Mathematica knows that:

$$\log(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{n^{(s - 1)}}\right)$$

The von Mangoldt function should then be:

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


So in the case of the logarithm of $2$, log(2), this is equal to a table (red and green) called $\zeta(s)$:

$k \mid n : 1$ else $0$.

matrix multiplied by a second table (red, green and yellow) called $(1 - 2^{1-s})$:

$n=k: 1$ else if $n=2 \cdot k: -2$ else $0$

Picture of tables $\zeta(s)$ and $(1 - 2^{1-s})$ [Image 1]:

![Zeta(s) and (1-2^(s-1)][1]

Taking the matrix product of the two tables above we get [Image 2]:

![Zeta(s) times (1-2^(s-1)][2]

The first column then has the numerators of the alternating series.

This is as said:

$$\log(2) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - 2^{(1 - s)}\right)$$

which with a sign change in the exponent is equal to:

$$\log(2) = \lim\limits_{s \rightarrow 1} \zeta(s)\left(1 - \frac{1}{2^{(s - 1)}}\right)$$

Before I get to the actual question I show the Dirichlet series for logarithm of n.

We had already:

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}... =\log(2)$$

which is the second column in:

Logarithms of $n$:

$$T_1 = \begin{bmatrix} 0&0&0&0&0&0&0 \\ 1&-1&1&-1&1&-1&1 \\ 1&1&-2&1&1&-2&1 \\ 1&1&1&-3&1&1&1 \\ 1&1&1&1&-4&1&1 \\ 1&1&1&1&1&-5&1 \\ 1&1&1&1&1&1&-6 \end{bmatrix}$$

It turns out that:

$$1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}... =\log(3)$$

and:

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

in general. That is same as the Mathematica known formula above.


The recurrence describing matrix $T_1$ is:

$$\displaystyle T_1(1,n)=0; n>1: T(n,k) = \sum\limits_{i=1}^{n-1} T(n,k-i)$$
 
It is natural to ask what this recurrence might do when applying it symmetrically,
like this:

$$\displaystyle T_2(n,1)=1, T_2(1,k)=1, n>=k: -\sum\limits_{i=1}^{k-1} T_2(n-i,k), n<k: -\sum\limits_{i=1}^{n-1} T_2(k-i,n)$$

von Mangoldt function matrix:

$$\displaystyle T_2 = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix}
$$

Here we then have $\log(2)$ in the second column and second row. Like wise $\log(3)$ in third row and third column.
 
By doing a Möbius inversion on each row or column one can conjecture that the von Mangoldt function is found as:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$
 
Alteratively by inspecting the numbers in the decimal digits one also finds conjecturally that:

$$\Lambda(k)=\sum\limits_{n=1}^{\infty} \frac{T_2(n,k)}{n}$$
 

Also, matrix $T_2$ is a matrix with the Greatest Common Divisor as lookup index:

$$T_2(n,k) = a(GCD(n,k))$$

where conjecturally:

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

which is also known as the Dirichlet inverse of the Euler Totient starting:

$$1,-1,-2,-1,-4,+2,-6,...$$

And as a side note we can by a result of Wolfgang Schramm calculate the Möbius function from this matrix as:

$$\displaystyle \mu(n) = \frac{1}{n} \sum\limits_{k=1}^{k=n} T_2(n,k) \cdot e^{i 2 \pi \frac{k}{n}}$$

So much for the elementary introduction.

Next we will look at the function:

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

in the complex plane where $$s=a+ib$$ is a complex number, and thereby get closer to the question, 

The plot of this I call the Riemann zeta function product. With a modification of Jeffrey Stopple's code;

    Show[Graphics[
      RasterArray[
       Table[Hue[
         Mod[3 Pi/2 + 
            Arg[Sum[Zeta[sigma + I t]*
               Total[1/Divisors[n]^(sigma + I t - 1)*
                  MoebiusMu[Divisors[n]]]/n, {n, 1, 30}]], 
           2 Pi]/(2 Pi)], {t, -30, 30, .1}, {sigma, -30, 30, .1}]]], 
     AspectRatio -> Automatic]

it looks like this:

Riemann zeta function product (30-th partial sum) [Image 3]:

![spectral zeta][3]

Again as a sidenote we can compare with the usual Riemann zeta function:

Normal or usual zeta [Image 4]:

![Normal zeta][4] 

which is plotted in the same (complex) region as Spectral Riemann zeta (-30 to +30 and -30i to +30i).

The critical strip of the Riemann zeta function product (Spectral Riemann zeta);

    scale = 50; (*scale = 5000 gives the plot below*)
    Print["Counting to 60"]
    Monitor[g1 = 
      ListLinePlot[
       Table[Re[
         Zeta[1/2 + I*k]*
          Total[Table[
            Total[MoebiusMu[Divisors[n]]/Divisors[n]^(1/2 + I*k - 1)]/(n*
               k), {n, 1, scale}]]], {k, 0 + 1/1000, 60, N[1/6]}], 
       DataRange -> {0, 60}, PlotRange -> {-0.15, 1.5}], Floor[k]]

looks like this:

Riemann zeta function product [Image 5]:

![zeta zero spectrum 1][5]

What a coincidence then that if we take the Fourier transform of the von Mangoldt function with [Heike][6]'s algorithm;


    Clear[f]
    scale = 100000;
    f = ConstantArray[0, scale];
    f[[1]] = N@HarmonicNumber[scale];
    Monitor[Do[
      f[[i]] = N@MangoldtLambda[i] + f[[i - 1]], {i, 2, scale}], i]
    xres = .002;
    xlist = Exp[Range[0, Log[scale], xres]];
    tmax = 60;
    tres = .015;
    Monitor[errList = 
       Table[(xlist^(1/2 + I k - 1).(f[[Floor[xlist]]] - xlist)), {k, 
         Range[0, 60, tres]}];, k]
    ListLinePlot[Im[errList]/Length[xlist], DataRange -> {0, 60}, 
     PlotRange -> {-.01, .15}]


where we have set the first term in the von Mangoldt function sequence:

    Table[Limit[Zeta[s]*Total[MoebiusMu[Divisors[n]]/Divisors[n]^(s - 1)], s -> 1], {n, 1, 32}]

$$\infty ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,$$

equal to a Harmonic number:

$$H_{\text{scale}} ,\log (2),\log (3),\log (2),\log (5),0,\log (7),\log (2),\log (3),0,\log (11),0,$$

and where scale is the _scale_ is the scale of the Fourier transform matrix and the number of terms used in the von Mangoldt function sequence, we get the same plot:

Fourier transform of von Mangoldt function [Image 6]:

![zeta zero spectrum 2][7]

This is unclear to me. But the parameter `scale` in the program anyways, is the value of the Harmonic Number $H_{\text{scale}}$.

Put simpler I have taken a matrix $A$:

$$ A = \left(
\begin{array}{cccccccccc}
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 2^{\frac{1}{2}-i t} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 3^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 4^{\frac{1}{2}-i t} & 2^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
 5^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
 6^{\frac{1}{2}-i t} & 3^{\frac{1}{2}-i t} & 2^{\frac{1}{2}-i t} & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
 7^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
 8^{\frac{1}{2}-i t} & 4^{\frac{1}{2}-i t} & 0 & 2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 \\
 9^{\frac{1}{2}-i t} & 0 & 3^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
 10^{\frac{1}{2}-i t} & 5^{\frac{1}{2}-i t} & 0 & 0 & 2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 1
\end{array}
\right)$$

calculated it's matrix inverse and multiplied by the Riemann zeta function:

$$B=\zeta(\frac{1}{2}-i t)\left(
\begin{array}{cccccccccc}
 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 -2^{\frac{1}{2}-i t} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 -3^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
 0 & -2^{\frac{1}{2}-i t} & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
 -5^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
 6^{\frac{1}{2}-i t} & -3^{\frac{1}{2}-i t} & -2^{\frac{1}{2}-i t} & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
 -7^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
 0 & 0 & 0 & -2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 1 & 0 & 0 \\
 0 & 0 & -3^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
 10^{\frac{1}{2}-i t} & -5^{\frac{1}{2}-i t} & 0 & 0 & -2^{\frac{1}{2}-i t} & 0 & 0 & 0 & 0 & 1
\end{array}
\right)$$

and then plotted the function that is matrix $B$

> So the question is if image 5 and image 6 are equivalent, differing only by a scale factor? Or as in the title: Is the Riemann zeta function product $f(t)$ (Spectral Riemann zeta):

$$f(t)=\sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(1/2+i \cdot t)\sum\limits_{d|n} \frac{\mu(d)}{d^{(1/2+i \cdot t-1)}}$$

similar or equal to the Fourier transform of:
 
$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$

times a scale factor?

In support of this claim is that if we take the Fourier transform of the Riemann zeta function product at the critical line (real part equal 1/2);

    Clear[n, k, t, A, nn, B]
    nn = 60
    A = Table[
      Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1, 
        nn}], {n, 1, nn}]; MatrixForm[A];
    B = FourierDCT[
       Table[Total[
         1/Table[n, {n, 1, nn}]*
          Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000, 
         600, N[1/6]}]];
    g1 = ListLinePlot[B[[1 ;; 700]]*Table[Sqrt[n], {n, 1, 700}], 
       DataRange -> {0, 60}, PlotRange -> {-60, 600}];
    mm = 11.35/Log[2];
    g2 = Graphics[
       Table[Style[Text[n, {mm*Log[n], 100 + 20*(-1)^n}], 
         FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 16}]];
    Show[g1, g2, ImageSize -> Large] 


we get:

Mobius function -> Dirichlet series -> Spectral Riemann zeta -> Fourier transform -> von Mangoldt function:

![from Mobius via spectral Riemann zeta to von Mangoldt][8]

which looks a lot like the von Mangoldt function in Heikes algorithm, in the way it is input there. Notice that the line in the last program above 

> B[[1 ;; 700]]*Table[Sqrt[n], {n, 1, 700}]

is not analytically correct. A larger image of the same plot: https://i.sstatic.net/02A1p.jpg

------------------------------------------------

The partial sums of the Riemann zeta function product:

12 first curves together or partial sums:

![12 curves][9]

and a larger plot: https://i.sstatic.net/5LirM.jpg

This whole question has previously been asked here:
http://math.stackexchange.com/questions/424530/is-this-similarity-to-the-fourier-transform-of-the-von-mangoldt-function-real

    Clear[n, k, t, A, nn, h]
    nn = 60;
    h = 2; (*h=2 gives log 2 operator, h=3 gives log 3 operator and so on*)
    A = Table[
       Table[If[Mod[n, k] == 0, 
         If[Mod[n/k, h] == 0, 1 - h, 1]/(n/k)^(1/2 + I*t - 1), 0], {k, 1, 
         nn}], {n, 1, nn}];
    MatrixForm[A];
    g1 = ListLinePlot[
       Table[Total[
         1/Table[n*t, {n, 1, nn}]*
          Total[Transpose[Re[Inverse[A]*Zeta[1/2 + I*t]]]]], {t, 1/1000, 
         nn, N[1/6]}], DataRange -> {0, nn}, PlotRange -> {-3, 7}];
    mm = N[2*Pi/Log[h], 12]
    g2 = Graphics[
       Table[Style[Text[n*2*Pi/Log[h], {mm*n, 1}], 
         FontFamily -> "Times New Roman", FontSize -> 14], {n, 1, 32}]];
    Show[g1, g2, ImageSize -> Large]

Matrix inverse of Riemann zeta times log 2 operator:

![2 Pi div log2 spectrum][10]

    Clear[n, k, t, A, nn, dd]
    dd = 220;
    Print["Counting to ", dd]
    nn = 20;
    A = Table[
       Table[If[Mod[n, k] == 0, 1/(n/k)^(1/2 + I*t - 1), 0], {k, 1, 
         nn}], {n, 1, nn}];
    Monitor[g1 = 
       ListLinePlot[
        Table[Total[
          1/Table[n*t, {n, 1, nn}]*
           Total[Transpose[
             Re[Inverse[
               IdentityMatrix[nn] + (Inverse[A] - IdentityMatrix[nn])*
                 Zeta[1/2 + I*t]]]]]], {t, 1/1000, dd, N[1/100]}], 
        DataRange -> {0, dd}, PlotRange -> {-7, 7}];, Floor[t]]
    mm = N[2*Pi/Log[2], 20]
    g2 = Graphics[
       Table[Style[Text[n, {mm*n, 1}], FontFamily -> "Times New Roman", 
         FontSize -> 14], {n, 1, 32}]];
    Show[g1, g2, ImageSize -> Large]

Matrix Inverse of matrix inverse times zeta function (on critical line):

![matrix inverse of matrix inverse as function of t][11]



  [1]: https://i.sstatic.net/GDmoU.png
  [2]: https://i.sstatic.net/c8KY7.png
  [3]: https://i.sstatic.net/2iXhr.jpg
  [4]: https://i.sstatic.net/hFtGB.jpg
  [5]: https://i.sstatic.net/SjS5c.jpg
  [6]: http://stackoverflow.com/a/8975710/1067753
  [7]: https://i.sstatic.net/QVWkQ.jpg
  [8]: https://i.sstatic.net/yj8Ef.jpg
  [9]: https://i.sstatic.net/HG7Jy.jpg
  [10]: https://i.sstatic.net/5aaE9.jpg
  [11]: https://i.sstatic.net/PVrdb.jpg