# Puzzling Prime Properties

Predicting the number of primes less than $n$ is a contentious mathematical subject. Here are some of my thoughts, from messing around with factorials and the natural log. Note that many of my arguments are not rigorous. Feel free to give feedback below.

By definition,

$N!=N\times(N-1)\times(N-2)\times...\times2\times1$

Prime-factor all of the terms on the right:

$N\times(N-1)\times(N-2)\times...\times2\times1 = \displaystyle\prod_{p\leq N} p^{\nu_p(N!)}$

where $\nu_p(x)$ is the "$p$-adic valuation of $x$" (exponent of $p$ in the prime factorization of $x$).

Take the natural logarithm of both sides:

$\ln N! = \sum_{p\leq N} \nu_p(N!) \ln p$

Recall Stirling's asymptotic formula for the factorial:

$N! \sim \sqrt{2\pi} \frac{N^{N+\frac{1}{2}}}{e^N}$

$\implies \ln N! \sim (N +\frac{1}{2})\ln N - N + \ln (\sqrt{2\pi})$

All the other terms on the right hand are small compared to $N\ln{N}$ when $N$ is very large, so

$\ln N! \sim N\ln N$

So:

$\displaystyle\sum_{p\leq N} \nu_p(N!) \ln p \sim N\ln N$

So far, we have managed to link the natural logarithm with the number of primes in the prime factorization of $N!$. Let's focus on $\nu_p{N!}$. One of Legendre's many theorems (1808) tells us that

$\nu_p(N!) = \displaystyle\sum_{1\leq k \leq \lfloor{\log_p{N}}\rfloor} \lfloor{\frac{N}{p^k}}\rfloor$

Now, $\displaystyle\lfloor{\frac{N}{p^k}}\rfloor \approx \frac{N}{p^k}$ Here's where the guessing comes in: as $\displaystyle\lfloor{\frac{N}{p^k}}\rfloor \approx \frac{N}{p^k}$, I conjecture that $\displaystyle\sum_{1\leq k \leq \lfloor{\log_p{N}}\rfloor} \lfloor{\frac{N}{p^k}}\rfloor \sim N\displaystyle\sum_{k=1}^N p^{-k}$, and in turn that $\displaystyle\sum_{k=1}^N p^{-k} \sim \sum_{k=1}^\infty p^{-k}$. I will assume from now on that this is true for large $N$. As such,

$\nu_p{N!} \sim \sum_{k=1}^\infty p^{-k} = N(\frac{1}{1-\frac{1}{p}}-1) = \frac{N}{p-1}$

We thus have

$\displaystyle\sum_{p\leq N} \frac{N}{p-1} \ln p \sim N\ln N$

$\implies\boxed{\sum_{p\leq N} \frac{\ln p}{p-1} \sim \ln N}$ $\sum_{p\leq N} \frac{\ln p}{p-1}$ (green points) v.s. $\ln N$ (red line) from $N=2$ to $N=20$. Created using Desmos Note by Andrei Li
1 year, 1 month ago

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## Comments

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I tried to find some more interesting results. See here

- 1 year, 1 month ago

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Cool! I wonder if we could go further than that...

Recall that $\ln N \sim N(N^{\frac{1}{N}}-1)$. If we substitute for my result, we get

$\sum_{p\leq N} p\frac{p^{\frac{1}{p}}-1}{p-1} \sim N(N^{\frac{1}{N}}-1)$ $\implies \sum_{p\leq N} p[\sum_{k=1}^{p-1} p^{\frac{k}{p}}]^{-1} \sim N(N^{\frac{1}{N}}-1)$

For your result:

$\sum_{p\leq N} (p^{\frac{1}{p}}-1) \sim N(N^{\frac{1}{N}}-1)$ $\implies \sum_{p\leq N} p^{\frac{1}{p}} - \sum_{p\leq N} 1 \sim N(N^{\frac{1}{N}}-1)$

The second term on the left hand is the prime counting function!

$\sum_{p\leq N} p^{\frac{1}{p}} - \pi(N) \sim N(N^{\frac{1}{N}}-1)$ $\sum_{p\leq N} p^{\frac{1}{p}} - \pi(N) \sim \ln{N}$

If only we could evaluate that pesky $\sum_{p\leq N} p^{\frac{1}{p}}$...

P.S. I used \lfloor and \rfloor to format the floor function.

- 1 year ago

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@Brilliant Mathematics, @Emma Wilson has posted a comment not related to mathematics.

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Staff - 1 year ago

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