Asymptotic formula for absolute difference of number of prime factors between consecutive integers For $ n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ define $\Omega(n)= \alpha_1+\cdots+\alpha_k$.
What is known about the asymptotic behavior as $N\rightarrow\infty$ for sums of the form
$$\sum_{n=1}^N |\Omega(n+1)-\Omega(n)| \quad ?$$
 A: I'm just filling in the details in a comment left above, which in turn builds off a comment left by Greg Martin. If I haven't made a mistake it works out that
$$
\sum_{n\leq x} |\Omega(n+1)-\Omega(n)| \sim \frac{2}{\sqrt{\pi}} x \sqrt{\log \log x}.
$$
This result is essentially due to Halberstam in "On the Distribution of Additive Number‐Theoretic Functions II"; a few small tricks take you from his result to this one. Halberstam's paper is at https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s1-31.1.1.
The most important result of Halberstam is Theorem 1 of that paper, but he uses this to deal almost exactly with the question you're after; as a special case of Theorem 2 (for $f(p)=1$), we get that
$$
\sum_{n\leq x} (\omega(n+1)-\omega(n))^k = (c_k+o(1)) (2 \log \log x)^{k/2},
$$
where
$$
c_k:=\int_{-\infty}^\infty t^k \frac{e^{-t^2/2}}{\sqrt{2\pi}}\, dt,
$$
and $\omega(n)$ is the number of prime factors of $n$ without multiplicity. 
Morally then this result just follows by 1) replacing the functions $t^k$ with $|t|$ above and evaluating the integral, and 2) showing that the transition from $\omega(n)$ to $\Omega(n)$ doesn't change anything since
$$
\sum_{n\leq x} \Omega(n)-\omega(n) = O(x).
$$
I'll leave 2) to you (but please ask if you have questions about it), but I'll expand a little on 1). This is essentially the method of moments in probability (see section 30 of Billingsley's Probability and Measure, and in particular Theorem 30.2 [I'm using the 3rd edition]). Using the abbreviation,
$$
A_{n,x}:= \frac{\omega(n+1)-\omega(n)}{\sqrt{2\log \log x}},
$$
this theorem from probability combined with Halberstam's tells us for any continuous and bounded function $g(x)$,
$$
\frac{1}{x} \sum_{n\leq x} g(A_{n,x}) = \int_{-\infty}^\infty g(t) \frac{e^{-t^2/2}}{\sqrt{2\pi}}\, dt + o(1).
$$
With just a little bit of extra work on the probability side, one can see that this claim holds more generally for functions $g(x)$ that are continuous and increase only polynomially, so in particular using $g(x) = |x|$,
$$
\frac{1}{x} \sum_{n\leq x} |A_{n,x}| \sim \int_{-\infty}^\infty |t| \frac{e^{-t^2/2}}{\sqrt{2\pi}}\, dt = \sqrt{\frac{2}{\pi}},
$$
and this gives us the result.
