I ask the following question.
Let $R$ denote the uniform distribution on $\{1,2,...,N\}$.
What is $\Pr_R(d(n+1)\ge d(n))$?
*Where d(n) is the divisor function.
The Erdös and Mirsky problem asks about $\Pr_R(d(n+1)= d(n))$, which is a $0$-density event, and should be much harder. However, I was yet to encounter any result regarding my question.
I conjecture it should be $\dfrac{1}{2} + o(1)$.
Apparently, a similar question was solved for the function $\omega(\cdot)$, which counts the number of distinct prime factors of a number $n$. For example, $\omega(2^k)=1$, whereas $d(2^k) = k+1$.
In this thread, some guy asks this question for the distinct prime factors function $\omega(\cdot)$, and is being answered by the almighty shiny phoenix of number theory.
https://mathoverflow.net/questions/171549/in-which-orders-can-the-numbers-of-prime-factors-of-consecutive-integers-be
I managed to prove my original claim using the fact that the random variables $$(\dfrac{\omega(n+1)-\log\log x}{\sqrt{\log\log x}}, \dfrac{\omega(n)-\log\log x}{\sqrt{\log\log x}})$$ have independent Gaussian limiting distribution. This is an Erdős-Kac like theorem that has been proven in a paper whose link can be found in the comments section of the above link.
The previous implies that $\omega(n+1)-\omega(n)$ has a Gaussian limit distribution with a $0$-mean and $\sqrt{2\log\log n}$-standard deviation. One can show that $\log(d(n))$ is very well approximated by $\omega(n)$. I.e., show that $\mathbb{E}|\log(d(n))-\omega(n)| = o(1)$. From this point the statement is trivial.
I will point out that this question seems so trivial, it must have a purely elementary solution. This is clearly not a proof from the book.