[Question]
  [
A palindromic number (in case you don't know) is a number which reads the same backwards and forwards (example, 11). The first 15 non-palindromic numbers are: `10, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26`. This is [A029742](http://oeis.org/A029742/list). I constantly need these numbers, but my sticky note pad is very small, so your code must be as short as possible.
## Rules
* Each submission must be a full program or function (e.g. in C, you can't just define a function without headers, but you can define a function WITH necessary headers
).
* If it is possible, provide a link to a site where your program can be tested.
* Your program must not write anything to `STDERR`.
* You can take input as an argument or from `STDIN` (or the closest alternative in your language).
* Programs are scored according to *bytes*. The usual character set is UTF-8, if you are using another please specify.
* [Standard loopholes](http://meta.codegolf.stackexchange.com/questions/1061/loopholes-that-are-forbidden-by-default) are forbidden.
## Test Cases
```
1
==> 10
-----
5
==> 15
-----
12
==> 23
```
## Scoring
This is [code-golf](/questions/tagged/code-golf "show questions tagged 'code-golf'"), so least bytes wins.
## Submissions
To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:
```
# Language Name, N bytes
```
where `N` is the size of your submission. If you improve your score, you *can* keep old scores in the headline, by striking them through. For instance:
```
# Ruby, <s>104</s> <s>101</s> 96 bytes
```
If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the *last* number in the header:
```
# Perl, 43 + 2 (-p flag) = 45 bytes
```
You can also make the language name a link which will then show up in the leaderboard snippet:
```
# [><>](http://esolangs.org/wiki/Fish), 121 bytes
```
## Leaderboard
Here is a Stack Snippet to generate both a regular leaderboard and an overview of winners by language.
```
/* Configuration */
var QUESTION_ID = 79251; // Obtain this from the url
// It will be like https://XYZ.stackexchange.com/questions/QUESTION_ID/... on any question page
var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";
var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 53406; // This should be the user ID of the challenge author.
/* App */
var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page;
function answersUrl(index) {
  return "https://api.stackexchange.com/2.2/questions/" +  QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER;
}
function commentUrl(index, answers) {
  return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER;
}
function getAnswers() {
  jQuery.ajax({
    url: answersUrl(answer_page++),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      answers.push.apply(answers, data.items);
      answers_hash = [];
      answer_ids = [];
      data.items.forEach(function(a) {
        a.comments = [];
        var id = +a.share_link.match(/\d+/);
        answer_ids.push(id);
        answers_hash[id] = a;
      });
      if (!data.has_more) more_answers = false;
      comment_page = 1;
      getComments();
    }
  });
}
function getComments() {
  jQuery.ajax({
    url: commentUrl(comment_page++, answer_ids),
    method: "get",
    dataType: "jsonp",
    crossDomain: true,
    success: function (data) {
      data.items.forEach(function(c) {
        if (c.owner.user_id === OVERRIDE_USER)
          answers_hash[c.post_id].comments.push(c);
      });
      if (data.has_more) getComments();
      else if (more_answers) getAnswers();
      else process();
    }
  });  
}
getAnswers();
var SCORE_REG = /<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;
var OVERRIDE_REG = /^Override\s*header:\s*/i;
function getAuthorName(a) {
  return a.owner.display_name;
}
function process() {
  var valid = [];
  
  answers.forEach(function(a) {
    var body = a.body;
    a.comments.forEach(function(c) {
      if(OVERRIDE_REG.test(c.body))
        body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>';
    });
    
    var match = body.match(SCORE_REG);
    if (match)
      valid.push({
        user: getAuthorName(a),
        size: +match[2],
        language: match[1],
        link: a.share_link,
      });
    
  });
  
  valid.sort(function (a, b) {
    var aB = a.size,
        bB = b.size;
    return aB - bB
  });
  var languages = {};
  var place = 1;
  var lastSize = null;
  var lastPlace = 1;
  valid.forEach(function (a) {
    if (a.size != lastSize)
      lastPlace = place;
    lastSize = a.size;
    ++place;
    
    var answer = jQuery("#answer-template").html();
    answer = answer.replace("{{PLACE}}", lastPlace + ".")
                   .replace("{{NAME}}", a.user)
                   .replace("{{LANGUAGE}}", a.language)
                   .replace("{{SIZE}}", a.size)
                   .replace("{{LINK}}", a.link);
    answer = jQuery(answer);
    jQuery("#answers").append(answer);
    var lang = a.language;
    if (/<a/.test(lang)) lang = jQuery(lang).text();
    
    languages[lang] = languages[lang] || {lang: a.language, user: a.user, size: a.size, link: a.link};
  });
  var langs = [];
  for (var lang in languages)
    if (languages.hasOwnProperty(lang))
      langs.push(languages[lang]);
  langs.sort(function (a, b) {
    if (a.lang > b.lang) return 1;
    if (a.lang < b.lang) return -1;
    return 0;
  });
  for (var i = 0; i < langs.length; ++i)
  {
    var language = jQuery("#language-template").html();
    var lang = langs[i];
    language = language.replace("{{LANGUAGE}}", lang.lang)
                       .replace("{{NAME}}", lang.user)
                       .replace("{{SIZE}}", lang.size)
                       .replace("{{LINK}}", lang.link);
    language = jQuery(language);
    jQuery("#languages").append(language);
  }
}
```
```
body { text-align: left !important}
#answer-list {
  padding: 10px;
  width: 290px;
  float: left;
}
#language-list {
  padding: 10px;
  width: 290px;
  float: left;
}
table thead {
  font-weight: bold;
}
table td {
  padding: 5px;
}
```
```
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b">
<div id="answer-list">
  <h2>Leaderboard</h2>
  <table class="answer-list">
    <thead>
      <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr>
    </thead>
    <tbody id="answers">
    </tbody>
  </table>
</div>
<div id="language-list">
  <h2>Winners by Language</h2>
  <table class="language-list">
    <thead>
      <tr><td>Language</td><td>User</td><td>Score</td></tr>
    </thead>
    <tbody id="languages">
    </tbody>
  </table>
</div>
<table style="display: none">
  <tbody id="answer-template">
    <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>
<table style="display: none">
  <tbody id="language-template">
    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
  </tbody>
</table>
```
  
[Answer]
# Pyth, 7 bytes
```
e.f!_I`
```
[Test suite](https://pyth.herokuapp.com/?code=e.f%21_I%60&test_suite=1&test_suite_input=1%0A5%0A10%0A100&debug=0)
**Explanation:**
```
e.f!_I`
e.f!_I`ZQ    Implicit variable introduction.
 .f     Q    Find the first Q numbers whether the following is truthy,
             starting at 1, where Q is the input.
      `Z     Convert the number to a string.
     _I      Check if it's the same when reversed.
    !        Logical not.
 e           Return the last element of the list.
```
[Answer]
## Haskell, 38 bytes
```
([x|x<-[1..],(/=)<*>reverse$show x]!!)
```
Uses 0-based index. `([x|x<-[1..],(/=)<*>reverse$show x]!!) 11` -> `23`.
The test whether to keep a number `(/=)<*>reverse$show x` translates to `(show x) /= (reverse (show x))`, i.e check if the string representation of the number does not equal the reverse of the string representation.
[Answer]
# [Brachylog](https://github.com/JCumin/Brachylog), ~~14~~ 11 bytes
```
;0{<≜.↔¬}ⁱ⁽
```
-3 bytes tanks to Fatalize
## Explanation
```
; {      }ⁱ⁽        --  Find the nth number
 0                  --      (starting with 0)
   <                --      which is bigger then the previous one
    ≜               --      make explicit (otherwise it fucks up)
      .             --      which is the output
       ↔            --      and if reversed
        ¬           --      is not the output
```
[Try it online!](https://tio.run/##ASUA2v9icmFjaHlsb2cy//87MHs84omcLuKGlMKsfeKBseKBvf//Nf9a "Brachylog – Try It Online")
[Answer]
# Jelly, 9 bytes
1 bytes thanks to [@Sp3000](https://codegolf.stackexchange.com/users/21487/sp3000).
```
ṚḌ_
0Ç³#Ṫ
```
[Try it online!](http://jelly.tryitonline.net/#code=4bma4biMXwoww4fCsyPhuao&input=&args=MTI)
[Test suite.](http://jelly.tryitonline.net/#code=4bma4biMXwoww4figbgj4bmqCsOH4oKs&input=&args=MSw1LDEy)
## Explanation
```
DUḌ_   Helper link. Check if x is not palindrome.
D      Convert to decimal.
 U     Reverse.
  Ḍ    Convert back to integer.
   _   Subtract x from the result above.
       For 23, this will yield 32-23 = 9.
       Only yield 0 (falsy) if x is palindrome.
       If x is not a palindrome,
       it will return a truthy number.
0Ç³#Ṫ  Main link.
0      Start from 0.
   #   Find the first         numbers:
  ³                   <input>
 Ç         where the above link returns a truthy number.
    Ṫ  Yield the last of the matches.
```
[Answer]
# [05AB1E](http://github.com/Adriandmen/05AB1E), 8 bytes
Code:
```
µNÂÂQ>i¼
```
Uses **CP-1252** encoding. [Try it online!](http://05ab1e.tryitonline.net/#code=wrVOw4LDglE-acK8&input=MTI).
[Answer]
# Clojure, 62 bytes
```
#(nth(for[i(range):when(not=(seq(str i))(reverse(str i)))]i)%)
```
0-indexed. Generate lazily infinite range of non-palindromic numbers using list comprehension and take `i`th one. See it online: <https://ideone.com/54wXI3>
[Answer]
## PowerShell v2+, 65 bytes
```
for(;$j-lt$args[0]){if(++$i-ne-join"$i"["$i".length..0]){$j++}}$i
```
Loops through numbers from `0` (implicit value for uninitialized `$i`) until we find input `$args[0]` many matches, then outputs the last one. Note that we don't initialize the loop, so `$j=0` is implicit.
Each iteration, we pre-increment `$i`, and check if it's not-equal to `$i` reversed. If so, that means we've found a non-palindrome, so increment `$j`. The loop then continues as many times as necessary.
### Examples
```
PS C:\Tools\Scripts\golfing> .\print-nth-palindromic-number.ps1 100
120
PS C:\Tools\Scripts\golfing> .\print-nth-palindromic-number.ps1 5
15
PS C:\Tools\Scripts\golfing> .\print-nth-palindromic-number.ps1 55
70
PS C:\Tools\Scripts\golfing> .\print-nth-palindromic-number.ps1 212
245
```
[Answer]
# Python 2, 60 bytes
```
f=lambda n,i=0,j=1:j>n and i-1or f(n,i+1,j+(`i`!=`i`[::-1]))
```
A one-indexed function that takes input of `n` via argument and returns the `n`th non-palindromic number.
**How it works**
This is an exhaustive recursive search, which consecutively tests integers `i` in the range `[1,∞)` until `n` non-palindromic numbers have been found; since `i` is pre-incremented, `i-1` is then returned. Testing whether a number is palindromic is performed by converting to a string, reversing, and then checking whether the original and reversed strings are equal.
The code is logically equivalent to:
```
def f(n,test=0,count=1):
    if count>n:
        return test
    elif str(test)!=reversed(str(test)):
        return f(n,test+1,count+1)
    else:
        return f(n,test+1,count)
```
which itself is essentially:
```
def f(n):
    test=0
    count=1
    while count<=n:
        if str(test)!=reversed(str(test)):
            count+=1
        test+=1
    return test-1
```
[Try it on Ideone](https://ideone.com/pkfUJC)
[Answer]
## Clojure, 62 bytes
```
#(nth(filter(fn[i](not=(seq i)(reverse i)))(map str(range)))%)
```
A quite different approach than the other answer, but equal length.
[Answer]
# [R](https://www.r-project.org/), ~~133~~ ~~117~~ ~~93~~ 76 bytes
-16 bytes thanks to JayCe. -41 bytes thanks to Giuseppe.
```
x=scan();while({F=F+any((D=T%/%10^(1:nchar(T)-1)%%10)!=rev(D));F<=x})T=T+1;T
```
[Try it online!](https://tio.run/##K/r/v8K2ODkxT0PTujwjMydVo9rN1k07Ma9SQ8PFNkRVX9XQIE7D0CovOSOxSCNEU9dQUxUopKloW5RapuGiqWntZmNbUasZYhuibWgd8t/0PwA "R – Try It Online")
[Answer]
# [Forth (gforth)](http://www.complang.tuwien.ac.at/forth/gforth/Docs-html/), 103 99 bytes
```
: f 9 swap 0 do begin 1+ dup 0 over begin 10 /mod >r swap 10 * + r> ?dup 0= until = 0= until loop ;
```
[Try it online!](https://tio.run/##RY1bDoIwFAW3Mt8SpP0EAq5FUopNkNtUHsuvrWL8uueeM8lYCeujnGw@MTZYal7H3aMwwjBObkEXmC03so/h1ymqpxj68MXTf6Eg9Nw@bMe2rG6m@8dZxNMmR6YVOgtcEl7PBR/f "Forth (gforth) – Try It Online")
### Explanation
Loop n times, each iteration finds the next non-palindromic number by incrementing a counter by 1 until the number doesn't equal itself reversed
### Ungolfed Code
Normally I wouldn't "ungolf" the code, but since this code is somewhat messy I figured it would help
```
: reverse ( s -- s )
    0 swap 
    begin 
        10 /mod
        >r swap
        10 * +
        r> ?dup 0=
    until 
; 
: f ( s -- s )
    9 swap 0
    0
    do
        begin
            1+ dup dup
            reverse =
        0= until
    loop
;
```
### Code Explanation
```
: f                \ start a new word definition
  9                \ start at 9, since all positive ints < 10 are palindromic
  swap 0           \ set up loop parameters from 0 to n-1
  do               \ start a counted loop       
    begin          \ start an indefinite loop
      1+ dup       \ increment counter and place a copy on the stack
      ( Reverse )
      0 over       \ add 0 to the stack (as a buffer) and copy the top counter above it
      begin        \ start another indefinite loop
        10 /mod    \ get the quotient and remainder of dividing the number by 10
        >r         \ store the quotient on the return stack
        swap 10 *  \ multiply the current buffer by 10
        +          \ add the remainder to the buffer
        r>         \ grab the quotient from the return stack
        ?dup       \ duplicate if not equal to 0
        0=         \ check if equal to 0
      until        \ end inner indefinite loop if quotient is 0
      ( End Reverse )
      = 0=         \ check if counter =/= reverse-counter            
    until          \ end the outer indefinite loop if counter =/= reverse-counter
  loop             \ end the counted loop
;                  \ end the word definition 
```
[Answer]
# [Perl 6](http://perl6.org), 29 bytes
```
{grep({$_!= .flip},^Inf)[$_]}
```
( uses 0 based index )
```
{         # The $_ is implied above
  grep(   # V
    { $_ != $_.flip }, # only the non-palindromic elements of
    ^Inf               # an Infinite list ( 0,1,2,3 ...^ Inf )
  )[ $_ ]              # grab the value at the given index
}
```
### Usage:
```
my &non-palindrome = {grep({$_!= .flip},^Inf)[$_]}
say non-palindrome 1  - 1; # 10
say non-palindrome 5  - 1; # 15
say non-palindrome 12 - 1; # 23
# this also works:
say non-palindrome 0..20;
# (10 12 13 14 15 16 17 18 19 20 21 23 24 25 26 27 28 29 30 31 32)
```
[Answer]
## Actually, 17 bytes
```
;τR9+;`$;R=Y`M@░E
```
[Try it online!](http://actually.tryitonline.net/#code=O8-EOStSO2AkO1I9WWBNQOKWkUU&input=MQ)
Values are 1-indexed. This could be easily changed to 0-indexed by replacing the first `R` with `r`. But, `R` is what I initially typed, so that's what I'm going with.
The nonpalindromic numbers satisfy `a(n) ≈ n + 10`, so `2n+9` is a sufficient upper bound.
Explanation:
```
;τR9+;`$;R=Y`M@░E
;τ9+R;             push n, range(1,(2*n)+10)
      `$;R=Y`M@░   take values that are not palindromic
                E  take nth element
```
[Answer]
## JavaScript (ES6), 54 bytes
Uses 1-based indexing. Only works up until the 7624th number.
```
d=(i,a=0)=>i?d(i-=++a!=[...''+a].reverse().join``,a):a
```
### Usage
```
d=(i,a=0)=>i?d(i-=++a!=[...''+a].reverse().join``,a):a
d(1)
10
d(123)
146
d(7624)
7800
d(7625)
// Uncaught RangeError: Maximum call stack size exceeded
```
---
## JavaScript (ES6), 59 bytes
Doesn't use recursion and so can handle much larger inputs.
```
i=>eval("for(a=9;i-=++a!=[...`${a}`].reverse().join``;);a")
```
### Usage
```
(i=>eval("for(a=9;i-=++a!=[...`${a}`].reverse().join``;);a"))(1)
10
(i=>eval("for(a=9;i-=++a!=[...`${a}`].reverse().join``;);a"))(7625)
7801
(i=>eval("for(a=9;i-=++a!=[...`${a}`].reverse().join``;);a"))(123456)
124579
```
[Answer]
## Javascript (using external library) (97 bytes)
```
n=>_.Sequence(n,i=>{i=_.From(i+"");if(!i.Reverse().SequenceEqual(i)){return i.Write("")}}).Last()
```
Link to lib: <https://github.com/mvegh1/Enumerable>
Code explanation: Library has static method called Sequence, where first param defines how many elements the sequence will guarantee to create, and the 2nd parameter is a predicate accepting the current iteration value, "i". The predicate converts the integer to a string, which gets converted to a char array by calling \_.From. The char array is compared against the reversal of the char array, and if they are not equal the char array is joined back into a string and returned. Otherwise, nothing is returned (i.e the result is undefined, which the library will always ignore). Finally, the last element of the sequence, i.e the Nth element is returned
[![enter image description here](https://i.stack.imgur.com/NfSo4.png)](https://i.stack.imgur.com/NfSo4.png)
[Answer]
## C, 84 bytes
Function `f(n)` takes integer `n` and returns `n-th` non-palindromic number (1 based).
```
g(n,r){return n?g(n/10,r*10+n%10):r;}s;f(n){for(s=9;n--;g(++s,0)==s&&s++);return s;}
```
Test it on [Ideone!](http://ideone.com/HD9e2s)
It's fairly trivial code, thus there is probably space for improvement.
[Answer]
# Ruby, 54 bytes
This function is 1-indexed and is partially based on [Dom Hastings's Javascript answer](https://codegolf.stackexchange.com/a/86605/47581). I think there's a way to golf this better, especially with that last ternary condition. Also, this function currently returns a string, which may need to be edited later. Any golfing suggestions are welcome.
```
f=->x,y=?9{x<1?y:(y.next!.reverse!=y)?f[x-1,y]:f[x,y]}
```
**Ungolfed:**
```
def f(x, y="9")
 if x<1
  return y
 else
  y = y.next
  if y.reverse != y
   return f(x-1, y)
  else
   return f(x, y)
  end
 end
end
```
[Answer]
## C++ (GCC), 148 bytes
It's 1-based and the algorithm is really naive
```
#import <iostream>
using namespace std;int n,i=1;string s;main(){cin>>n;while(s=to_string(i+1),(n+=equal(begin(s),end(s),s.rbegin()))-i++);cout<<i;}
```
[Answer]
# APL NARS 35 chars
```
r←v a;c
r←c←0
A:r+←1⋄c+←r≠⍎⌽⍕r⋄→A×⍳c<a
```
it is the function v; "⍎⌽⍕"r traslate number r in string, reverse that string,
traslate from string to number. Test and help functions:
```
  ⍝ return the one string for the basic types ('Char', 'Int', 'Float', 'Complex or Quaternion or Oction')
  ⍝ or one string for composite types ('Tensor 3' 'Tensor 4' etc 'Matrix', 'List', 'String')
  ⍝ follow the example in: https://codegolf.stackexchange.com/a/39745
  type←{v←⍴⍴⍵⋄v>2:'Tensor ',⍕v⋄v=2:'Matrix'⋄(v=1)∧''≡0↑⍵:'String'⋄''≡0↑⍵:'Char'⋄v=1:'List'⋄⍵≢+⍵:'Complex or Quaternion or Oction'⋄⍵=⌈⍵:'Int'⋄'Float'}
  h←{'Int'≢type ⍵:¯1⋄(⍵<1)∨⍵>2e5:¯1⋄v ⍵} 
  h 1
10
  h 1.32
¯1
  h 7878
8057
  h¨3 5 12
13 15 23 
  h 6 7 8
¯1
  h '123'
¯1
  h '1'
¯1
  h 1.0
10
  h 1.0003
¯1
  h ¯2
¯1
  h 0
¯1
  h 200000
201200
  h 200001
¯1
  
```
[Answer]
# [Husk](https://github.com/barbuz/Husk), 6 bytes
Yay for `↔` :)
```
!fS≠↔N
```
[Try it online!](https://tio.run/##yygtzv7/XzEt@FHngkdtU/z@//9vaAQA "Husk – Try It Online")
### Explanation
```
 f   N  -- filter the naturals by:
  S≠    --   is it not equal to..
    ↔   --   ..itself reversed
!       -- index into that list
```
[Answer]
# [Perl 5](https://www.perl.org/), 33 + 1 (`-p`) = 34 bytes
```
map{1until++$\!=reverse$\}1..$_}{
```
[Try it online!](https://tio.run/##K0gtyjH9/z83saDasDSvJDNHW1slRtG2KLUstag4VSWm1lBPTyW@tvr/f0Ojf/kFJZn5ecX/dQsA "Perl 5 – Try It Online")
[Answer]
# C# 7, 89 bytes
```
n=>{int i=9;for(;n-->0;)if(Enumerable.SequenceEqual(++i+"",(""+i).Reverse()))i++;return i;}
```
1 indexed
[Try on Repl.It](https://repl.it/@ryzngard/SuperiorFormalDiscussions)
```
n=>
  int i = 9;                                  | Start at 9. Iterate exactly n times. Assume n >= 1      
  for(;n-->0;)                                | Iterate n times
  if(EnumerableSequenceEqual(                 | Compare two sequences
  ++i+"",(""+i).Reverse())                    | Generate the forward and backward strings, which behave like char sequences for Linq
  i++                                         | If the sequences are equal, the number is a palindrome. Increment i to skip
  return i;                                   | Return the number after the for loop exits
```
I don't think this uses any language features from c# 7, but I put there since that's what I tested against
[Answer]
# Java 8, ~~117~~ ~~95~~ 94 bytes
```
n->{int r=10;for(;n-->0;)if((++r+"").contains(new StringBuffer(r+"").reverse()))r++;return r;}
```
0-indexed
**Explanation:**
[Try it here.](https://tio.run/##hVDBasMwDL33K0RPMiYhgd1Me9i9vfQ4evBcebhL5SA7GaXk2zOn6XUMJIHeE@89dLWjrWJPfL18z66zKcHBBn5sAAJnEm8dwXFZnwA4XCYrU5CpdKmUbQ4OjsCwg5mr/WM5kV3bGB8FDVfVvjEqeEStRW@3qnaRc3FJyPQDpyyBv94H70lw5YVGkkSolBKtjVAehEHMNJtNceyHz644vozHGC5wK3K4Kn2cwao18emeMt3qOOS6L1TuGLl22Khn/j/5t3/4tlWvB0zzLw)
```
n->{             // Method with integer as both parameter and return-type
  int r=10;      //  Result-integer, starting at 10
  for(;n-->0;)   //  Loop an amount of times equal to the input
    if((++r+"")  //   First raise `r` by 1, and then check if `r`
               .contains(new StringBuffer(r+"").reverse()))
                 //   is the same as `r` reversed (and thus a palindrome)
      r++;       //    And if it is: raise `r` by 1 again
  return r;}     //  Return result-integer
```
[Answer]
# [Japt](https://github.com/ETHproductions/japt), ~~14~~ ... ~~9~~ 7 [bytes](https://en.wikipedia.org/wiki/ISO/IEC_8859-1)
0-indexed.
```
È¦sw}iU
```
[Try it](https://petershaggynoble.github.io/Japt-Interpreter/?v=1.4.6&code=yKZzd31pVQ&input=MA)
```
È¦sw}iU     :Implicit input of integer U
È           :Function taking an integer X as an argument
 ¦          :  Test for inequality with
  s         :  Convert to string
   w        :  Reverse
    }       :End function
     iU     :Get the Uth integer that returns true
```
[Answer]
## TCC, 11 bytes
```
?>!~<>;i;'T
```
[Try it online!](https://www.ccode.gq/projects/tcc.html?code=%3F%3E!~%3C%3E%3Bi%3B%27T&input=12)
```
            | Printing is implicit
?>          | Find n-th number for which the following is "T":
  !~        | If not equal...
    <>;     | reverse. No value specified, so input is assumed.
       i;   | Input, since double semicolons are ignored
         'T | ... print string "T"
```
]