Is this irregular curve asymptotic to $\log (2) (\log (x)-\log (2))^2-\frac{\log (2)}{2}$, or is the asymptotic something else?

Question

Let $a(n)$ be the Dirichlet inverse of the Euler totient function:

$$a(n) = \sum\limits_{d|n} d \cdot \mu(d) \tag{1}$$

And let the matrix $T$ be: $$T(n,k)=a(\gcd(n,k)) \tag{2}$$

which has the property that the row sums are the von Mangoldt function:

$$\Lambda(n) = \sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k} \tag{3}$$

which has been proven by both joriki and GH from MO for $n>1$.

The Möbius transform of the Harmonic numbers is then:

$$\sum\limits_{d|n} \mu(d) H_{n/d}=\sum_{k=1}^{k=n}\frac{T(n,k)}{k} \tag{4}$$

Because the Möbius transform of the Harmonic numbers is similar to the von Mangoldt function:

$$\sum\limits_{d|n} \mu(d) H_{n/d} \sim \Lambda(n) \tag{5}$$

the partial sums of the left hand side of $(5)$ should also have a Riemann hypothesis.

For the right hand side the Riemann hypothesis is:

$$|\psi(x)-x|\le\frac{\sqrt x\,\log^2 x}{8\pi} \tag{6}$$

Seeking to relate the factor $\log^2 x$ with the expansion of the Möbius transform of the Harmonic numbers, we will look at a signed version of matrix $T$:

$$T_s(n,k)=(-1)^{n k} a(\gcd (n,k)) \tag{7}$$

$$\sum _{k=1}^n \frac{T_s(n,k)}{k}=\sum _{k=1}^n \frac{(-1)^{n k} a(\gcd (n,k))}{k} \tag{8}$$

Question:

Do the partial sums of $(8)$ have the asymptotic below?

$$\sum _{n=1}^x \sum _{k=1}^n \frac{(-1)^{n k} a(\gcd (n,k))}{k} \sim \log (2) (\log(x)-\log (2))^2-\frac{\log (2)}{2} \tag{9}$$ Or is it something else?

Relation to $\frac{2 \sqrt{n+1}}{\sqrt{c}} $:

https://mathoverflow.net/a/387056/25104

Associated Mathematica program for plots below:

"Mathematica start"
Clear[nn, t, n, k]; b = {1, -1}; nn = 500; 
a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]];
Monitor[g1 = 
   ListLinePlot[
    Accumulate[
     Table[Sum[(-1)^(n*k)*a[GCD[n, k]]/k, {k, 1, n}], {n, 1, nn}]]], 
  n];
g2 = Plot[(Log[n] - Log[2])^2*Log[2] - Log[2]/2, {n, 1, nn}, 
  PlotStyle -> Red, ImageSize -> Large]; Show[g1, g2]

Clear[nn, t, n, k]; b = {1, -1}; nn = 1000; 
a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]];
Monitor[g1 = 
   ListLinePlot[
    Accumulate[
     Table[Sum[(-1)^(n*k)*a[GCD[n, k]]/k, {k, 1, n}], {n, 1, nn}]]], 
  n];
g2 = Plot[(Log[n] - Log[2])^2*Log[2] - Log[2]/2, {n, 1, nn}, 
  PlotStyle -> Red, ImageSize -> Large]; Show[g1, g2]

Clear[nn, t, n, k]; b = {1, -1}; nn = 5000; 
a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]];
Monitor[g1 = 
   ListLinePlot[
    Accumulate[
     Table[Sum[(-1)^(n*k)*a[GCD[n, k]]/k, {k, 1, n}], {n, 1, nn}]]], 
  n];
g2 = Plot[(Log[n] - Log[2])^2*Log[2] - Log[2]/2, {n, 1, nn}, 
  PlotStyle -> Red, ImageSize -> Large]; Show[g1, g2]
    
    Clear[nn, t, n, k]; b = {1, -1}; nn = 10000; 
    a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]];
    Monitor[g1 = 
       ListLinePlot[
        Accumulate[
         Table[Sum[(-1)^(n*k)*a[GCD[n, k]]/k, {k, 1, n}], {n, 1, nn}]]], 
      n];
    g2 = Plot[(Log[n] - Log[2])^2*Log[2] - Log[2]/2, {n, 1, nn}, 
      PlotStyle -> Red, ImageSize -> Large];
    Show[g1, g2]
    "Mathematica end"

"Mathematica start"
Clear[nn, t, n, k, a, b];
nn = 1000;
a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]];
Monitor[g1 = 
   ListLinePlot[
    b = Accumulate[
      Table[Sum[(-1)^(n*k)*a[GCD[n, k]]/(k), {k, 1, n}], {n, 1, 
        nn}]]], n];
ListLinePlot[Re[b]/Table[Log[n]^2, {n, 1, nn}], PlotRange -> {0, 1}]

$$\sum _{n=1}^x \sum _{k=1}^n \frac{(-1)^{n k} a(\gcd (n,k))}{k}\sim c*\log(x)^{2}$$

Algebra to be solved:
Substitute minus with ""[m] and plus with ""[p] in the Mathematica code:

"For comparison: Substitute m=-1. Compare this program:"
Clear[nn, t, n, k, a];
nn = 5;
a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]];
Monitor[g1 = 
   ListLinePlot[
    accumulate = 
     Table[Sum[
        If[n >= k, (""[m]*1)^(n*k)*a[GCD[n, k]]/k, 0], {k, 1, n}]*
       Sqrt[ToString[n]], {n, 1, nn}]], n];
accumulate // Column

"And substitute p = +1. With this program:"
Clear[nn, t, n, k, a];
nn = 5;
a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]];
Monitor[g2 = 
   ListLinePlot[
    accumulate = 
     Table[Sum[
       If[n >= k, (""[p] 1)^(n*k)*a[GCD[n, k]]/k, 0], {k, 1, n}], {n, 
       1, nn}]], n];
accumulate // Column

This gives the expressions where $m=-1$:

$$\begin{array}{l} \text{}(m) \\ \left(\text{}(m)^2-\frac{\text{}(m)^4}{2}\right) \\ \left(-\frac{1}{3} 2 \text{}(m)^9+\frac{\text{}(m)^6}{2}+\text{}(m)^3\right) \\ \left(-\frac{1}{4} \text{}(m)^{16}+\frac{\text{}(m)^{12}}{3}-\frac{\text{}(m)^8}{2}+\text{}(m)^4\right) \\ \left(-\frac{1}{5} 4 \text{}(m)^{25}+\frac{\text{}(m)^{20}}{4}+\frac{\text{}(m)^{15}}{3}+\frac{\text{}(m)^{10}}{2}+\text{}(m)^5\right) \end{array}$$

times a constant $c$, times square roots.

to be compared with the expression where $p=+1$:

$$\begin{array}{l} \text{}(p) \\ \text{}(p)^2-\frac{\text{}(p)^4}{2} \\ -\frac{1}{3} 2 \text{}(p)^9+\frac{\text{}(p)^6}{2}+\text{}(p)^3 \\ -\frac{1}{4} \text{}(p)^{16}+\frac{\text{}(p)^{12}}{3}-\frac{\text{}(p)^8}{2}+\text{}(p)^4 \\ -\frac{1}{5} 4 \text{}(p)^{25}+\frac{\text{}(p)^{20}}{4}+\frac{\text{}(p)^{15}}{3}+\frac{\text{}(p)^{10}}{2}+\text{}(p)^5 \end{array}$$