1, 2021 at 0:56

1 Answer 1

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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.

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    $\begingroup$ I don't understand the last comment. Let $S$ be a set of integers and let $f(n)$ be the number of such integers less than $n$. Then if $S$ satisfies Benford's law, we have $f( (k+1) \cdot 10^n)= (1+o(1)) f( k \cdot 10^n)$ for any $1 \leq k \leq 9$, since otherwise the frequency of numbers with leading digit $k$ would be different at the two points. Combining this for $k$ from $1$ to $n$, we get $f( 10^{n+1}) = (1+o (1)) f(10^n)$, i.e. $f(n)$ grows slower than $n^{\alpha}$ for any $\alpha$. So $\alpha<1$ or the lack of a good asymptotic is not really enough. $\endgroup$
    – Will Sawin
    Commented Sep 1, 2021 at 2:15
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    $\begingroup$ @WillSawin You are absolutely correct. I convinced myself when I made that satisfying Benford's law implied that one was growing no more than $x^\alpha$ for any $\alpha$. I did not have a set in mind that satisfied it. You are correct to point out that the much stronger statement about what the set needs to satisfy is in fact true. $\endgroup$
    – JoshuaZ
    Commented Sep 1, 2021 at 2:24
  • $\begingroup$ Many thanks for your answer, and your attention and the attention of yours colleagues (professors) of this post and my recently deleted post about mathematicians (now I was disconnected). I need some days to read and understand the comment thread of this post and your excellent answer. $\endgroup$
    – user142929
    Commented Sep 1, 2021 at 8:02
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    $\begingroup$ The fraction $R_d(x)/R(x)$ should oscillate, with minima near $d10^t$ and maximal near $(d+1)10^t$. I suspect density might exist and satisfy Benford's law in logarithmic density sense. $\endgroup$
    – Wojowu
    Commented Sep 1, 2021 at 12:17

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