Let $\Omega(n)$ be the total number of prime factors of $n$ (e.g $\Omega(12)=\Omega(2\*2\*3)=3$).


Consider these probabilities (probably they have a name):


* $P\_{>}(n)$ the probability that $\Omega(n+1) > \Omega(n)$
* $P\_{<}(n)$ the probability that $\Omega(n+1) < \Omega(n)$
* $P\_{=}(n)$ the probability that $\Omega(n+1) = \Omega(n)$


And the same probabilities, with an increment of 2; e.g.:


* $P'\_{=}(n)$ the probability that $\Omega(n+2) = \Omega(n)$


What are the values of these limits (if known):


$\lim\_{n \to \infty} P\_{=}(n)$ i.e. the probability that a successor of a number $n$ have the same number of factors of $n$


$\lim\_{n \to \infty} P'\_{=}(2n)$ i.e. the the probability that the even successor of an even number $n$ have the same number of factors of $n$


$\lim\_{n \to \infty} P'\_{=}(2n-1)$ i.e. the the probability that the odd successor of an odd number $n$ have the same number of factors of $n$


Are they all zero?



Probability of an equal number of factors of a successor
Loosely speaking, $\Omega(n)$ is approximately normal with mean and variance $\log\log n$. (See <https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem>) We expect $\Omega(n)$ and $\Omega(n+1)$ to behave like independent random variables. So we expect that all your probabilities containing an equal sign have a limit of zero, while the probabilities containing $>$ or $<$ are expected to approach $1/2$.


