Of the first $N$ natural numbers, we select two different numbers at random. We'll call the greater one $A$ and the lesser one $B$. What is the probability $P$ that the sum of $A$'s prime factors is less than the sum of $B$'s prime factors?
For instance, if $A = 20$ and $B = 11$, $A$'s prime factors are $2$, $2$, and $5$ (sum $= 9$), and $B$'s prime factors are $11$ (sum $= 11$). So this satisfies the condition that $A > B$ (as $20 > 11$), but the sum of prime factors of $A$ is less than the sum of prime factors of $B$ (as $9 < 11$).
I presume that this question or something near it has been answered before, but I suck at searching through mathematics literature. So I just arduously calculated this probability for $N = 2$ through $300$. When $N = 10$, the probability's about $4.4\%$, but it slowly moves upward to:
$10\%$ at $N = 20$; $19.2\%$ at $N = 50$; $24.5\%$ at $N = 100$; $28.2\%$ at $N = 200$; $30.1\%$ at $N = 300$. (I can provide more complete data if necessary.)
The growth seems to slow logarithmically (prime number theorem hunch), so I did some division. And for some reason, $\ln N/P$ tends toward somewhere around $19.04$, with little oscillation once $N$ is above $100$.
My question: anyone know why? Why $19.04$?