Puzzling Prime Properties

Predicting the number of primes less than nn 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×(N1)×(N2)×...×2×1N!=N\times(N-1)\times(N-2)\times...\times2\times1

Prime-factor all of the terms on the right:

N×(N1)×(N2)×...×2×1=pNpνp(N!)N\times(N-1)\times(N-2)\times...\times2\times1 = \displaystyle\prod_{p\leq N} p^{\nu_p(N!)}

where νp(x)\nu_p(x) is the "pp-adic valuation of xx" (exponent of pp in the prime factorization of xx).

Take the natural logarithm of both sides:

lnN!=pNνp(N!)lnp\ln N! = \sum_{p\leq N} \nu_p(N!) \ln p

Recall Stirling's asymptotic formula for the factorial:

N!2πNN+12eNN! \sim \sqrt{2\pi} \frac{N^{N+\frac{1}{2}}}{e^N}

    lnN!(N+12)lnNN+ln(2π)\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 NlnNN\ln{N} when NN is very large, so

lnN!NlnN\ln N! \sim N\ln N

So:

pNνp(N!)lnpNlnN\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!N!. Let's focus on νpN!\nu_p{N!}. One of Legendre's many theorems (1808) tells us that

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

Now, NpkNpk\displaystyle\lfloor{\frac{N}{p^k}}\rfloor \approx \frac{N}{p^k} Here's where the guessing comes in: as NpkNpk\displaystyle\lfloor{\frac{N}{p^k}}\rfloor \approx \frac{N}{p^k}, I conjecture that 1klogpNNpkNk=1Npk\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 k=1Npkk=1pk\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 NN. As such,

νpN!k=1pk=N(111p1)=Np1\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

pNNp1lnpNlnN\displaystyle\sum_{p\leq N} \frac{N}{p-1} \ln p \sim N\ln N

    pNlnpp1lnN\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 pNlnpp1\sum_{p\leq N} \frac{\ln p}{p-1} (green points) v.s. lnN\ln N (red line) from N=2N=2 to N=20N=20. Created using Desmos

Note by Andrei Li
4 months, 2 weeks ago

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@Andrei Li- I also find a proof for prime number theorem

Zakir Husain - 3 months, 2 weeks ago

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I tried to find some more interesting results. See here

Zakir Husain - 4 months, 1 week ago

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

Recall that lnNN(N1N1)\ln N \sim N(N^{\frac{1}{N}}-1). If we substitute for my result, we get

pNpp1p1p1N(N1N1)\sum_{p\leq N} p\frac{p^{\frac{1}{p}}-1}{p-1} \sim N(N^{\frac{1}{N}}-1)     pNp[k=1p1pkp]1N(N1N1)\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:

pN(p1p1)N(N1N1)\sum_{p\leq N} (p^{\frac{1}{p}}-1) \sim N(N^{\frac{1}{N}}-1)     pNp1ppN1N(N1N1)\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!

pNp1pπ(N)N(N1N1)\sum_{p\leq N} p^{\frac{1}{p}} - \pi(N) \sim N(N^{\frac{1}{N}}-1) pNp1pπ(N)lnN\sum_{p\leq N} p^{\frac{1}{p}} - \pi(N) \sim \ln{N}

If only we could evaluate that pesky pNp1p\sum_{p\leq N} p^{\frac{1}{p}}...

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

Andrei Li - 3 months, 4 weeks ago

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

A Former Brilliant Member - 3 months, 2 weeks ago

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Brilliant Mathematics Staff - 3 months, 2 weeks ago

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