Consider the nine consecutive natural numbers starting from $1584614377$.
n = 1584614377 no. of divisors: 2
n = 1584614378 no. of divisors: 4
n = 1584614379 no. of divisors: 8
n = 1584614380 no. of divisors: 12
n = 1584614381 no. of divisors: 14
n = 1584614382 no. of divisors: 16
n = 1584614383 no. of divisors: 32
n = 1584614384 no. of divisors: 40
n = 1584614385 no. of divisors: 48
Their number of divisors is strictly increasing. This is an examples of a increasing sequence of the number of divisors of length $9$. For every positive integer $n \ge 1$, we calculate far we can go before the sequence of increasing divisors breaks. For some $n$ the sequence will break immediately at $n+1$ while for others it might go longer. The longest such sequence I have found so far is the above sequence of length $9$.
On an average how far we can go before the sequence breaks? Let $I_n$ be the average of length of increasing sequence for the first $n$ natural numbers.
Update 12-12-21: Experimental data shows that $I_{10^6} \approx 1.537$ and $I_{1.2 \times 10^{10}} \approx 1.5537$. It will be nice if it converges to well known constant such as $\pi/2$.
Question 1: What is $\lim_{n \to \infty}I_n$?
Instead of an increasing sequence of divisors, let is consider a decreasing sequence of divisors and let $D_n$, be the average of length of decreasing sequence for the first $n$ natural numbers.
Question 2: Is it true that $\lim_{n \to \infty}I_n = \lim_{n \to \infty}D_n$?
Update 20-Jan-22: Posted in MO since it is unanswered in MSE.