10
$\begingroup$

I was first investigating the sequence (A111000) defined by the function $$a_n = \lfloor\zeta(\zeta(n))\rfloor$$ where $\zeta(n)$ is the Riemann Zeta Function and $$\lfloor x \rfloor = \max\{m \in \mathbb{Z} \mid m \leq x\}$$ also known as the floor function. The sequence begins, $a_n = 2, 5, 12, 27, 58, 120, ... \text{ for } n = 2, 3, 4, 5, 6, 7, ...$

I then tried investigating other functions relating to the Zeta Function that would satisfy the requirement that the positive infinite limit would tend toward infinity. This led me to $$b_n = \left\lfloor\frac{1}{\zeta(n)-1}\right\rceil$$ where $\lfloor x \rceil = \left\lfloor x + \frac{1}{2} \right\rfloor$. And secondly, $$c_n = \lceil\Gamma(\zeta(n)-1)\rceil$$ where $\Gamma(n)$ is the Gamma Function and $$\lceil x \rceil = \min\{n \in \mathbb{Z} \mid n \geq x\}$$ I found that for some reason, the generated sequences were very similar. $$a_n = 2, 5, 12, 27, 58, 120, 245, 498, 1006, 2024, 4064, 8149, 16327, 32692, ...$$ $$b_n = 2, 5, 12, 27, 58, 120, 245, 498, 1005, 2024, 4064, 8149, 16327, 32692, ...$$ $$c_n = 2, 5, 12, 27, 58, 120, 245, 498, 1005, 2023, 4064, 8149, 16327, 32692, ...$$ The seemed to be the same for almost all $n$. I tested this up to $n = 1000$ and found the following observations to be true up to that point. $$a_n \geq b_n \geq c_n \text{, and by Squeeze Theorem, if } a_n = c_n \text{ then } a_n = b_n = c_n$$ $$a_n - c_n \leq 1$$ I am curious why these values are so similar. Is there some equality or near equality I do not know about. Is it possible to prove my observations? How?

$\endgroup$
  • $\begingroup$ You should expect the first two to be close; $\zeta(n)$ will be very close to $1$ as $n\to\infty$ (in fact, $\approx 1+\frac1{2^n}$, so it converges pretty quickly to $1$!), and it's a classical result that $\zeta(1+\epsilon) \approx \epsilon^{-1}$, which is really the core content of $a_n\approx b_n$. $\endgroup$ – Steven Stadnicki May 1 '18 at 6:48
2
$\begingroup$

The closeness of these terms follows form the Stieltjes series expansion of the Riemann Zeta function and the series expansion of the gamma fucntion. We have

$$ \zeta(s) = \frac{1}{s-1} + \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\gamma_n(s-1)^n $$

where $\gamma_n$ are the Stieltjes constants.

$\endgroup$

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.