Prove that this sequence of integers is on average equal to zero. Consider the sequence $\{a(n)\}_{n\in\mathbb{N}^*}$ that is defined by the Dirichlet series:
$$\zeta (s)^2\cdot\left(1-\frac{1}{2^{s-1}}-\frac{1}{3^{s-1}}+\frac{1}{6^{s-1}}\right)=\sum_{n\geq 1}\frac{a(n)}{n^s}$$
Prove or disprove that on average it is equal to zero.
The sequence starts:
$$a(n)=1,0,-1,-1,2,0,2,-2,-3,0,2,1,2,0,-2,-3,2,0,2,-2,-2,0,2,2,3,0,-5,-2,2,0,2,-4,...$$
To be more precise:
Prove or disprove that:
$$\lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^{N}a(n)=0.$$
Edit: To generate this sequence:
1. Multiply the red and green matrix $\zeta(s)$ with the matrix on the left (matrix multiplication).
2. Repeat.
3. The sequence $a(n)$ is found in the first column in the last matrix.  

Edit:
After reading the answer below I tried to verify this in a Mathematica program:
a = {+1, -1, -2, -1, +1, +2}
Monitor[a = 
  Table[Sum[
     Sum[If[Mod[n, k] == 0, a[[1 + Mod[k - 1, 6]]], 0], {k, 1, 
       n}], {n, 1, kk}]/kk, {kk, 1, 2000}], kk]
ListLinePlot[a]

This is the plot from the program:

which looks as if it is converging to zero based on the scale on the y-axis. 2000 terms used.
To confirm that the program generates the sequence $a(n)$:
a = {+1, -1, -2, -1, +1, +2}
Monitor[a = 
  Table[Sum[
    If[Mod[n, k] == 0, a[[1 + Mod[k - 1, 6]]], 0], {k, 1, n}], {n, 1, 
    32}], n]

{1, 0, -1, -1, 2, 0, 2, -2, -3, 0, 2, 1, 2, 0, -2, -3, 2, 0, 2, -2, \
-2, 0, 2, 2, 3, 0, -5, -2, 2, 0, 2, -4}
aNumberThatIsAsBigAsPossible = 1000; a = {+1, -1, -2, -1, +1, +2};
Monitor[a = 
   Table[Sum[
      Sum[If[Mod[n, k] == 0, a[[1 + Mod[k - 1, 6]]], 0], {k, 1, 
        n}], {n, 1, kk}]/kk, {kk, 1, 
     aNumberThatIsAsBigAsPossible}];, kk]; Show[
 ListLinePlot[a, PlotRange -> {-0.5, +0.5}], 
 Graphics[Line[{{0, 0.3088626596}, {aNumberThatIsAsBigAsPossible, 
     0.3088626596}}]]]

The black line is at the value 0.3088626596


We have:
$$ \zeta(s)=\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1},\qquad \zeta(s)^2=\sum_{n\geq 1}\frac{d(n)}{n^s} $$
hence, assuming that the sequence $\{a(n)\}_{n\in\mathbb{N}^*}$ is the sequence of coefficients of the Dirichlet series associated with $f(s)$:
$$ f(s)=\zeta(s)^2\left(1-\frac{1}{2^{s-1}}\right)\left(1-\frac{1}{3^{s-1}}\right)=\sum_{n\geq 1}\frac{a(n)}{n^s}$$
we have:
$$ \zeta(s)^2\left(1-\frac{2}{2^{s}}\right)\left(1-\frac{3}{3^{s}}\right)=\sum_{n\geq 1}\frac{d(n)}{n^s}-\sum_{n\geq 1}\frac{2 d(n)}{(2n)^s}-\sum_{n\geq 1}\frac{3 d(n)}{(3n)^s}+\sum_{n\geq 1}\frac{6 d(n)}{(6n)^s} $$
so:
$$ \sum_{n=1}^{N}a(n) = \sum_{n=1}^{N}d(n)-2\sum_{\substack{n=1\\2\mid n}}^{N} d(n/2)-3\sum_{\substack{n=1\\3\mid n}}^{N} d(n/3)+6\sum_{\substack{n=1\\6\mid n}}^{N} d(n/6).\tag{1}$$
Since the average order of the divisor function is given by:
$$ \sum_{n=1}^{N} d(n) = N \log N + (2\gamma-1)N + O(\sqrt{N}) \tag{2}$$
by the Dirichlet hyperbola method, we have:
$$ \frac{1}{N}\sum_{n=1}^{N}a(n) = O\left(\frac{1}{\sqrt{N}}\right)\tag{3}$$
so the limit average is zero.
