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JoshuaZ
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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.

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.

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.

Comment about what would be needed for a set of satisfy it.
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JoshuaZ
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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) $$R_9(x) \geq R(x)) - R(\frac{9}{10}x) = \frac{x}{2 \log x} - \frac{\frac{9}{10}x}{2 \log x} = \frac{x}{10 \log x} .$$$$R_9(x) \geq R(x)) - R(\frac{9}{10}x) = \frac{x}{2 \log x} - \frac{\frac{9}{10}x}{2 \log x} = \frac{x}{10 (2)\log x} .$$

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 $b$$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.

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) $$R_9(x) \geq R(x)) - R(\frac{9}{10}x) = \frac{x}{2 \log x} - \frac{\frac{9}{10}x}{2 \log x} = \frac{x}{10 \log x} .$$

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 $b$ 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.)

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.

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JoshuaZ
  • 7k
  • 2
  • 27
  • 59

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) $$R_9(x) \geq R(x)) - R(\frac{9}{10}x) = \frac{x}{2 \log x} - \frac{\frac{9}{10}x}{2 \log x} = \frac{x}{10 \log x} .$$

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 $b$ 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.)