It's a well known result that the Dirichlet series of the Liouville function $ \lambda(n) $ is given by

$$ \sum_{k=1}^{\infty} \frac{\lambda(k)}{k^s} = \frac{\zeta(2s)}{\zeta(s)} $$

If we use Perron's formula on this, we get some expression like

$$ L(x) = \sum_{k \leq x} \lambda(k) = \frac{x^{1/2}}{\zeta(1/2)} + \sum_{\rho} \frac{x^\rho}{\rho} \frac{\zeta(2 \rho)}{\zeta'(\rho)} + O(1) $$

where the sum runs over the nontrivial zeroes (of multiplicity one) of the Riemann zeta function. Since $ \zeta(1/2) = -1.46 \ldots $, if we assume the Riemann hypothesis this suggests that the function $ L $ should have a bias to be negative, and this is indeed true: it's the subject of the Polya conjecture, for instance. Moreover, the first term of this expression also predicts the magnitude of the bias quite well: just to give an example, $ L(10^4) = -94 $, and based just on the first term of the above expression we would predict it's around $ -100/1.46 \approx -68 $, which is not far off from the actual value.

My question is this: often we can get "elementary intuition" into the behavior of these kinds of sums by pseudorandomness heuristics, appropriately modified. For example, the Riemann hypothesis is (roughly) equivalent to the central limit theorem bound on the function $ L(x) $. Is there such an intuition for why the partial sums of the Liouville function would be predominantly negative? Any intuition would probably have to contrast the Liouville function with the Mobius function, since the absence of the $ \zeta(2s) $ on the numerator from the Dirichlet series of $ \mu(k) $ means that the partial sums of the Mobius function show no such bias to be negative.

I would be happy with any answer which doesn't appeal to some analytic number theoretic machinery: Perron's formula, Mellin transforms, *et cetera*. Ideally I'm looking for something of the form "there are slightly more primes that are $ 3 \pmod{4} $ than primes $ 1 \pmod{4} $ because the congruence class $ 1 \pmod{4} $ has all of the odd squares in it".

**Edit:** There has been another question on this subject a decade ago, but it got no answer of the nature I'm looking for here.

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