Consider $t_n$ as the Thue-Morse sequence. Let $m$ be a positive integer and $s$ a complex number, and recall that the Odiuos numbers are the indices of nonzero entries in the Thue-Morse sequence. Now consider the sequence of functions below:


This is the zeta function valid for $\mathrm{Real}(s)>1$.


This is the alternating zeta function valid for $\mathrm{Real}(s)>0$.

$$f(3,s)=1-2^{-s}-3^{-s}+4^{-s}+5^{-s}-6^{-s}-7^{-s}+8^{-s}+\dotsb = 4^{-s} (\zeta(s,1/4) - \zeta(s,2/4) - \zeta(s,3/4) + \zeta(s,4/4) ) $$

( $\zeta(s,a)$ is Hurwitz zeta )

I'm not sure if this has an official name yet but it clear that it is valid for $\mathrm{Real}(s)>-1$. This sequence of functions is constructed in the similar way the Thue-Morse sequence is constructed.

$$\begin{align} &\vdots\\ f(\infty,s)&= \sum (-1)^{t_n} n^{-s} \end{align}$$

This is a nice generalization/variant of the Riemann Zeta function and the Dirichlet eta or Dirichlet $L$-functions. It follows that $f(m, s)$ is valid for $\mathrm{Real}(s)>-m+2$. Now there are two logical questions analogue to the questions about the Riemann Zeta function:

  1. What are the functional equations for $f(m,s)$?

  2. Call the $N^\text{th}$ zero $Z_n(m)$. Are all the zero's of $f(m,s)$ for any $m$ with $0<\mathrm{Real}(s)<1$ on the critical line $(\mathrm{Real}(Z_N(m))=1/2)$ ?

  3. Is clearly a generalizations of the Riemann Hypothesis. And I think it might be true! (I made some plots that were convincing but the accuracy was low.)

I wonder if these functions have a name yet and what the answers to the 2 logical questions are. I also invite the readers to make more conjectures and variants with this.


Update :


User reuns “ improved “ the definition and has solved the problem of finding the functional equation :

Quote :

The standard methods for Dirichlet L-functions apply.

  • Let $$h_k(t) = t\prod_{m=0}^{k-1}(1 - t^{2^m}) = \sum_{n=1}^{2^k} a_k(n)t^n$$ $$F_k(s) = \sum_{n=1}^\infty a_k(n \bmod 2^k) n^{-s} $$

$$f_k(x)=\sum_{n=1}^\infty a_k(n \bmod 2^k) e^{- n x}= \frac{h_k(e^{-x})}{1-e^{-2^k x}}$$ Note $h_k(1) = 0$ so $f_k$ is $C^\infty(\mathbb{R})$.

  • For $\Re(s) > 0$ $$\Gamma(s) F_k(s) = \int_0^\infty x^{s-1}\frac{h_k(e^{-x})}{1-e^{-2^k x}}dx$$ For $\Re(s) > -K-1$ $$\Gamma(s) F_k(s) = \sum_{r=0}^K \frac{f_k^{(r)}(0)}{r!} \frac1{s+r}+ \int_0^\infty x^{s-1}(\frac{h_k(e^{-x})}{1-e^{-2^k x}}-1_{x < 1}\sum_{r=0}^K \frac{f_k^{(r)}(0)}{r!} x^r) dx$$ Thus $\Gamma(s) F_k(s)$ is meromorphic everywhere with simple poles at negative integers and $F_k(s)$ is entire.

  • Functional equation : Poisson summation formula, same method as for Dirichlet L-functions and $\sum_n \chi(n) e^{-\pi n^2 x}$.

    Let $\sum_{n=0}^{2^k-1} a_k(n \bmod 2^k) e^{2i \pi mn/2^k}= h_k(e^{2i \pi m/2^k})$ the discrete Fourier transform of $a_k(n \bmod 2^k)$. Then

$$\sum_{n=1}^\infty a_k(n \bmod 2^k) e^{- \pi n^2 x} = (2^k x)^{-1/2} \sum_{m=1}^\infty \frac{h_k(e^{2i \pi m/2^k})}{2^k} e^{- \pi m^2 2^k/ x}$$

$$F_k(s)\Gamma(s/2)\pi^{-s/2}2^{sk/2}= \int_0^\infty x^{s/2-1} \sum_{n=1}^\infty a_k(n \bmod 2^k) e^{- \pi n^2 x/2^k}dx$$ $$= \int_1^\infty (x^{s/2-1} \sum_{n=1}^\infty a_k(n \bmod 2^k) e^{- \pi n^2 x/2^k}+x^{(1-s)/2-1}\sum_{m=1}^\infty \frac{h_k(e^{2i \pi m/2^k})}{2^{k/2}} e^{- \pi m^2 x/2^k}) dx$$

So $F_k(s)$ is a Dirichlet series with functional equation. The standard tools apply, density of zeros, explicit formula for $\log F_k, 1/F_k, F_k'/F_k$ and their Dirichlet series coefficients in term of the non-trivial zeros. But since the $a_k(n \bmod 2^k)$ aren't multiplicative, no Euler product, no Riemann hypothesis.

  • the limit $F_\infty(s) = \lim_{k \to \infty}F_k(s)$. Some properties of the $F_k$ are preserved (the analytic continuation), some are not (functional equation, density of zeros). Asking about a Riemann hypothesis for $F_\infty$ doesn't really make sense.

—- End quote —-

So we are only left with the positions of the zero’s. In particular $f(3,s),f(4,s),f(oo,s)$ interest me. Plots are also nice.

There are however 2 other open question ; “3),4) “ in the related thread :

See comment

Is this zeta-type function meromorphic?


Update :

I used 3 approximating methods to find nonreal zero’s of $f(3,s)$.

And they kinda suggest the same thing.

Those methods are

1) contour type integrals such as the argument principle. 2) riemann mapping a rectangle to a circle and then consider the appropriate taylor series that converges in that circle. 3) truncated dirichlet series like e.g. $\sum_{n=1}^{80} a_n n^{-s} = 0 $

Numerical precision is pretty low and iT deels pretty hard to compute without a computer , assuming no closed forms for the integrals in method 1).

Anyway these are suggested :

$f(3,s)$ has it non-real zero’s always close to the lines $Re(z) = {-1,0,\frac{1}{5},\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{1}{3},\frac{2}{3},\frac{4}{3},1}$

In particular the lines $Re(z) = {-1,0,1,\frac{2}{3}} $ seems very attracting.

Someone proved it for $Re(s) = 1$.

The zero’s i found with confidence are

$$ s = \frac{2}{3} + \frac{62}{9} i , s = \frac{2}{3} + \frac{467}{18} i $$

Appproximately. ( or exact !? )

Notice there are also real zero’s.

Perhaps $f(m,s) $ always has infinitely Many zero’s near $Re(s) = \frac{m-1}{m}$.


  • 2
    $\begingroup$ why would $f(3,s)$ be related with any Riemann hypothesis function ? It has much more chances to be related with the Hurwitz $\zeta$ having no Riemann hypothesis $\endgroup$ – reuns Jun 11 '16 at 6:58
  • $\begingroup$ Maybe plotting zero's would be insightful .. $\endgroup$ – mick Aug 5 '16 at 10:59
  • $\begingroup$ Finding zero’s for these functions with pencil and paper is a nightmare !! Lol. Or maybe I am missing something. $\endgroup$ – mick Nov 28 '18 at 20:57
  • $\begingroup$ Related : math.stackexchange.com/questions/3018712/… $\endgroup$ – mick Dec 3 '18 at 3:12
  • $\begingroup$ I would advise being less readily convined by a plot! There's a perpetual all-world project to find the zeros of the ordinary Riemann zeta function that has verified that all the zeros up to some diabolickally large number lie on the critical line ... and mathematicians still are not convinced! Some say things like that if it could be verified for the first 10^1000 zeros, they might begin to feel some conviction! $\endgroup$ – AmbretteOrrisey Dec 3 '18 at 20:43

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