Another interesting fact is that the sum of 1/p for every prime p diverges. It shows that the set of primes is "large" in a certain sense. It would probably be much easier to factorize integers if it wasn't.
> 50% of integers are have a factor under 3
every other number is divisible by 2, that is literally half.
> 73% - under 6
Well I think about it this way: every other number is divisible by two, every third number is divisible by 3 and every fifth number is divisible by 5. It just so happens that some of the numbers that are divisible by 2 are also divisible by 3 or 5 such as 10 for instance. Thus the 73% is nor rly surprising.
1/2 + 1/3rd of what remains + 1/5 of what remains - the intersections!
> but then it slows down to a trickle. 88% under 100, and just 92% under 1000.
I honestly expected it to be a bit lower than 92 but I am clearly wrong about this. Here is an interesting quick visualization I made. Blue are primes, whites are composites!
Edit: you can clearly notice a diagonal shape in the pic above. If you however change the table from 5 cols to any N cols, you will get a distinct shape for each N. It's pretty cool to play with actually. There are a ton of shapes that have been known and used for centuries. I'd highly recommend googling around and reading a bit about them on wikipedia!
I used a chrome plugin called "Full Page Screen Capture". You can use similar extension for FF if you are using that!
The table itself is a long html page! :D
apologies, I don't know your background nor do I have an insight as to what goes on in your head. All I wanted to point out is that it's actually not a surprising growth rate. And while yes, you are right about the "speed" of the function. However that is not "unexpected". At least I don't find that to be the case personally!
f_n = (1 - f_(n-1)) * P_n^(-1) + f_(n-1), where f_n is the fraction divisible by the first n primes and P_n is the nth prime.
So we can see the faster f_n approaches 1, the faster the first term approaches 0, and the faster f_n ~ f_(n-1). Perhaps doubly so, as if f_n approaches 1 faster, then P_n is smaller and it's reciprocal bigger as well. (We could plug in an estimate for P_n if we were converting to a closed form, to work out the asymptotic implications.)