19
$\begingroup$

I found the following formula in another math group:

$$\large\color{blue}{\pi\left( \dfrac{\left( \pi!\right)!-\lceil \pi \rceil \pi! }{{\pi}^{\sqrt \pi}-\pi!}\right)=2020}$$

It actually looks very "elegant". But, then I used Wolfram because I was in doubt.. The result shows that this formula is wrong.

I wrote these steps:

$${\qquad \quad \color{red}{\pi\left( \dfrac{\left( \pi!\right)!-\lceil \pi \rceil \pi! }{{\pi}^{\sqrt \pi}-\pi!}\right)=\color{blue}{\dfrac {\pi \Gamma(1 + \Gamma(1 + \pi))-4 \pi \Gamma(1 + \pi))}{{\pi}^{\sqrt \pi} - \Gamma(1 + \pi)}}\color{red}{\approx55221,71}}\color{blue}{\neq2020}}$$

My questions are:

  • I wonder if there is a small typo in the formula?
  • Or is the formula far from accurate in any case?
$\endgroup$
32
$\begingroup$

The outer $\pi$ is the prime-counting function.

$\endgroup$
  • 5
    $\begingroup$ Or, $\pi(17577.6)=2020$ $\endgroup$ – QC_QAOA Jan 1 at 21:13
  • 1
    $\begingroup$ In Mathematica, PrimePi[17577.61614280863]=2020 $\endgroup$ – Pixel Jan 1 at 21:13
  • 3
    $\begingroup$ @Henry so we have "two different pi" in the same formula? Do I understand correct? $\endgroup$ – Elementary Jan 1 at 21:28
  • 2
    $\begingroup$ @Elementary in this case, yes, different meanings are intended. From a table of primes: 2018 17551 2019 17569 2020 17573 2021 17579 2022 17581 oeis.org/A000040/b000040.txt $\endgroup$ – Will Jagy Jan 1 at 21:40
  • 3
    $\begingroup$ From Wolfram MathWorld: "The notation pi(n) for the prime counting function is slightly unfortunate because it has nothing whatsoever to do with the constant pi=3.1415.... This notation was introduced by number theorist Edmund Landau in 1909 and has now become standard. In the words of Derbyshire (2004, p. 38), "I am sorry about this; it is not my fault. You'll just have to put up with it." $\endgroup$ – DJohnM Jan 1 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.