Palindromic Residue

Question

Today, as I'm writing this, is March 31st. In the US, this is 3/31. I was playing around with 331 as a number to come up with a challenge, and found that its residues (modulo small numbers) is palindromic. 331%2=1, 331%3=1, 331%4=3, 331%5=1, 331%6=1 (11311).

Your challenge here is, when given an integer n > 2, output the first n positive numbers that have palindromic residue when taken modulo [2,n].

For example, for input 7, the output should be 1, 42, 43, 140, 182, 420, 421. Here's the chart explaining why that's the case:

        mod
num | 2 3 4 5 6 7
-----------------
  1 | 1 1 1 1 1 1
 42 | 0 0 2 2 0 0
 43 | 1 1 3 3 1 1
140 | 0 2 0 0 2 0
182 | 0 2 2 2 2 0
420 | 0 0 0 0 0 0
421 | 1 1 1 1 1 1

Input

A single positive integer n with n > 2 in any convenient format.

Output

The resulting array/list of the first n palindromic residues, as outlined above. Again, in any suitable format.

Rules

• For n > 10, assume the residue list to be flattened before checking whether it's a palindrome. That is, [1, 10, 11] is palindromic, but [1, 10, 1] is not.
• Either a full program or a function are acceptable. If a function, you can return the output rather than printing it.
• If possible, please include a link to an online testing environment so other people can try out your code!
• Standard loopholes are forbidden.
• This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.

Examples

[input]
[output]

3
[1, 6, 7]

4
[1, 4, 5, 8]

5
[1, 50, 60, 61, 110]

6
[1, 30, 31, 60, 61, 90]

7
[1, 42, 43, 140, 182, 420, 421]

8
[1, 168, 169, 336, 337, 504, 505, 672]

9
[1, 2520, 2521, 5040, 5041, 7560, 7561, 10080, 10081]

10
[1, 280, 281, 560, 1611, 1890, 1891, 2170, 2171, 2241]

11
[1, 22682, 27720, 27721, 50402, 55440, 55441, 78122, 83160, 83161, 105842]

Answer 1

Haskell, 57 bytes

f n=take n[x|x<-[1..],(==)=<<reverse$show.mod x=<<[2..n]]

Usage example: f 4 -> [1,4,5,8]. Try it online!

The first =<< is in function context and translates to the lambda \x -> reverse x == x and the second =<< is in list context and equivalent to concatMap, i.e. map-and-flatten-one-list-level.

Answer 2

05AB1E, 12 bytes

µN2¹Ÿ%JÂQD½–

Try it online!

Explanation

µ              # until counter equals input do:
 N             # push current iterations number
     %         # modulus each in
  2¹Ÿ          # range [2 ... input]
      J        # joined to string
       ÂQ      # equals it's reverse
         D     # duplicate
          ½    # if true, increase counter
           –   # if true print iteration number

Answer 3

Mathematica, 79 bytes

NestList[#+1//.x_/;!PalindromeQ[ToString/@Mod[x,Range@n+1]<>""]:>x+1&,1,n=#-1]&

Answer 4

JavaScript (ES6), 104 bytes

f=(n,x=(k=--n,2))=>k?([...Array(n)].map(_=>(r=x%++i+r,x%i),i=1,r='').join``==r?k--&&x+' ':'')+f(n,x+1):1

Demo

NB: Because of the numerous recursive calls, this will crash for n > 8 on Firefox or n > 10 on Chrome.

f=(n,x=(k=--n,2))=>k?([...Array(n)].map(_=>(r=x%++i+r,x%i),i=1,r='').join``==r?k--&&x+' ':'')+f(n,x+1):1

for(n = 2; n < 9; n++) {
  console.log(n, '=>', f(n));
}

Answer 5

Python 2, 98 97 bytes

n=input();i=j=0
while i<n:
 j+=1;l=''
 for k in range(2,n+1):l+=`j%k`
 if l==l[::-1]:i+=1;print j

Try it Online!

Answer 6

MATL, 19 bytes

Thanks to @AdmBorkBork for pointing out a mistake in an earlier version of the code, now corrected

`@Gq:Q\VXztP=?@]NG<

Try it online!

Explanation

`        % Do...while
  @      %   Push iteration index, starting at 1
  Gq:Q   %   Push [2 3 ... n], where n is the input
  \      %   Modulo, element-wise
  V      %   Convert to string. Numbers are separated by spaces
  Xz     %   Remove spaces
  tP     %   Duplicate, flip
  =      %   Equal? (element-wise)
  ?      %   If all results were true
    @    %     Push current iteration index. It is one of the sought numbers
  ]      %   End
  N      %   Push number of elements in stack
  G      %   Push input n
  <      %   Less than? This is the loop condition
         % End (implicit). Display (implicit)

Answer 7

Scala, 90 86 82 bytes

(n:Int)=>Stream.from(1)filter{i=>val d=(2 to n)map(i%)mkString;d.reverse==d}take(n)

Explanation

Stream.from(1)                              // From an infinite Stream starting from 1,
    filter ( i => {                         // keep only elements matching the next condition :
        val d=(2 to n)map(i%)mkString;      // Generate residues and convert to String,
        d.reverse==d                        // return true if palindrom, false otherwise
    })take(n)                               // Finally, take the n first elements matching the condition

Test cases

val f = (n:Int)=>...    // assign function
(3 to 11).foreach { i =>
    println(i + "\n" + f(i).mkString(", ") + "\n")
}

Results

3
1, 6, 7

4
1, 4, 5, 8

5
1, 50, 60, 61, 110

6
1, 30, 31, 60, 61, 90

7
1, 42, 43, 140, 182, 420, 421

8
1, 168, 169, 336, 337, 504, 505, 672

9
1, 2520, 2521, 5040, 5041, 7560, 7561, 10080, 10081

10
1, 280, 281, 560, 1611, 1890, 1891, 2170, 2171, 2241

11
1, 22682, 27720, 27721, 50402, 55440, 55441, 78122, 83160, 83161, 105842

Edits

#1 (90 => 86)

• anonymous function

#2 (86 => 82)

• remove useless dot characters after parenthesis or bracket (ex. : (2 to n).map(%i) => (2 to n)map(%i)

Answer 8

Jelly, 12 bytes

%ЀḊDFŒḂ
1ç#

How?

1ç# - Main link: n
1   - initialise "i" at 1
  # - increment i and yield a list of the first n truthful results of:
 ç  -     last link (1) as a dyad

%ЀḊDFŒḂ - Link 1, test a value "i" for mod [2,n] being palindromic: i, n
 Ѐ      - for each, mapped over the right argument, i.e. for j = 1 to n:
%        -     i modulo j
   Ḋ     - dequeue, i.e. discard the modulo 1 result
    D    - convert to decimal list (vectorises)
     F   - flatten into one list
      ŒḂ - is palindromic?

Try it online!

Answer 9

CJam, 28 bytes

0ri:N{{)_N),2>f%s_W%#}g_p}*;

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Explanation

0          e# Push 0, the value we'll repeatedly increment to search for valid outputs.
ri:N       e# Read input, convert to integer, store in N.
{          e# Run this block N times...
  {        e#   Run this block until the condition is true, which will find the next
           e#   number with palindromic residues...
    )_     e#     Increment and duplicate.
    N),2>  e#     Push [2 3 ... N].
    f%     e#     Take the current value modulo each of these.
    s      e#     Flatten them into a single string.
    _W%    e#     Duplicate and reverse.
    #      e#     Try to find the reverse in the original. A common way to compute
           e#     "not equal" for strings of the same length.
  }g
  _p       e#   Print a copy of the result.
}*
;          e# Discard the final result to prevent printing it twice.

Answer 10

PHP, 93 Bytes

for(;$x<$a=$argn;$s="")for($i=1,++$n;$i++<$a;)if($i==$a&strrev($s.=$n%$i)==$s)echo$n._.!++$x;

Online Version 2 Loops Output as string

Expanded

for(;$x<$a=$argn;$s="") 
for($i=1,++$n;$i++<$a;)
    if($i==$a&strrev($s.=$n%$i)==$s)echo$n._.!++$x; 

PHP 130 Bytes

for(;count($r)<$a=$argn;$s=[])for($i=1,++$n;$i++<$a;){$s[]=$n%$i;if(count($s)==$a-1&strrev($j=join($s))==$j)$r[]=$n; }print_r($r);

Online Version 2 Loops

Expanded

for(;count($r)<$a=$argn;$s=[])
for($i=1,++$n;$i++<$a;){
    $s[]=$n%$i;
    if(count($s)==$a-1&strrev($j=join($s))==$j)$r[]=$n; 
}
print_r($r);

PHP, 139 Bytes with 1 loop

for($i=$n=1;count($r)<($a=$argn)&$i++<$a;){$s[]=$n%$i;if(count($s)==$a-1){if(strrev($j=join($s))==$j)$r[]=$n;$n++;$s=[];$i=1;}}print_r($r);

Online Version 1 Loop

Run with

echo '<string>' | php -nR '<code>'

Expanded

for($i=$n=1;count($r)<($a=$argn)&$i++<$a;){
    $s[]=$n%$i;
    if(count($s)==$a-1){
        if(strrev($j=join($s))==$j)$r[]=$n;
        $n++;
        $s=[];
        $i=1;
    }
}
print_r($r);

Answer 11

QBIC, 48 bytes

:{A=G[2,a|A=A+!q%b$]~A=_fA||h=h+1?q]q=q+1~h=a|_X

Beats Mathematica! Sample run:

Command line: 10
 1 
 280 
 281 
 560 
 1611 
 1890 
 1891 
 2170 
 2171 
 2241 

Explanation:

:{          Get 'a' from the command line, start an inf. loop
A=G         Clear out whatever's in A$
[2,a|       For each of the numbers we want to modulo
A=A+        Add to A$ 
     q%b       our current number MODULO te loop iterator
    !   $      cast to string
]           NEXT
~A=_fA|     If the string of remainders is a palindrome (_f ... | is Reverse())
|h=h+1      THEN h=h+1 (h starts at 0) - this counts how many hits we've had
 ?q            also, print the number with the palindromic remainder
]           END IF
q=q+1       Test the next number
~h=a|_X     If we've had 'a' hits, quit.
            The last IF and the infinite loop are closed implicitly.

Answer 12

Japt, 26 bytes

L³o fR{C=Uò2@R%Xì ¥CwÃj1U

Try it online! Takes a few seconds on all inputs, so be patient please.

This would be considerably shorter (and faster) if there were a built-in to get the first N numbers satisfying some condition:

R{C=Uò2@R%Xì ¥Cw}aU

Answer 13

Jelly, 8 bytes

1%ⱮḊŒḂɗ#

Try it online!

How it works

1%ⱮḊŒḂɗ# - Main link. Takes n on the left
1        - Set the return value to 1
       # - Count up i = 1, 2, ..., and return the first n which return true:
      ɗ  -   Combine the previous three links into a dyad
             with i on the left and n on the right:
  Ɱ      -     For each integer m in 1, 2, ..., n:
 %       -       i % m
   Ḋ     -     Remove i % 1
    ŒḂ   -     Is palindromic?
           

Answer 14

Japt, 13 bytes

@õ!uX ÅêS}jU1

Try it or run all test cases

@õ!uX ÅêS}jU1     :Implicit input of integer U
@                 :Function taking an integer X as argument
 õ                :  Range [1,U]
  !uX             :  Mod X by each
      Å           :  Slice off the first
       êS         :  Is palindrome?
         }        :End function
          jU1     :Starting with 1, return the first U integers that return true