BIU numbers (or Sumdivized numbers) How to spot them
Take a positive integer k. Find its divisors. Find the distinct prime factors of each divisor. Sum all these factors together. If this number (sum) is a divisor of k (if the sum divides k) then, this number k, is a BIU number
Examples
Let's take the number 54
Find all the divisors: [1, 2, 3, 6, 9, 18, 27, 54]
Find the distinct prime factors of each divisor
NOTE: For the case of 1 we take as distinct prime factor 1
1  -&gt; 1  
2  -&gt; 2  
3  -&gt; 3  
6  -&gt; 2,3  
9  -&gt; 3  
18 -&gt; 2,3  
27 -&gt; 3  
54 -&gt; 2,3 

Now we take the sum of all these prime factors
1+2+3+2+3+3+2+3+3+2+3=27
27 divides 54 (leaves no remainder)
So, 54 is a BIU number.
Another (quick) example for k=55
Divisors: [1,5,11,55]
Sum of distinct prime factors: 1+5+11+5+11=33
33 is NOT a divisor of 55, that's why 55 is NOT a BIU number.
BIU numbers
Here are the first 20 of them:

1,21,54,290,735,1428,1485,1652,2262,2376,2580,2838,2862,3003,3875,4221,4745, 5525,6750,7050...

but this list goes on and there are many BIU numbers that are waiting to be descovered by you!
The Challenge
Given an integer n&gt;0 as input, output the nth BIU number
Test Cases
Input-&gt;Output
1-&gt;1  
2-&gt;21   
42-&gt;23595  
100-&gt;118300    
200-&gt;415777    
300-&gt;800175    

This is codegolf.Shortest answer in bytes wins!
Jelly, 16 15 bytes

ÆDÆfQ€SS‘ḍ
1Ç#Ṫ


Try it online!

Woohoo for builtins (but they mysteriously hide from me sometimes so -1 byte thanks to @HyperNeutrino)

How it Works

ÆDÆfQ€SS‘ḍ - define helper function: is input a BIU number?
ÆD             - divisors
  Æf           - list of prime factors
    Q€         - now distinct prime factors
      SS       - sum, then sum again ('' counts as 0)
        ‘      - add one (to account for '')
         ḍ     - does this divide the input?

1Ç#Ṫ - main link, input n
  #     - starting at 
1          - 1
        - get the first n integers which meet:
 Ç         - helper link
   Ṫ    - tail