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Riemann Zeta Function:

$$\frac{1}{1^{s}}+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\frac{1}{6^{s}}+\frac{1}{7^{s}}+\frac{1}{8^{s}}+\frac{1}{9^{s}}+\frac{1}{10^{s}}+\dots$$

"Reverse $2$" Riemann Zeta Function:

$$\frac{1}{1^{s}}+\color{red}{\frac{1}{\sqrt2^{s}}}+\frac{1}{3^{s}}+\color{red}{\frac{1}{2^{s}}}+\frac{1}{5^{s}}+\color{red}{\frac{1}{\sqrt{18}^{s}}}+\frac{1}{7^{s}}+\color{red}{\frac{1}{\sqrt{8}^{s}}}+\frac{1}{9^{s}}+\color{red}{\frac{1}{\sqrt{50}^{s}}}+\dots$$

"Reverse $3$" Riemann Zeta Function:

$$\frac{1}{1^{s}}+\frac{1}{2^{s}}+\color{blue}{\frac{1}{\sqrt3^{s}}}+\frac{1}{4^{s}}+\color{blue}{\frac{1}{\sqrt{35}^{s}}}+\color{blue}{\frac{1}{\sqrt{12}^{s}}}+\color{blue}{\frac{1}{\sqrt{35}^{s}}}+\frac{1}{8^{s}}+\color{blue}{\frac{1}{3^{s}}}+\frac{1}{10^{s}}+\dots$$

"Reverse $10$" Riemann Zeta Function:

$$\frac{1}{1^{s}}+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\frac{1}{6^{s}}+\frac{1}{7^{s}}+\frac{1}{8^{s}}+\frac{1}{9^{s}}+\color{green}{\frac{1}{\sqrt{10}^{s}}}+\dots$$

In general, a Reverse $x$ Riemann zeta function takes $n=1,2,3\dots$ and multiplies it with its reverse value (reverse digits) in number base $x$, then takes the square root of the product. After that, raise it to the power of $-s$ as you would for zeta(s).

As you can see, if $n$ is a palindrome in number base $x$, then it remains $n$. Otherwise, it becomes the $\sqrt{n\times\overline{n}}$ where $\overline{n}$ is the number $n$, but its digits are reversed when written in number base $x$.

The value of "Reverse Zeta $x$" of $s$ tends to $\zeta{(s)}$ as $x$ grows bigger. Suppose we take the limit $x\to\infty$, then we have a normal zeta function. It's because that makes the number base a base where each number is a single symbol (one digit), meaning all $n$s reverse back to $n$.


  • Does something on this already exist somewhere or is this the first time this is mentioned?

  • How can we check when this will converge for a given $x$?
    I suppose the bound slightly varies based on $x$ and converges to that the real part of $s$ must be $\gt1$ as $x$ tends to $\infty$?

  • Can one find closed forms for some $x$ and some $s$?
    For example, what is the (is there a) closed form of this when $x=2$ and $s=2$ ?


Can we express the "reverse $x$ zeta function" in the terms of the zeta function?

Also, I couldn't find anything on the reciprocal sums of palindromic numbers.

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    $\begingroup$ Is there any motivation behind studying these? $\endgroup$
    – Alex R.
    Commented Jun 4, 2017 at 17:47
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    $\begingroup$ I think you're missing a term in the second sum. I doubt this has been looked at - mathematicians usually don't like things that depend on a choice of base. But - I admire the creativity $\endgroup$ Commented Jun 4, 2017 at 17:51
  • $\begingroup$ en.wikipedia.org/wiki/… Since in every base the number of ways that $n$ can be represented as the product of some other number and its mirror is trivially $O(n)$, I think grouping like denominators together shows that it is defined and analytic on $Re(s) \gt 4$? And if the number of ways is $O(1)$ then that should mean the same for $Re(s) \gt 2$. $\endgroup$ Commented Jun 4, 2017 at 18:41

1 Answer 1

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Note: Not a complete answer, but here is some Mathematica code that can evaluate this sum and is good to play around with the behavior of your sum.

P[n_, x_] := If[PalindromeQ[IntegerDigits[n, x]], n, Sqrt[n * FromDigits[Reverse[IntegerDigits[n, x]], x]]]
Table[Plot[Sum[1 / (P[n, b])^s, {n, 1, x}], {x, 1, 50}, PlotLabel -> StringForm["s = `1`, base = `2`", s, b]], {s, 1, 5}, {b, 2, 4, 1}]

It's most likely not optimized, since I'm fairly new to Mathematica, but nonetheless. This will generate a table of plots of $\zeta_b(s)$ from $s = 1 \dots 5$ and $b_{base} = 2 \dots 4$, where $\zeta_b$ denotes the reverse base zeta with base $b$. To change some of the parameters: In the second line {s, 1, 5} is {s, <begin_index>, <end_index>} so you can change that to change the increment of $s$. Same goes with the base denoted as $b$. There's a free online Wolfram Lab you can use to test this if you don't have Mathematica. Below is a table of plots generated by this code, where the $x$-axis serves as the bound of the summation:

table of plots

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  • $\begingroup$ Where is this Wolfram Lab? I doubt it's Wolfram Alpha, because I already know it would choke on your code. $\endgroup$ Commented Jun 4, 2017 at 19:40
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    $\begingroup$ Just type wolfram lab into google. It's the first thing for me. You might have to make a free account first. However, it's definitely there cause that's how I ran it and got these images. $\endgroup$
    – Dando18
    Commented Jun 4, 2017 at 19:45

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