# Hostile Divisor Numbers

Some divisors of positive integers really hate each other and they don't like to share one or more common digits.

Those integers are called Hostile Divisor Numbers (HDN)

Examples

Number 9566 has 4 divisors: 1, 2, 4783 and 9566
(as you can see, no two of them share the same digit).
Thus, 9566 is a Hostile Divisor Number

Number 9567 is NOT HDN because its divisors (1, 3, 9, 1063, 3189, 9567) share some common digits.

Here are the first few HDN

1,2,3,4,5,6,7,8,9,23,27,29,37,43,47,49,53,59,67,73,79,83,86,87,89,97,223,227,229,233,239,257,263,267,269,277,283,293,307,337...


The above list goes on and your task is to find the nth HDN

Input

A positive integer n from 1 to 4000

Output

The nth HDN

Test Cases

here are some 1-indexed test cases.

input -> output
1        1
10       23
101      853
1012     26053
3098     66686
4000     85009


This is , so the lowest score in bytes wins.

EDIT

Good news! I submitted my sequence to OEIS and...
Hostile Divisor Numbers are now OEIS A307636

• I think square numbers would be the least hostile of numbers. May 3 '19 at 17:59
• @JoeFrambach That I do not understand. There are perfect-square HDN. For a somewhat large example, 94699599289, the square of 307733, has divisors [1, 307733, 94699599289] which shows it is a HDN. Seems hostile to me. May 5 '19 at 17:21
• @JeppeStigNielsen For a much smaller example, why not just 49? Factors [1, 7, 49] qualifies as hostile... Or, well, 4: [1, 2, 4]... May 6 '19 at 13:27
• @DarrelHoffman Not to mention, the square number 1 with divisor list [1]. (Maybe large HDN are more interesting?) May 6 '19 at 18:09
• I interpreted 49 as having divisors [7, 7], which not only share digits but are the same digits. 49 has factors [1, 7, 49] May 6 '19 at 19:59

# 05AB1E, 12 10 bytes

µNNÑ€ÙSDÙQ


-2 bytes thanks to @Emigna.

1-indexed

Try it online or verify most test cases (last two test cases are omitted, since they time out).

Explanation:

µ           # Loop while the counter_variable is not equal to the (implicit) input yet:
N          #  Push the 0-based index of the loop to the stack
NÑ        #  Get the divisors of the 0-based index as well
#   i.e. N=9566 → [1,2,4783,9566]
#   i.e. N=9567 → [1,3,9,1063,3189,9567]
€Ù      #  Uniquify the digits of each divisor
#   → ["1","2","4783","956"]
#   → ["1","3","9","1063","3189","9567"]
S     #  Convert it to a flattened list of digits
#   → ["1","2","4","7","8","3","9","5","6"]
#   → ["1","3","9","1","0","6","3","3","1","8","9","9","5","6","7"]
D    #  Duplicate this list
Ù   #  Unique the digits
#   → ["1","2","4","7","8","3","9","5","6"]
#   → ["1","3","9","0","6","8","5","7"]
Q  #  And check if it is still equal to the duplicated list
#   → 1 (truthy)
#   → 0 (falsey)
#  And if it's truthy: implicitly increase the counter_variable by 1
# (After the loop: implicitly output the top of the stack,
#  which is the pushed index)

• You beat me to it this time. I had µNNÑ€ÙSDÙQ for 10. May 3 '19 at 14:05
• @Emigna Ah, I was just working on an alternative with µ, so you save me the trouble. ;) May 3 '19 at 14:06
• this is poetically eloquent May 8 '19 at 23:45

# Python 2, 104 bytes

n=input()
x=1
while n:
x=i=x+1;d={0};c=1
while i:m=set(i*(x%i<1));c*=d-m==d;d|=m;i-=1
n-=c
print x


Try it online!

0-indexed.

# JavaScript (ES6), 78 bytes

1-indexed.

n=>eval("for(k=0;n;n-=!d)for(s=d=++k+'';k%--d||d*!s.match([${s+=d,d}]););k")  Try it online! ### Faster version, 79 bytes n=>{for(k=0;n;n-=!d)for(s=d=++k+'';k%--d||d*!s.match([${s+=d,d}]););return k}


Try it online!

### How?

Given an integer $$\k>0\$$, we build the string $$\s\$$ as the concatenation of all divisors of $$\k\$$.

Because $$\k\$$ is always a divisor of itself, $$\s\$$ is initialized to $$\k\$$ (coerced to a string) and the first divisor that we try is $$\d=k-1\$$.

For each divisor $$\d\$$ of $$\k\$$, we test whether any digit of $$\d\$$ can be found in $$\s\$$ by turning $$\d\$$ into a character set in a regular expression.

Examples

• $$\s=\text{"}956647832\text{"}\$$, $$\d=1\$$"956647832".match(/[1]/) is falsy
• $$\s=\text{"}9567\text{"}\$$, $$\d=3189\$$"9567".match(/[3189]/) is truthy

### Commented

This is the version without eval(), for readability

n => {                   // n = input
for(                   // for() loop:
n;                   //   go on until n = 0
n -= !d              //   decrement n if the last iteration resulted in d = 0
)                      //
for(                 //   for() loop:
s =                //     start by incrementing k and
d = ++k + '';      //     setting both s and d to k, coerced to a string
k % --d ||         //     decrement d; always go on if d is not a divisor of k
d *                //     stop if d = 0
!s.match(          //     stop if any digit of d can be found in s

# Perl 6, 53 bytes

{(grep {/(.).*$0/R!~~[~] grep$_%%*,1..$_},^∞)[$_]}


Try it online!

1-indexed.

/(.).*$0/ matches any number with a repeated digit. grep$_ %% *, 1 .. $_ returns a list of all divisors of the number $_ currently being checked for membership in the list.

[~] concatenates all of those digits together, and then R!~~ matches the string on the right against the pattern on the left. (~~ is the usual match operator, !~~ is the negation of that operator, and R is a metaoperator that swaps the arguments of !~~.)

• 50 bytes
– Jo King
May 8 '19 at 21:43
• Ah wait, this checks for digits within the same divisor. The regex should be /(\d).*\s.*$0/, where you don't concatenate the divisors – Jo King Sep 12 '20 at 11:11 # Python 2 (PyPy), 117 114 bytes Uses 1-indexing k=input();n=0;r=range while k:n+=1;k-=1-any(set(a)&set(b)for a in r(1,n+1)for b in r(1,a)if n%a<1>n%b) print n  Try it online! # Wolfram Language 103 bytes Uses 1-indexing. I'm surprised it required so much code. (k=1;u=Union;n=2;l=Length;While[k<#,If[l[a=Join@@u/@IntegerDigits@Divisors@#]==l@u@a&@n,k++];n++];n-1)&  • Can you please add a TIO link so that everybody can check your answer? May 3 '19 at 16:40 • 95 bytes: (n=t=1;While[t<=#,If[!Or@@IntersectingQ@@@Subsets[IntegerDigits@Divisors@n,{2}],t++];n++];n-1)& I am not planning to post an answer so I will leave this here May 3 '19 at 16:45 • @J42161217, I've been trying to get the code to work in TIO without success. There must be some trick I'm missing. May 3 '19 at 20:55 • @J42161217, Your code seems to work but takes 3 times the runtime. You can submit it as your own. (Maybe I'll learn how to implement TIO from your example.) May 3 '19 at 20:57 • 72 bytes using For instead of While, E!=##& to test for repeated digits – att May 3 '19 at 22:23 # PowerShell, 112 bytes for($a=$args[0];$a-gt0){$z=,0*10;1..++$n|?{!($n%$_)}|%{"$_"|% t*y|sort -u|%{$z[+"$_"]++}};$a-=!($z|?{$_-ge2})}$n  Try it online! Takes 1-indexed input $args[0], stores that into $a, loops until that hits 0. Each iteration, we zero-out a ten-element array $z (used to hold our digit counts). Then we construct our list of divisors with 1..++$n|?{!($n%$_)}. For each divisor, we cast it to a string "$_", cast it toCharArray, and sort those digits with the -unique flag (because we don't care if a divisor itself has duplicate digits). We then increment the appropriate digit count in $z. Then, we decrement $a only if $z contains 0s and 1s (i.e., we've found an HDN). If we've finished our for loop, that means we found the appropriate number of HDNs, so we leave $n on the pipeline and output is implicit.

• you could save some bytes: $a-=!($z-ge2) instead $a-=!($z|?{$_-ge2}) May 4 '19 at 6:40 • a bit golfed May 4 '19 at 8:13 # Python 3, 115 bytes 1-indexed f=lambda n,x=1,s="",l="",d=1:n and(d>x+1and f(n-1,x+1)or{*s}&{*l}and f(n,x+1)or f(n,x,s+l,(1-x%d)*str(d),d+1))or~-x  Try it online! This uses a lot of recursion; even with increased recursion limit, it can't do f(30). I think it might be golfable further, and I tried finding something to replace the (1-x%d) with, but couldn't come up with anything (-~-x%d has the wrong precedence). Any bytes that can be shaved off are greatly appreciated. # How it works # n: HDNs to go # x: Currently tested number # s: String of currently seen divisor digits # l: String of digits of last tried divisor if it was a divisor, empty string otherwise # d: Currently tested divisor f=lambda n,x=1,s="",l="",d=1:n and( # If there are still numbers to go d>x+1and f(n-1,x+1)or # If the divisors have been # exhausted, a HDN has been found {*s}&{*l}and f(n,x+1)or # If there were illegal digits in # the last divisor, x isn't a HDN f(n,x,s+l,(1-x%d)*str(d),d+1) # Else, try the next divisor, and # check this divisor's digits (if # if is one) in the next call )or~-x # Else, return the answer  # Brachylog (v2), 14 bytes ;A{ℕfdᵐc≠&}ᶠ⁽t  Try it online! Function submission; input from the left, output to the right. (The TIO link contains a command-line argument to run a function as though it were a full program.) ## Explanation "Is this a hostile divisor number?" code: ℕfdᵐc≠ ℕ number is ≥0 (required to match the question's definition of "nth solution") f list of all factors of the number ᵐ for each factor d deduplicate its digits c concatenate all the deduplications with each other ≠ the resulting number has no repeated digits  This turned out basically the same as @UnrelatedString's, although I wrote it independently. "nth solution to a " wrapper: ;A{…&}ᶠ⁽t & output the successful input to { }ᶠ the first n solutions of the problem ⁽ taking <n, input> as a pair ;A form a pair of user input and a "no constraints" value t take the last solution (of those first n)  This is one of those cases where the wrapper required to produce the nth output is significantly longer than the code required to test each output in turn :-) I came up with this wrapper independently of @UnrelatedString's. It's the same length and works on the same principle, but somehow ends up being written rather differently. It does have more potential scope for improvement, as we could add constraints on what values we were looking at for free via replacing the A with some constraint variable, but none of the possible constraint variables save bytes. (If there were a "nonnegative integer" constraint variable, you could replace the A with it, and then save a byte via making the the ℕ unnecessary.) • It’s 2-indexed? May 9 '19 at 10:03 # Java 10, 149139138126125120 119 bytes n->{int r=0,i,d;for(;n>0;n-=d){var s="1";for(r+=d=i=1;i++<r;)if(r%i<1){d=s.matches(".*["+i+"].*")?0:d;s+=i;}}return r;}  -10 bytes by using .matches instead of .contains per digit, inspired by @Arnauld's JavaScript answer. -5 bytes thanks to @ValueInk -1 byte thanks to @ceilingcat 1-indexed Try it online. Explanation: n->{ // Method with integer as both parameter and return-type int r=0, // Result-integer, starting at 0 i, // Index integer d; // Decrement integer for(;n>0; // Loop until the input n is 0: n-=d){ // After every iteration: decrease n by the decrement integer d var s="1"; // Create a String s, starting at "1" for(r+=d=i=1; // (Re)set the decrement and index integers to 1, // and increase the result by 1 as well i++<r;) // Inner loop i in the range [2, r]: if(r%i<1){ // If r is divisible by i: d=s.matches(".*["+i+"].*")? // If string s contains any digits also found in integer i: 0 // Set the decrement integer d to 0 :d; // Else: leave d unchanged s+=i;}} // And then append i to the String s return r;} // After the loops, return the result r  # Brachylog, 16 bytes g{∧0<.fdᵐc≠∧}ᵘ⁾t  Try it online! Very slow, and twice as long as it would be if this was a . 1-indexed.  The output t is the last ᵘ⁾ of a number of unique outputs, g where that number is the input, { } from the predicate declaring that: . the output < which is greater than 0 zero ∧ (which is not the empty list) f factorized ᵐ with each factor individually d having duplicate digits removed ≠ has no duplicate digits in c the concatenation of the factors ∧ (which is not the output).  • If you just read that explanation as a sentence though... May 4 '19 at 11:24 • I try to write my explanations like plain English, which typically ends up just making them harder to read May 4 '19 at 20:25 # Wolfram Language (Mathematica), 74 bytes Nest[1+#//.a_/;!Unequal@@Join@@Union/@IntegerDigits@Divisors@a:>a+1&,0,#]&  Try it online! # Japt v2.0a0, 17 bytes _=â ®sâÃ¬UµZ¶â}f1  Try it Port of this Brachylog answer. Credit: 4 bytes savings total thanks to Shaggy who also suggested there was a better solution leading to many more bytes :) Original answer 28 byte approach: Èâ¬rÈ«è"[{Y}]" ©X+Y}Xs)«U´Ãa  Try it Port of this JavaScript answer. • 28 bytes May 7 '19 at 17:04 • Nice - I hadn't used the « shortcut before :) I figure if Shaggy is only improving on my score by a handful of bytes, I must be getting (somewhat) decent at this? – dana May 7 '19 at 17:42 • It can be done in 20 (maybe less) b7 employing a slightly different method. May 7 '19 at 18:43 • Hah - I guess I spoke too soon :) yeah, some of the other golfing lang's have much shorter solutions. – dana May 7 '19 at 19:29 • 17 bytes May 8 '19 at 17:56 # Perl 5-p, 66 bytes map{1while(join$",map{$\%$_==0&&$_}1..++$\)=~/(\d).* .*\1/}1..$_}{  Try it online! 1 indexed # Husk, 10 bytes !fȯS=uṁdḊN  Try it online! Same method as Jelly. # Husk, 19 bytes !f(Λo¬Fnfo¬Eπ2mdḊ)N  Try it online! The more manual cehcker. # Icon, 123 bytes procedure f(n) k:=m:=0 while m<n do{ k+:=1 r:=0 s:="" every k%(i:=1 to k)=0&(upto(i,s)&r:=1)|s++:=i r=0&m+:=1} return k end  Try it online! 1-indexed. Really slow for big inputs. # Perl 6, 74 bytes {(grep {!grep *>1,values [(+)] map *.comb.Set,grep$_%%*,1..$_},1..*)[$_]}


0-indexed. Only the first three cases are listed on TIO since it's too slow to test the rest.

Try it online!

• 57 bytes
– Jo King
May 8 '19 at 6:07

# Ruby, 1109792 84 bytes

-13 bytes by leveraging @Arnauld's JavaScript regex check.

-5 bytes for swapping out the times loop for a decrementer and a while.

-8 bytes by ditching combination for something more like the other answers.

->n{x=0;n-=1if(s='';1..x+=1).all?{|a|x%a>0||(e=/[#{a}]/!~s;s+=a.to_s;e)}while n>0;x}


Try it online!

# J, 87 59 bytes

-28 bytes thanks to FrownFrog

0{(+1,1(-:~.)@;@(~.@":&.>@,i.#~0=i.|])@+{.)@]^:(>{:)^:_&0 0


Try it online!

# J, 87 bytes

[:{:({.@](>:@[,],([:(-:~.)[:-.&' '@,/~.@":"0)@((]#~0=|~)1+i.)@[#[)}.@])^:(#@]<1+[)^:_&1


Try it online!

Yikes.

This is atrociously long for J, but I'm not seeing great ways to bring it down.

## explanation

It helps to introduce a couple helper verbs to see what's happening:

d=.(]#~0=|~)1+i.
h=. [: (-:~.) [: -.&' '@,/ ~.@":"0

• d returns a list of all divisors of its argument
• h tells you such a list is hostile. It stringifies and deduplicates each number ~.@":"0, which returns a square matrix where shorter numbers are padded with spaces. -.&' '@,/ flattens the matrix and removes spaces, and finally (-:~.) tells you if that number has repeats or not.

With those two helpers our overall, ungolfed verb becomes:

[: {: ({.@] (>:@[ , ] , h@d@[ # [) }.@])^:(#@] < 1 + [)^:_&1


Here we maintain a list whose head is our "current candidate" (which starts at 1), and whose tail is all hostile numbers found so far.

We increment the head of the list >:@[ on each iteration, and only append the "current candidate" if it is hostile h@d@[ # [. We keep doing this until our list length reaches 1 + n: ^:(#@] < 1 + [)^:_.

Finally, when we're done, we return the last number of this list [: {: which is the nth hostile number.

• 66 May 9 '19 at 9:21
• 62 May 9 '19 at 9:41
• This is great, many thanks. Will go over it and update tonight May 9 '19 at 12:39
• 59 May 9 '19 at 23:54