I wrote almost the same but the last cycle used enumerate

Could you please explain the following line: "for j in range(1,int(i**.5)+1):" ?

I just wanna say this is a beautiful and clever implementation , 100x faster than mine lol.
I love it how you came up with the range for the 2nd for loop to be up to i**0.5 this is so efficient.
So, if i is 1024, you only loop from 1 to 32 and then you get the other numbers by dividing 1024 by (1,32).
In my case I would go all the way to 1024 in the loop, lol.
When i tested my solution for (1,30 000 ) it takes 44s, lol, where as yours takes 0.4s :D