# The Möbius function as eigenvalues

Let the $$N$$ by $$N$$ matrix $$A$$ be defined by the tetration:

$$\Large \text{If } \gcd(n, k)=1 \text{ then } A(n,k)= \underbrace{e^{e^{\cdot^{\cdot^{e^{\Re\left(\frac{1}{n^s}\right)}}}}}}_m \text{ else } A(n,k)=0$$

where:
$$n=1..N$$ $$k=1..N$$.

This a matrix starting:

$$A=\left( \begin{array}{cccccccc} e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & e^{e^{e^e}} & \dots\\ e^{e^{e^{e^{2^{-s}}}}} & 0 & e^{e^{e^{e^{2^{-s}}}}} & 0 & e^{e^{e^{e^{2^{-s}}}}} & 0 & e^{e^{e^{e^{2^{-s}}}}} \\ e^{e^{e^{e^{3^{-s}}}}} & e^{e^{e^{e^{3^{-s}}}}} & 0 & e^{e^{e^{e^{3^{-s}}}}} & e^{e^{e^{e^{3^{-s}}}}} & 0 & e^{e^{e^{e^{3^{-s}}}}} \\ e^{e^{e^{e^{4^{-s}}}}} & 0 & e^{e^{e^{e^{4^{-s}}}}} & 0 & e^{e^{e^{e^{4^{-s}}}}} & 0 & e^{e^{e^{e^{4^{-s}}}}} \\ e^{e^{e^{e^{5^{-s}}}}} & e^{e^{e^{e^{5^{-s}}}}} & e^{e^{e^{e^{5^{-s}}}}} & e^{e^{e^{e^{5^{-s}}}}} & 0 & e^{e^{e^{e^{5^{-s}}}}} & e^{e^{e^{e^{5^{-s}}}}} \\ e^{e^{e^{e^{6^{-s}}}}} & 0 & 0 & 0 & e^{e^{e^{e^{6^{-s}}}}} & 0 & e^{e^{e^{e^{6^{-s}}}}} \\ e^{e^{e^{e^{7^{-s}}}}} & e^{e^{e^{e^{7^{-s}}}}} & e^{e^{e^{e^{7^{-s}}}}} & e^{e^{e^{e^{7^{-s}}}}} & e^{e^{e^{e^{7^{-s}}}}} & e^{e^{e^{e^{7^{-s}}}}} & 0 \\ \vdots & & & & & & & \ddots \end{array} \right)$$

Let $$\lambda(n)$$ be the sequence of eigenvalues of the matrix $$A$$.

Show that $$m$$ pieces of natural logarithms of the eigenvalues $$\lambda(n)$$:

$$\text{sgn}(\lambda(n))\underbrace{\log(\log(...\log(}_m |\lambda(n)|)...))=\Re\left(\frac{\mu(n)}{n^s}\right)$$

converge to the real part of the Möbius function divided by $$n^s$$ for $$s$$ a complex number, as $$m \rightarrow \infty$$.

Mathematica program to demonstrate the conjecture:

(*start*)
Clear[A, s, mm, nn, i];
s = 1/2 + I;
mm = 7;
Do[A = Table[
Table[If[GCD[n, k] == 1, Exp[Exp[Exp[Exp[Re[1/n^s]]]]], 0], {k, 1,
nn}], {n, 1, nn}];
a = Eigenvalues[A];
Print[b =
N[Table[Sign[a[[i]]] If[a[[i]] == 0, 0,
Log[Log[Log[Log[Abs[Re[a[[i]]]]]]]]], {i, 1, nn}], 6]], {nn, 1,
mm}]
MatrixForm[A]
Total[b] - Sum[MoebiusMu[n]/n^s, {n, 1, Length[b]}]
(*end*)


The original program with a simpler claim:

(*start*)
Clear[A, s, mm, nn, i];
s = 1;
mm = 7;
Do[A = Table[
Table[If[GCD[n, k] == 1, Exp[Exp[Exp[Exp[1/n^s]]]], 0], {k, 1,
nn}], {n, 1, nn}];
a = Eigenvalues[A];
Print[b =
N[Table[Sign[a[[i]]] If[a[[i]] == 0, 0,
Log[Log[Log[Log[Abs[a[[i]]]]]]]], {i, 1, nn}], 6]], {nn, 1, mm}]
MatrixForm[A]
Total[b] - Sum[MoebiusMu[n]/n^s, {n, 1, Length[b]}]
(*end*)


And its output:

{1.00000,-0.500000,-0.333333,-0.200000,0.166667,-0.142857,0}

This question has gone unanswered for a few years now, as part of an earlier question on Mathematics stack exchange here.

Edit 27.11.2019:

This is what I had in mind:

$$\text{Fourier Transform of } \Lambda(n) \sim \Re\left(\sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n}\frac{\mu(d)}{d^{(1/2+i \cdot t-1)}}\right)$$

Where the real part of the Möbius function divided by $$d^s$$ is part of the function. In other words, the eigenvalues, with the magnitudes logarithmized $$m$$ times, accentuate the Riemann zeta zeros when multiplied with the Riemann zeta function.

More about the zeta zero spectrum

• Is $m=4$? $\,\,\,\,\,\,\,\,\,\,\,$ Nov 13, 2019 at 18:10
• In the program I used $m=4$ but $m \rightarrow \infty$. Nov 13, 2019 at 18:19
• What is the size of $A$?
– user6976
Nov 15, 2019 at 1:14
• Sounds like it could be related to Redheffer matrix and it's variants. See en.m.wikipedia.org/wiki/Redheffer_matrix
– lcv
Dec 5, 2019 at 16:14
• $$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$ Jan 2, 2020 at 19:23