Probability that a number has $m$ indistinct factors I just discovered Matlab's factor()-function, and I randomly typed in 20081294819, and to my surprise it only had two factors (5099 and 3938281)! I had expected many more factors for such a big number and wondered how normal this was. The following is what I've done in my investigation so far.
In the first figure below, I have plotted the probability that some number below $10^k$ has $m$ indistinct prime factors. 
The second figure shows the evolution of the probability of finding a number with $m$ indistinct prime factors as the numbers one is looking at grows, i.e. it is just another way to visualize the same data. I have left out all points of zero probability.
If one looks at the first part of the figure, it seems that the peak of the curve for some $k$ shifts toward higher $m$ as $k$ grows. 
My first question is this: Does this behavior (i.e. the peak shifting toward higher $m$ as $k$ grows) continue indefinitely, or will the peak stabilize at some point? The latter seems unlikely but I can't be sure.
My second question is this: Approximately how fast does the peak-value $m_{peak}$ move as $k$ grows?
I briefly looked at the Erdõs-Kac Theorem, but this involves distinct primes and I don't really know much number theory, so any help would be greatly appreciated!

 A: The results for distinct and indistinct primes are virtually identical. This is because $$\sum_{n\leq x}|\Omega(n)-\omega(n)|=o\left(\sum_{n\leq x} \Omega(n)\right)$$ where $\Omega(n),\omega(n)$ count the number of prime divisors with and without multiplicity, respectively. Consequently the Erdos-Kac theorem will apply equally to your case, and we see that the number of prime factors of the integers in the interval $[1,x]$, counted with multiplicity, converges weakly to a normal distribution with mean $\log \log x$ and standard deviation $\sqrt{\log \log x}$. Specifically, for any fixed $a<b$ we have that $$\lim_{x\rightarrow \infty} \frac{1}{x}\#\left(n\leq x:\ a\leq \frac{\Omega(n)-\log \log x}{\sqrt{\log \log x}}\leq b\right)=\frac{1}{\sqrt{2\pi }}\int_a^b e^{-x^2/2}dx.$$  For a fixed value of $k$, that is for $k$ not growing with $x$ we can count the number of integers with $k$ prime factors (counted with multiplicity) and find that it will be asymptotic to $$\sum_{n\leq x:\Omega(n)=k}1\sim \frac{x(\log \log x)^{k-1}}{(k-1)!\log x}.$$ (Note that this asymptotic takes on a different form if we left $k$ go to infinity with $x$, and I will not quote that result here) The prime number theorem, that is the statement $\sum_{p\leq x}1\sim \frac{x}{\log x}$ is when $k=1$. This generalization can be obtained from the prime number theorem by using repeated partial summation, and this asymptotic describes the different kind of peaks that you have been observing in your data. The Erdos-Kac theorem tells us that the peak of the distribution of the number of prime factors will be at $k\sim\log \log x$.
