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

Prime-factor all of the terms on the right:

N×(N−1)×(N−2)×...×2×1=∏p≤Npνp(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:

ln⁡N!=∑p≤Nνp(N!)ln⁡p\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}

  ⟹  ln⁡N!∼(N+12)ln⁡N−N+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 Nln⁡NN\ln{N} when NN is very large, so

ln⁡N!∼Nln⁡N\ln N! \sim N\ln N

So:

∑p≤Nνp(N!)ln⁡p∼Nln⁡N\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!)=∑1≤k≤⌊log⁡pN⌋⌊Npk⌋\nu_p(N!) = \displaystyle\sum_{1\leq k \leq \lfloor{\log_p{N}}\rfloor} \lfloor{\frac{N}{p^k}}\rfloor

Now, ⌊Npk⌋≈Npk\displaystyle\lfloor{\frac{N}{p^k}}\rfloor \approx \frac{N}{p^k} Here's where the guessing comes in: as ⌊Npk⌋≈Npk\displaystyle\lfloor{\frac{N}{p^k}}\rfloor \approx \frac{N}{p^k}, I conjecture that ∑1≤k≤⌊log⁡pN⌋⌊Npk⌋∼N∑k=1Np−k\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=1Np−k∼∑k=1∞p−k\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=1∞p−k=N(11−1p−1)=Np−1\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

∑p≤NNp−1ln⁡p∼Nln⁡N\displaystyle\sum_{p\leq N} \frac{N}{p-1} \ln p \sim N\ln N

  ⟹  ∑p≤Nln⁡pp−1∼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 ∑p≤Nln⁡pp−1\sum_{p\leq N} \frac{\ln p}{p-1} (green points) v.s. ln⁡N\ln N (red line) from N=2N=2 to N=20N=20. Created using Desmos

#NumberTheory

Note by Andrei Li
1 year, 1 month ago

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Comments

I tried to find some more interesting results. See here

Zakir Husain - 1 year ago

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

Recall that ln⁡N∼N(N1N−1)\ln N \sim N(N^{\frac{1}{N}}-1). If we substitute for my result, we get

∑p≤Npp1p−1p−1∼N(N1N−1)\sum_{p\leq N} p\frac{p^{\frac{1}{p}}-1}{p-1} \sim N(N^{\frac{1}{N}}-1)   ⟹  ∑p≤Np[∑k=1p−1pkp]−1∼N(N1N−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:

∑p≤N(p1p−1)∼N(N1N−1)\sum_{p\leq N} (p^{\frac{1}{p}}-1) \sim N(N^{\frac{1}{N}}-1)   ⟹  ∑p≤Np1p−∑p≤N1∼N(N1N−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!

∑p≤Np1p−π(N)∼N(N1N−1)\sum_{p\leq N} p^{\frac{1}{p}} - \pi(N) \sim N(N^{\frac{1}{N}}-1) ∑p≤Np1p−π(N)∼ln⁡N\sum_{p\leq N} p^{\frac{1}{p}} - \pi(N) \sim \ln{N}

If only we could evaluate that pesky ∑p≤Np1p\sum_{p\leq N} p^{\frac{1}{p}}...

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

Andrei Li - 1 year ago

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

A Former Brilliant Member - 1 year ago

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