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Let $p_n$ be the $n$-th prime and $d_n$ be the integer in the prime gap $(p_n,p_{n+1})$ such that $d_n$ has more divisors than any other integer in that gap. If two or more integers in the gap have the largest number of divisors then we take $d_n$ to be the smallest among them. Experimental data shows that the sequence of ratios $\frac{p_{n+1} - d_n}{p_{n+1} - p_n}$ approaches uniform distribution in $(0,1)$ as $n \to \infty$.

Motivation: This is somewhat counter intuitive in the sense that bigger number tend to have more divisors so my naïve guess that that the distribution of $d_n$ should be skewed towards $p_{n+1}$ but experimental data showed otherwise.

Experimental verification: To test this conjecture, I evaluated the limit below for $n = 10^{10}$ and for different functions $f(x) $and observed that its value agreed with the integral on the RHS.

$$ \lim_{n \to \infty} \frac{1}{n}\sum_{r=1}^n f \left(\frac{d_r - p_r}{p_{r+1} - p_r}\right) = \lim_{n \to \infty} \frac{1}{n}\sum_{r=1}^n f \left(\frac{p_{r+1} - d_r}{p_{r+1} - p_r}\right) = \int_{0}^1 f(x)dx $$

Question: Does $\frac{p_{n+1} - d_n}{p_{n+1} - p_n}$ approach uniform distribution in $(0,1)$ as $n \to \infty$?

Related: Riemann sum formula for definite integral using of prime numbers

Nilotpal Sinha
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  • Numbers without small prime factors are not so rare as one might think. The probability that a huge random number has no prime factor below $10^{10}$ is $1:41$ – Peter Aug 03 '23 at 23:43
  • "Bigger numbers tend to have more divisors"—this is only true on average over very large intervals, not tiny intervals like the ones consecutive primes create. – Greg Martin Sep 05 '23 at 05:42

1 Answers1

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I can give intuition but not full solution.

Its highly like that $\prod_{k \in S} m_{k}$ which lies between $p_n$ and $p_{n+1}$ has the greatest number of divisors. where $m_i$ is the $i^{th}$ primorial number.

See https://en.wikipedia.org/wiki/Primorial $m_i \approx e^{i \log(i)}$

See https://en.wikipedia.org/wiki/Prime_number_theorem $p_n \approx n \log(n)$

Further A conjecture in https://en.wikipedia.org/wiki/Prime_gap $p_{n+1} - p_n \approx C \times \log(n)^2$

So $\frac{p_{n+1}-\prod_{k=j}^i m_k}{p_{n+1}-p_n} \approx \frac{(n+1) \log(n+1) - \prod_{k \in S} e^{k \log(k)}}{ C \times \log(n)^2}$

So we want the set $S$ such that $p_n \leq \prod_{k \in S} m_k \leq p_{n+1} \implies n \log(n) \leq \prod_{k \in S} e^{k \log(k)} \leq (n+1) \log(n+1)$.

I think you need more accurate estimates of $p_n$ in the above to get a good estimate of gap $p_{n+1}-d_n$.

See if any idea in the above helps.