# 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  -> 1
2  -> 2
3  -> 3
6  -> 2,3
9  -> 3
18 -> 2,3
27 -> 3
54 -> 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>0 as input, output the nth BIU number

# Test Cases

Input->Output

1->1
2->21
42->23595
100->118300
200->415777
300->800175


This is .Shortest answer in bytes wins!

• But 1 is not prime... – Stephen Sep 27 '17 at 0:38
• @Stephen thats why I said "For the case of 1 we take as distinct prime factor 1". This is my challenge and this is one of the rules of this challenge.I didn't say that 1 is prime. – user72253 Sep 27 '17 at 0:42
• Why are the numbers called "BIU"? – Misha Lavrov Sep 27 '17 at 1:08
• I'm not sure but I think that it has to do with bisexual intellectual unicorns using them in their everyday life (not in our universe of course...) – user72253 Sep 27 '17 at 1:13
• Downvoters, don't be shy. Share your thoughts with the rest of us. – user72253 Sep 27 '17 at 13:17

# 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:
Ṫ    - tail

• -1 byte using ÆfQ€ instead of ÆFḢ€€ – HyperNeutrino Sep 27 '17 at 0:56
• but they mysteriously hide from me sometime "Jelly is a game of atom hide and programmer seek" ~ i cri everytim – HyperNeutrino Sep 27 '17 at 1:14
• I think you can save 1 byte with ÆDÆFSSḢ‘ḍ. – Arnauld Sep 27 '17 at 16:11

# 05AB1E, 9 bytes

µNNÑfOO>Ö


Uses teh 05AB1E encoding. Try it online!

# Husk, 13 bytes

!fṠ¦ö→ΣṁoupḊN


Try it online!

### Explantaion

  Ṡ¦ö→ΣṁoupḊ    Predicate: returns 1 if BIU, else 0.
Ḋ    List of divisors
ṁ        Map and then concatenate
oup     unique prime factors
Σ         Sum
Ṡ¦            Is the argument divisible by this result
f          N   Filter the natural numbers by that predicate
!               Index


# Mathematica, 85 bytes

If[#<2,1,n=#0[#-1];While[Count[(d=Divisors)@++n,1+Tr@Cases[d/@d@n,_?PrimeQ,2]]<1];n]&


# Actually, 16 bytes

u⌠;÷♂y♂iΣu@%Y⌡╓N


Try it online!

Explanation:

u⌠;÷♂y♂iΣu@%Y⌡╓N
u⌠;÷♂y♂iΣu@%Y⌡╓   first n+1 numbers x starting with x=0 where
÷                divisors
♂y              prime factors of divisors
♂iΣu          sum of prime factors of divisors, plus 1
;       @%        x mod sum
Y       is 0
N  last number in list


# Pyth, 22 bytes

e.f|qZ1!%Zhssm{Pd*M{yP


Try it here!

This is my first ever Pyth solution, I began learning it thanks to the recommendations of some very kind users in chat :-)... Took me about an hour to solve.

## Explanation

e.f|qZ1!%Zhssm{Pd*M{yP  - Whole program. Q = input.

.f                     - First Q integers with truthy results, using a variable Z.
qZ1                - Is Z equal to 1?
|                    - Logical OR.
{yP  - Prime factors, powerset, deduplicate.
*M     - Get the product of each. This chunck and ^ are for divisors.
m}Pd      - Get the unique prime factors of each.
ss           - Flatten and sum.
h             - Increment (to handle that 1, bah)
%Z               - Modulo the current integer by the sum above.
!                 - Logical negation. 0 -> True, > 0 -> False.
e                       - Last element.


All of the list comprehensions here can probably be golfed down, but I'm not sure how. Golfing suggestions welcome! Try it online!

x!y=rem x y<1
b n=[a|a<-[1..],a!(1+sum[sum[z|z<-[2..m],m!z,and[not\$z!x|x<-[2..z-1]]]|m<-[x|x<-[2..a],a!x]])]!!(n-1)


Ungolfing

This answer is actually three functions mashed together.

divisors a = [x | x <- [2..a], rem a x == 0]
sumPrimeDivs m = sum [z | z <- [2..m], rem m z == 0, and [rem z x /= 0 | x <- [2..z-1]]]
biu n = [a | a <- [1..], rem a (1 + sum [sumPrimeDivs m | m <- divisors a]) == 0] !! (n-1)


# Japt, 2221 17 bytes

I feel like the g function method should lead to a shorter solution, but I can't figure out how it works! Nope! Wrong method! But, in my defense, the i method didn't exist at the time.

1-indexed.

ÈvXâ mk câ x Ä}iU


Try it

ÈvXâ mk câ x Ä}iU     :Implicit input of integer U
È                     :Function taking an integer X as argument
v                    :  Test X for divisibility by
Xâ                  :    Divisors of X
m                :    Map
k               :      Prime factors
c             :    Flatten after
â            :      Deduplicating