[murbard2]

If you think of the covariance matrix, entry i,j for i ≠ j will be

   floor(n / (p[i]*p[j])) / n - floor(n / p[i]) * floor(n/p[j]) / n^2

and the ith diagonal entry will be

   floor(n / p[i]) / n - ( floor(n / p[i]) / n )^2

for n large, you approximately get a diagonal matrix with diagonal entries / eigenvalues 1/p[i] - 1/p[i]^2.

[mturmon]

Smart observation. Another way to say it is that, for distinct primes p1 and p2, the events “p1 divides n”, and “p2 divides n”, are approximately statistically independent. So you get a near-diagonal covariance with entries as you wrote.