Htykuut

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:

$$f(1,s)=1+2^{-s}+3^{-s}+4^{-s}+dotsb$$

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

$$f(2,s)=1-2^{-s}+3^{-s}-4^{-s}+dotsb$$

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.

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Update :

https://math.stackexchange.com/users/276986/reuns

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

$$f_k(x)=sum_{n=1}^infty a_k(n bmod 2^k) e^{- n x}= frac{h_k(e^{-x})}{1-e^{-2^k x}}$$
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.

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End quote
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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?

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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 $Re(s) = frac{m-1}{m}$.

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