Divisor Rich and Poor Numbers

Question

Introduction

In the strange world of integer numbers, divisors are like assets and they use to call "rich" the numbers having more divisors than their reversal, while they call "poor" the ones having less divisors than their reversal.

For example, the number \$2401\$ has five divisors : \$1,7,49,343,2401\$, while its reversal, \$1042\$, has only four : \$1,2,521,1042\$.
So \$2401\$ is called a rich number, while \$1042\$ a poor number.

Given this definition, we can create the following two integer sequences of rich and poor numbers :

(here we list the first 25 elements of the sequences)

 Index | Poor | Rich
-------|------|-------
     1 |   19 |   10
     2 |   21 |   12
     3 |   23 |   14
     4 |   25 |   16
     5 |   27 |   18
     6 |   29 |   20
     7 |   41 |   28
     8 |   43 |   30
     9 |   45 |   32
    10 |   46 |   34
    11 |   47 |   35
    12 |   48 |   36
    13 |   49 |   38
    14 |   53 |   40
    15 |   57 |   50
    16 |   59 |   52
    17 |   61 |   54
    18 |   63 |   56
    19 |   65 |   60
    20 |   67 |   64
    21 |   69 |   68
    22 |   81 |   70
    23 |   82 |   72
    24 |   83 |   74
    25 |   86 |   75
   ... |  ... |  ...

Notes :

• as "reversal" of a number we mean its digital reverse, i.e. having its digits in base-10 reversed. This means that numbers ending with one or more zeros will have a "shorter" reversal : e.g. the reversal of 1900 is 0091 hence 91
• we intentionally exclude the integer numbers having the same number of divisors as their reversal i.e. the ones belonging to OEIS:A062895

Challenge

Considering the two sequences defined above, your task is to write a program or function that, given an integer n (you can choose 0 or 1-indexed), returns the n-th poor and n-th rich number.

Input

• An integer number (>= 0 if 0-indexed or >= 1 if 1-indexed)

Output

• 2-integers, one for the poor sequence and one for the rich sequence, in the order you prefer as long as it is consistent

Examples :

INPUT          |   OUTPUT
----------------------------------
n (1-indexed)  |   poor    rich
----------------------------------
1              |   19      10
18             |   63      56
44             |   213     112
95             |   298     208
4542           |   16803   10282
11866          |   36923   25272
17128          |   48453   36466
22867          |   61431   51794
35842          |   99998   81888

General rules:

• This is code-golf, so shortest answer in bytes wins.
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
• Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
• Default Loopholes are forbidden.
• If possible, please add a link with a test for your code (i.e. TIO).
• Also, adding an explanation for your answer is highly recommended.

Answer 1

05AB1E, 16 bytes

∞.¡ÂÑgsÑg.S}¦ζsè

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---

0-indexed [rich,poor]:

∞                # Push infinite list.
 .¡        }     # Split into three lists by...
   ÂÑgsÑg.S      # 1 if rich, 0 if nothing, -1 if poor.
            ¦ζ   # Remove list of nothings, and zip rich/poor together.
              sè # Grab nth element.

Maybe someone can explain why this version doesn't seem to terminate, but when I click "cancel execution" on TIO it finishes with the correct answer, or if you wait 60 seconds you get the correct answer. For a version that terminates "correctly" you could use: T+nL.¡ÂÑgsÑg.S}¦ζsè +3 bytes

Answer 2

JavaScript (ES6),  121 115 113  111 bytes

Input is 1-indexed. Outputs as [poor, rich].

x=>[(s=h=(n,i=x)=>i?h(++n,i-=(g=(n,k=n)=>k&&!(n%k)*~-s+g(n,k-1))(n)>g([...n+''].reverse().join``)):n)``,h(s=2)]

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Commented

Helper function

g = (n,                   // g is a helper function taking n and returning either the
        k = n) =>         // number of divisors or its opposite; starting with k = n
  k &&                    // if k is not equal to 0:
    !(n % k)              //   add either 1 or -1 to the result if k is a divisor of n
    * ~-s                 //   use -1 if s = 0, or 1 if s = 2
    + g(n, k - 1)         //   add the result of a recursive call with k - 1

Main

x => [                    // x = input
  ( s =                   // initialize s to a non-numeric value (coerced to 0)
    h = (n,               // h is a recursive function taking n
            i = x) =>     // and using i as a counter, initialized to x
      i ?                 // if i is not equal to 0:
        h(                //   do a recursive call ...
          ++n,            //     ... with n + 1
          i -=            //     subtract 1 from i if:
            g(n)          //       the number of divisors of n (multiplied by ~-s within g)
            >             //       is greater than
            g(            //       the number of divisors of the reversal of n obtained ...
              [...n + ''] //         ... by splitting the digits
              .reverse()  //             reversing them
              .join``     //             and joining back
            )             //       (also multiplied by ~-s within g)
        )                 //   end of recursive call
      :                   // else:
        n                 //   we have reached the requested term: return n
  )``,                    // first call to h for the poor one, with n = s = 0 (coerced)
  h(s = 2)                // second call to h for the rich one, with n = s = 2
]                         // (it's OK to start with any n in [0..9] because these values
                          // are neither poor nor rich and ignored anyway)

Answer 3

Jelly, 22 bytes

ṚḌ;⁸Æd
Ç>/$Ɠ©#żÇ</$®#Ṫ

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A full program taking 1-indexed \$n\$ on stdin and returning a list of the n-th poor and rich integers in that order.

Explanation

ṚḌ;⁸Æd | Helper link: take an integer and return the count of divisors fof its reverse and the original number in that order

Ṛ      | Reverse
 Ḍ     | Convert back from decimal digits to integer
  ;⁸   | Concatenate to left argument
    Æd | Count divisors


Ç>/$Ɠ©#żÇ</$®#Ṫ | Main link

    Ɠ©          | Read and evaluate a line from stdin and copy to register
   $  #         | Find this many integers that meet the following criteria, starting at 0 and counting up
Ç               | Helper link
 >/             | Reduce using greater than (i.e. poor numbers)
       ż        | zip with
           $®#  | Find the same number of integers meeting the following criteria
        Ç       | Helper link
         </     | Reduce using less than (i.e. rich numbers)
              Ṫ | Finally take the last pair of poor and rich numbers

Answer 4

Wolfram Language (Mathematica), 152 bytes

(a=b=k=1;m=n={};p=AppendTo;While[a<=#||b<=#,#==#2&@@(s=0~DivisorSigma~#&/@{k,IntegerReverse@k++})||If[#<#2&@@s,m~p~k;a++,n~p~k;b++]];{m[[#]],n[[#]]}-1)&

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If the conjecture is true, then this 140 bytes solution also works

(a=k=1;m=n={};p=AppendTo;While[a<=#,#==#2&@@(s=0~DivisorSigma~#&/@{k,IntegerReverse@k++})||If[#<#2&@@s,m~p~k;a++,n~p~k]];{m[[#]],n[[#]]}-1)&   

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Here is poor vs rich plot

Answer 5

Perl 6, 81 bytes

{(*>*,* <*).map(->&c {grep
{[[&c]] map {grep $_%%*,1..$_},$_,.flip},1..*})»[$_]}

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• * > * is an anonymous function that returns true if its first argument is greater than its second. Similarly for * < *. The former will select numbers that belong to the rich sequence, the latter will select those that belong to the poor sequence.
• (* > *, * < *).map(-> &c { ... }) produces a pair of infinite sequences, each based on one of the comparator functions: the rich sequence and the poor sequence, in that order.
• »[$_] indexes into both of those sequences using $_, the argument to the top-level function, returning a two-element list containing the $_th member of the rich sequence and the $_th member of the poor sequence.
• grep $_ %% *, 1..$_ produces a list of the divisors of $_.
• map { grep $_ %% *, 1..$_ }, $_, .flip produces a two-element list of the divisors of $_, and the divisors of $_ with its digits reversed ("flipped").
• [[&c]] reduces that two-element list with the comparator function &c (either greater-than or less-than), producing a boolean value indicating whether this number belongs to the rich sequence of the poor sequence.

Answer 6

Python 2, 142 141 bytes

f=lambda i,n=1,a=[[],[]]:zip(*a)[i:]or f(i,n+1,[a[j]+[n]*(cmp(*[sum(x%y<1for y in range(1,x))for x in int(`n`[::-1]),n])==1|-j)for j in 0,1])

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---

---

Non-recursive alternative (very similar to the other Python answers)

Python 2, 143 bytes

i=input()
a=[[],[]];n=1
while~i+len(zip(*a)):([[]]+a)[cmp(*[sum(x%i<1for i in range(1,x))for x in int(`n`[::-1]),n])]+=n,;n+=1
print zip(*a)[i]

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Answer 7

Python 2, 158 153 bytes

^{-2 bytes thanks to shooqie}

n=input()
p=[];r=[];c=1
while min(len(p),len(r))<=n:[[],r,p][cmp(*[sum(x%-~i<1for i in range(x))for x in c,int(str(c)[::-1])])]+=[c];c+=1
print p[n],r[n]

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Input is 0-indexed. Outputs as poor rich.

Answer 8

Ruby, 128 bytes

Input is zero-indexed. Outputs as [poor, rich].

->n,*a{b=[];i=0;d=->z{(1..z).count{|e|z%e<1}};(x=d[i+=1];y=d[i.digits.join.to_i];[[],b,a][x<=>y]<<i)until a[n]&b[n];[a[n],b[n]]}

Explanation

->n,*a{                             # Anonymous function, initialize poor array
       b=[];i=0;                    # Initialize rich array and counter variable
    d=->z{(1..z).count{|e|z%e<1}};  # Helper function to count number of factors
    (                               # Start block for while loop
     x=d[i+=1];                     # Get factors for next number
     y=d[i.digits.join.to_i];       # Factors for its reverse
                                    # (digits returns the ones digit first, so no reversing)
     [[],b,a][x<=>y]                # Fetch the appropriate array based on
                                    #  which number has more factors
                    <<i             # Insert the current number to that array
    )until a[n]&b[n];               # End loop, terminate when a[n] and b[n] are defined
    [a[n],b[n]]                     # Array with both poor and rich number (implicit return)
}                                   # End function

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Answer 9

Perl 6, 76 bytes

{classify({+(.&h <=>h .flip)},^($_*3+99)){-1,1}[*;$_]}
my&h={grep $_%%*,^$_}

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I didn't see Sean's Perl 6 answer, but this works in a different way. Note that I've hardcoded the upperbound as n*3+99, which is probably not strictly correct. However, I could replace the *3 with ³ for no extra bytes, which would make the program far less efficient, if more correct.

Answer 10

Python 2, 152 bytes

def f(n):
 a,b=[],[];i=1
 while not(a[n:]and b[n:]):[[],b,a][cmp(*[sum(m%-~i<1for i in range(m))for m in i,int(`i`[::-1])])]+=[i];i+=1
 return a[n],b[n]

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Ends up being pretty similar to Rod's answer. Returns zero-indexed nth poor, rich tuple.

Answer 11

Icon, 180 175 bytes

procedure f(n)
a:=[];b:=[];k:=1
while{s:=t:=0 
i:=1to k+(m:=reverse(k))&(k%i=0&s+:=1)|(m%i=0&t+:=1)&\x
s>t&n>*a&push(a,k)
s<t&n>*b&push(b,k)
k+:=1;n>*a|n>*b}
return[!b,!a];end

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Answer 12

APL (Dyalog Unicode), 34 bytes

{⍵⌷⍉1↓⊢⌸{×-/{≢∪⍵∨⍳⍵}¨⍵,⍎⌽⍕⍵}¨⍳1e3}

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Thanks to Adám and ngn for helping me golf this monstrosity.

TIO times out for the bigger indices (which require ⍳1e5 or ⍳1e6) , but given enough time and memory the function will terminate correctly.

Answer 13

R, 152 137 bytes

-12 bytes thanks to Giuseppe -3 bytes thanks to digEmAll

n=scan()
F=i=!1:2
`?`=sum
while(?n>i)if(n==?(i[s]=i[s<-sign((?!(r=?rev((T=T+1)%/%(e=10^(0:log10(T)))%%10)*e)%%1:r)-?!T%%1:T)]+1))F[s]=T
F

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T is the integer currently being tried; the latest poor and rich numbers are stored in the vector F.

The shortest way I could find of reversing an integer was converting it to digits in base 10 with modular arithmetic, then converting back with powers of 10 inverted, but I expect to be outgolfed on this and other fronts.

Explanation (of previous, similar version):

n=scan() # input
i=0*1:3  # number of poor, middle class, and rich numbers so far
S=sum
while(S(n>i)){ # continue as long as at least one of the classes has less than n numbers
  if((i[s]=i[
    s<-2+sign(S(!(           # s will be 1 for poor, 2 for middle class, 3 for rich
      r=S((T<-T+1)%/%10^(0:( # reverse integer T with modular arithmetic
        b=log10(T)%/%1       # b is number of digits
        ))%%10*10^(b:0)) 
      )%%1:r)-               # compute number of divisors of r
      S(!T%%1:T))            # computer number of divisors of T
    ]+1)<=n){                # check we haven't already found n of that class
    F[s]=T
  }
}
F[-2] # print nth poor and rich numbers

Answer 14

JavaScript (Node.js), 190 180 bytes

Outputs as [poor, rich].

n=>{let p,r,f=h=i=0;while(f<n){k=d(i),e=d(+(i+"").split``.reverse().join``);if(k<e){p=i;f++}if(k>e&&h<n){r=i;h++}i++}return[p,r]}
d=n=>{c=0;for(j=1;j<=n;j++)if(n%j==0)c++;return c}

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Explanation

d(n) Function

This helper finds the number of factors that a number has.

d=n=>{              // Equivalent to `function d(n) {
  c=0;              // Counter
  for(j=1;j<=n;j++) // Check every integer from 1 to n
    if(n%j==0)c++;  // If the number is a factor, add 1 to the counter
  return c
};

Main Function

n=>{ 
  let p,r,f=h=i=0; // p -> the poor number, r -> the rich number, f -> the number of poor numbers found, h -> the number of rich numbers found, i -> the current number being checked
  while(f<n){ // While it's found less than n poor numbers (assumes that it will always find the rich number first)
    k=d(i),        // k -> number of factors of i
    e=d(+((i+"").split``.reverse().join``)); // e -> number of factors of reversed i
    if(k<e){p=i;f++}  // If it hasn't found enough poor numbers and i is poor, save it and count it
    if(k>e&&h<n){r=i;h++}  // If it hasn't found enough rich numbers and i is rich, save it and count it
    i++
  };
  return[p,r]
}

Answer 15

C# (Visual C# Interactive Compiler), 221 bytes

n=>{int g,h,i,j=9,k,l,m,o,r,p;m=o=r=p=0;while(m<n||o<n){k=g=h=0;l=++j;while(l!=0){k=k*10+l%10;l/=10;}for(i=2;i<j+k;i++){if(j%i<1&&i<j)g++;if(k%i<1&&i<k)h++;}if(g<h&&o<n){o++;p=j;}if(g>h&&m<n){m++;r=j;}}return new{n,p,r};}

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