Composite Factor Soup Per the fundamental theorem of arithmetic, for a given number \$n\$, it is possible to find it's prime factors, and they are unique. Let's imagine we talk only of \$n\$ that is non-prime (composite). 

We can also find the factors of all the composite numbers smaller than \$n\$. For example if \$n\$ is 10, then it has factors 5 and 2. 9 has 3 and 3. 8 has 2, thrice. 6 has 3 and 2. 4 has 2 and 2. So for the number 10, all the prime factors of all the composite numbers smaller than 10 would be listed as 2,3, and 5. 

Now if you put a lot of vegetables in a pot for soup, often the largest will rise to the top. So if we put all of these factors in a big pot, which one will be the largest and rise to the top? For 10 soup, that answer is 5.

"Silly", you might think, "the number \$n\$ itself will have the largest factors, larger than the factors of numbers smaller than \$n\$". But this is where you are wrong my friend!

For example, the factors of 16 are all 2, repeated four times. The factors of 15 are 5 and 3, now, I don't have to be a mathematician to tell that 15 is smaller than 16, but 5 is bigger than 2!

Your challenge is to explore how this works for bigger \$n\$. For any number given input number \$n\$, assumed natural, composite, and less than 2^32, find out which is the "largest factor" of all the prime factors of all the composite numbers less than or equal to \$n\$.

Good luck and have fun!

Smallest byte count wins.
Neim, 3 bytes

ᚺ&gt;:


Explanation:

ᚺ   halve
 &gt;  increment
  : previous prime


Try it online!