The Dirichlet inverse of the Euler totient function is: $$\varphi^{-1}(n) = \sum_{d \mid n} \mu(d)d \tag{1}$$ and the von Mangoldt function can be expanded/computed as: $$\Lambda(n) = \sum\limits_{k=1}^{\infty}\frac{\varphi^{-1}(\gcd(n,k))}{k} \tag{2}$$ Consider the sequences: $$a(n)=\sum _{k=1}^{j} \frac{\varphi^{-1}(\gcd (n,k))}{k}$$ $$b(n)=\sum _{k=1}^{j+1} \frac{\varphi^{-1}(\gcd (n,k))}{k}$$

**Question:**

For what values of $j$ does $b(n)$ eventually correlate better than $a(n)$ with the von Mangoldt function $\Lambda(n)$ as $n \rightarrow \infty$?

The exceptions of $j$ when $b(n)$ correlates worse than $a(n)$ appear to be:

$$j=7, 15, 24, 26, 31$$ which when added with $1$ gives initially a sequence of powers of some sort:

$$j+1=8, 16, 25, 27, 32$$

A very slow Mathematica program that computes the sequences $a(n)$ and $b(n)$ and compares their Pearson correlation with the von Mangoldt function, is:

```
Clear[a, n, k, start, end]
nnn = 200;
a[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Do[column = j;
earlier =
Table[Correlation[
Table[Sum[a[GCD[n, k]]/k, {k, 1, column}], {n, 1, nn}],
Table[N[MangoldtLambda[n]], {n, 1, nn}]], {nn, 2, nnn}];
later = Table[
Correlation[
Table[Sum[a[GCD[n, k]]/k, {k, 1, column + 1}], {n, 1, nn}],
Table[N[MangoldtLambda[n]], {n, 1, nn}]], {nn, 2, nnn}];
sign = Sign[later - earlier];
Print[If[Last[sign] == -1, j, 0]], {j, 2, 32}]
```

`nnn = 200;`

is considered a large number that serves as the substitute for $n \rightarrow \infty$ in the program.

Edit 10.7.2021:

Correlations of partial sums:

```
Clear[a, n, k, start, end]
nnn = 400;
a[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Do[column = j;
earlier =
Table[Correlation[
Accumulate[
Table[Sum[a[GCD[n, k]]/k, {k, 1, column}], {n, 1, nn}]],
Accumulate[Table[N[MangoldtLambda[n]], {n, 1, nn}]]], {nn, 2,
nnn}];
later = Table[
Correlation[
Accumulate[
Table[Sum[a[GCD[n, k]]/k, {k, 1, column + 1}], {n, 1, nn}]],
Accumulate[Table[N[MangoldtLambda[n]], {n, 1, nn}]]], {nn, 2,
nnn}];
sign = Sign[later - earlier];
Print[{earlier, Count[sign, 1], sign, j}], {j, 2, 32}]
```