Can the Riemann hypothesis be relaxed to say that this matrix A consists of square roots?

Question

I realize that asking this question is like presenting to a patent attorney a wheel-less skateboard while asking to patent a hoverboard.

Anyways. Lagarias version of the Riemann hypothesis sets a bound on the sum of divisors:

$$\sigma(n) \le H_n + e^{H_n} \ln H_n$$

where $H_n$ is a harmonic number.

The von Mangoldt function matrix, as I call it, can be generated from the matrix product of two matrices:

$$A = \left( \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 1 & \sqrt{2} & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & \sqrt{3} & 0 & 0 & 0 & 0 \\ 1 & \sqrt{2} & 0 & \sqrt{4} & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & \sqrt{5} & 0 & 0 \\ 1 & \sqrt{2} & \sqrt{3} & 0 & 0 & \sqrt{6} & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & \sqrt{7} \\ \vdots&&&&&&&\ddots \end{array} \right)$$

which is equal to $A(n,k)=\sqrt{k}$ if $k$ divides $n$, else $A(n,k)=0$

The matrix inverse of $A$ is by its terms essentially equal to the matrix:

$$B = \left( \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots \\ 0 & -\sqrt{2} & 0 & -\sqrt{2} & 0 & -\sqrt{2} & 0 \\ 0 & 0 & -\sqrt{3} & 0 & 0 & -\sqrt{3} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -\sqrt{5} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \sqrt{6} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -\sqrt{7} \\ \vdots&&&&&&&\ddots \end{array} \right)$$

which is equal to $B(n,k)=\mu(n)\sqrt{n}$ if $n$ divides $k$, else $A(n,k)=0$

where $\mu(n)$ is the Möbius function

defined by:

$$\mu(n)=\begin{cases} (-1)^{\omega(n)}=(-1)^{\Omega(n)} &\text{if }\; \omega(n) = \Omega(n)\\ 0&\text{if }\;\omega(n) \ne \Omega(n).\end{cases}$$

or as in the Wikipedia page:

• $\mu(n) = 1$ if $n$ is a square-free positive integer with an even number of prime factors.
• $\mu(n) = -1$ if $n$ is a square-free positive integer with an odd number of prime factors.
• $\mu(n) = 0$ if $n$ has a squared prime factor.

The von Mangoldt function matrix is then the matrix product $A$ times $B$:

$$T = A.B = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

And the von Mangoldt function is then:

$$\Lambda(n) = \sum\limits_{k=1}^{k=\infty}\frac{T(n,k)}{k}$$

as proven by joriki here.

or as the Dirichlet generating functions of the columns as proven here by GH from MO at Mathoverflow:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$

Here comes the hoverboard / wheel-less skateboard:

Since according to the explicit formula, the von Mangoldt function is a sum of logarithmic square root waves as follows:

$$\sum_{n=1}^{n=k} \Lambda(n) = \Re\left(-\sum _{j=1}^{\infty} \left(\frac{x^{1-\rho _j}}{1-\rho _j}+\frac{x^{\rho _j}}{\rho _j}\right)-\frac{1}{2} \log \left(1-\frac{1}{x^2}\right)+x-\log (2 \pi )\right)$$

or as a Mathematica one-liner from Alex Kontorovich web page:

Plot[Re[X - Log[2 Pi] - Log[1 - 1/X^2]/2 - 
   Sum[X^(N[ZetaZero[j]])/(N[ZetaZero[j]]) + 
     X^(1 - N[ZetaZero[j]])/(1 - N[ZetaZero[j]]), {j, 1, 30}]], {X, 
  1.1, 30}]

Can the Riemann hypothesis be relaxed/be made precise to say that the so called von Mangoldt function matrix $T$ is a matrix product $T=A.B$ as in the example above?

(*Matrix T Mathematica  8*)
nn = 32;
A = Table[
   Table[If[Mod[n, k] == 0, k^(ZetaZero[k]), 0], {k, 1, nn}], {n, 1, 
    nn}];
B = Table[
   Table[If[Mod[k, n] == 0, MoebiusMu[n]*n^(ZetaZero[-n]), 0], {k, 1, 
     nn}], {n, 1, nn}];
MatrixForm[T=N[A.B]]

It appears to work for any complex number sequence in the exponents as long as the sum of the two matrices $A$ and $B$'s respective real parts is equal to 1, and the imaginary parts are each others negatives. In other words a condition that applies to any two complex number sequences of that form, of which the zeta zeros are a subset, so no progress.

To demonstrate this I have made this variant of the program above:

(*Matrix T Mathematica 8 start*)nn = 32;
a = Table[RandomComplex[], {n, 1, 32}]
A = Table[
   Table[If[Mod[n, k] == 0, k^(a[[k]]), 0], {k, 1, nn}], {n, 1, nn}];
B = Table[
   Table[If[Mod[k, n] == 0, MoebiusMu[n]*n^(1 - a[[n]]), 0], {k, 1, 
     nn}], {n, 1, nn}];
MatrixForm[T = Chop[N[A.B]]]
(*end*)

which produces matrix $T$. This probably has to do with the elementary fact that:

$$n=n^{a} n^{1-a}$$

$n$ is here a substitute for the terms in matrix $T$.

So for some arbitrary complex number sequence $a$ like for example:

$$a=0.771518+0.640552I,0.192739+0.923147I,0.931096+0.758704I,...$$

or the non-trivial Riemann zeta zeros:

$$a=0.5 + 14.1347 I, 0.5 + 21.022 I, 0.5 + 25.0109 I,...$$ we have in general the matrices:

$$A = \left( \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 1 & 2^{a_2} & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 3^{a_3} & 0 & 0 & 0 & 0 \\ 1 & 2^{a_2} & 0 & 4^{a_4} & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 5^{a_5} & 0 & 0 \\ 1 & 2^{a_2} & 3^{a_3} & 0 & 0 & 6^{a_6} & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 7^{a_7} \\ \vdots&&&&&&&\ddots \end{array} \right)$$

$$B = \left( \begin{array}{ccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots \\ 0 & -2^{1-a_2} & 0 & -2^{1-a_2} & 0 & -2^{1-a_2} & 0 \\ 0 & 0 & -3^{1-a_3} & 0 & 0 & -3^{1-a_3} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -5^{1-a_5} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 6^{1-a_6} & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -7^{1-a_7} \\ \vdots&&&&&&&\ddots \end{array} \right)$$

Which have the same property as the earlier matrices $A$ and $B$ producing matrix $T$ as the matrix product:

$$T = A.B = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$