# The Rien Number

The Champernowne constant is a number that is constructed by concatenating the first n numbers, with n tending to infinity. It looks something like this:

0.123456789101112131415161718192021222324252627282930...


Now, I will describe to you the Rien number. It can be thought of as a minimization of the Champernowne constant as an integer. I will refer to the Rien number with the first n digits as Ri(n). This is how to formulate it:

1. The first n natural numbers (the sequence {1,2,3,...}) are concatenated.
2. This result is then sorted, according to the digit value. So 1..12 would look like 011111223456789.
3. Since the Rien number cannot have leading zeroes, we move all 0s so that they are significant, whilst keeping the number minimized, resulting in, say, 101111223456789. This is Ri(n), in this case, Ri(12).

Here are some results for Ri(n):

n    Ri(n)
1    1
2    12
3    123
7    1234567
9    123456789
10   10123456789
15   101111111223344556789
34   10001111111111111222222222222223333333334444555666777888999
42   100001111111111111122222222222222233333333333333444444455556666777788889999
45   100001111111111111122222222222222233333333333333344444444444555556666777788889999
55   10000011111111111111122222222222222223333333333333333444444444444444455555555555566666777778888899999
100  100000000000111111111111111111112222222222222222222233333333333333333333444444444444444444445555555555555555555566666666666666666666777777777777777777778888888888888888888899999999999999999999
999  100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000011111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222222333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444444555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555555666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666666777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777777888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888888999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999


Objective Given a number 1 ≤ n < 10000 as input (via arguments, STDIN, or hardcoding if your language does not support conventional input), output/return Ri(n).

This is a , so the shortest code in bytes wins. You may use a language that was made after this contest, as long as it was not made for answering this challenge. (Of course, you can use it, if it provides an interesting solution, but mark your answer as non-competing.)

# Reference implementation

I tested this in IE, so there really shouldn't be a problem. If there is a problem, there's an easy solution: get a sane browser.

function min(n) {
var seq = [];
for(var i = 1; i <= n; i++) seq.push(i);
seq = seq.join("").split("").map(Number);
var to;
if(seq.indexOf(1) >= 0) to = seq.splice(seq.indexOf(1), 1);
seq.sort(function(a, b) {
return a - b;
});
if(to) seq = to.concat(seq);
return seq.join("");
}
t.onchange = t.onkeyup = function() {
h.innerHTML = min(this.value)
}
* {
font-family: Consolas, monospace;
}
input {
border: 2px dotted #aaaaaa;
margin: 10px;
}
<input id="t" type="number">
<div id="h">

The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

## 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 snippet:

## [><>](http://esolangs.org/wiki/Fish), 121 bytes


var QUESTION_ID=68685,OVERRIDE_USER=44713;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){return e.lang>s.lang?1:e.lang<s.lang?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}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>

• I always thought 0 is the rien number. – flawr Jan 6 '16 at 19:36
• Not sure if I'm missing something, but we can just move one of the 1s in front of the 0s, yeah? – FryAmTheEggman Jan 6 '16 at 19:40
• @FryAmTheEggman You are correct. – Conor O'Brien Jan 6 '16 at 19:40
• @MartinBüttner \o/ you found it. – Conor O'Brien Jan 6 '16 at 20:02
• @CᴏɴᴏʀO'Bʀɪᴇɴ Nope that one was different. That one only allowed permutations of the entire numbers, not of their individual digits. – Martin Ender Jan 6 '16 at 20:05

# Pyth, 8

+1SjktSQ


Makes a list [1, .. , input] then removes the leading one, joins and sorts, then prepends a 1.

Test Suite

## Perl, 44424133 31 bytes

Yaaay, first post ever !

Thanks to primo for the 2 bytes save.

print 1,sort"@{[2..<>]}"=~/\d/g


As others did, removing 1 and prepending it manually does the work.

Try it online

• Welcome to PPCG, and congrats on your first post (though it seems strange having me welcome you, since you joined before I did ...). Two things -- I count 43 bytes ... second, your "Try it online" link looks to point to an older/different revision of your code. Manually copy-pasting your code into Ideone works, but not your link. – AdmBorkBork Jan 6 '16 at 20:49
• Thank you for you comment ! I used TextWrangler for the byte count, guess I went 1 too far... (plus a space wasn't needed, so that reduces the whole bytecount to 42). The Ideone should be fixed now. – Paul Picard Jan 6 '16 at 20:52
• Oh, didn't know about that. Also works on my Perl on my Mac (5.18) Thanks ! – Paul Picard Jan 6 '16 at 20:54
• Crossed out 44 is still regular 44 ;( – AdmBorkBork Jan 6 '16 at 20:56
• Saved two bytes by getting rid of the split/join: print 1,sort"@{[2..<>]}"=~/\d/g – primo Jan 7 '16 at 11:32

# Japt, 14 12 bytes

1+2o°U ¬¬n ¬


Try it online!

### How it works

1+2o°U ¬¬n ¬  // Implicit: U = input integer
2o°U        // Generate the range of integers from 2 to U, inclusive.
¬¬     // Join, then split into chars.
n    // Sort.
1+         ¬  // Join again, and add a 1 to the beginning.
// Implicit: output last expression

• o_o you golfed it in the 5 minute grace period? fastest-gun-in-the-west – Conor O'Brien Jan 6 '16 at 19:41
• @CᴏɴᴏʀO'Bʀɪᴇɴ I'm guessing you don't want to post new code golfs after that :D – Pierre Arlaud Jan 7 '16 at 9:30

## Retina, 78 bytes

Time to show off some new Retina features (it's still not very competitive, but before today this would probably have been closer to 300 bytes).

.+
$0$*1
\B
$ (1)+$#1
^1|

.
1$0$*1
+r(1+\2) (1+)\b
$2$1
1(1)*
$#1 ^ 1  Try it online. ### Explanation While it's possible to convert between decimal and unary quite conveniently now, this is still quite long because I have to convert back and forth several times because some operations are more doable in decimal than in unary and vice-versa. .+$0$*1  Let's start by converting the input to unary. This works by matching the input and then using $*1 which repeats 1 that many times (this repetition feature is new as of today).

\B
$  Next, we generate a range from 1 to N in unary. I've explained why this works in my FizzBuzz answer. (1)+$#1


We convert each number in the range back to decimal so we can work with the decimal digits. This is done by matching each of the unary numbers such that each 1 generates a separate capture. Then we replace that with the number of captures of group one, using the new capture count syntax $#1. ^1|  This removes the leading 1 as well as all spaces from the string so we're left with only the digits (except for a single 1). . 1$0$*1  We convert back to unary and add 1 to each digit (to ensure that even 0 is a non-empty). We also insert a space in front of each digit to ensure that they are separated. +r(1+\2) (1+)\b$2 $1  We repeatedly match a small number preceded by a larger number and swap them. That's bubble sort in Retina. :)  1(1)*$#1


Aaaand back to decimal.

^
1


Finally, we insert a single 1 at the front to account for the one we've removed earlier.

• What new features do ye speak of? – Conor O'Brien Jan 6 '16 at 19:58
• @CᴏɴᴏʀO'Bʀɪᴇɴ I'll add an explanation later. This one uses new substitution syntax for counting captures and repeating characters which can be used for reasonably short decimal/unary conversion. Here is the full changelog: github.com/mbuettner/retina/blob/master/CHANGELOG.md – Martin Ender Jan 6 '16 at 19:59
• @MartinBüttner nice. Does this mean Retina can no longer claim this exemption? – Digital Trauma Jan 6 '16 at 20:17
• @DigitalTrauma I think it's still the most natural form of input. Also I never understood how that kind of language-ism ever got so much support on this site. – Martin Ender Jan 6 '16 at 20:18

import Data.List
f n='1':sort(show=<<[2..n])


Unfortunately sort is in Data.List, that's 17 bytes!

# JavaScript (ES6), 656254 52 bytes

Saved 3 bytes thanks to edc65

x=>eval("for(b='';x>1;)1+[...b+=x--].sort().join")


Builds a string of all numbers from 2 to x, then splits, sorts, joins, and adds a 1 to the beginning. This may still be golfable; suggestions welcome!

• I don't have ES6 on hand, but can't you just use (new) Array(x-1).map((_,y)=>y+2)? – Conor O'Brien Jan 6 '16 at 19:57
• @CᴏɴᴏʀO'Bʀɪᴇɴ map skip empty array elements, so you should use Array(n).fill().map(... (see ES6 tips) – edc65 Jan 6 '16 at 20:55
• Too late to post mine, but a hint for you: n=>1+[...[...Array(n-1)].map(_=>++n,n=1).join].sort().join  (1 byte shorter, it's the split) – edc65 Jan 6 '16 at 20:56
• @edc65 and Cᴏɴᴏʀ O'Bʀɪᴇɴ Thanks for the tips! I wondered why using .split() felt so odd... – ETHproductions Jan 6 '16 at 21:44
• @edc65 Huh, is my solution the same length as yours: n=>1+[...[...Array(n+1).keys()].slice(2).join].sort().join – Neil Jan 7 '16 at 12:59

# CJam, 9

1qi),2>s$ Try it online Alternatively: qi,:)(\s$


# Mathematica, 52 bytes

"1"<>ToString/@Sort[Join@@IntegerDigits[2~Range~#]]&


Once again, string processing happened...

• I think IntegerDigits threads over lists so you don't need map it. – Martin Ender Jan 6 '16 at 22:44
• Oh right, precedence. – Martin Ender Jan 7 '16 at 11:24

# APL (17)

'1',∆[⍋∆←1↓∊⍕¨⍳⎕]


Explanation:

'1',∆[⍋∆←1↓∊⍕¨⍳⎕]
-----------------
⎕   read a number from the keyboard
⍳    get the natural numbers up to and including that number
⍕¨    get the string representation for each number
∊       flatten the array (giving a string of digits)
1↓        remove the first digit (which is always 1)
∆←          store the result in ∆
⍋            get a permutation to sort ∆ upwards
∆[           ] rearrange ∆ so that it is sorted
'1',               add a 1 to the front


Test:

      '1',∆[⍋∆←1↓∊⍕¨⍳⎕]
⎕:
1
1
'1',∆[⍋∆←1↓∊⍕¨⍳⎕]
⎕:
10
10123456789
'1',∆[⍋∆←1↓∊⍕¨⍳⎕]
⎕:
55
10000011111111111111122222222222222223333333333333333444444444444444455555555555566666777778888899999

• Nice! Much better than mine. :D I didn't realize that ∊ would flatten the array and give you a string. That's good to know. – Alex A. Jan 7 '16 at 20:50

## Python 2, 60 bytes

_="".join;print"1"+_(sorted(_(map(str,range(2,input()+1)))))


# ClojureScript, 48 bytes

#(apply str"1"(sort(apply str(range 2(inc %)))))


Same as all of the others, pretty much. REPL available here.

# Ruby, 48 bytes

An anonymous function. Basically just Rubyfied some of the other answers here..

->n{n>1?(?1+[*2..n].join.chars.sort*'').to_i: n}


# Brachylog, 76 41 bytes

1 .;{,1:.e?}?:1fcbZlL,ZoOlM,10^(L-M)=:Oc.


Takes a number as input.

This solution works with the few changes I made to the built-in Findall predicate f. OP is apparently OK with using languages older than the answer so I think this is fine (the changes I made were intented a long time ago, I just motivated myself to do it because of this challenge).

### Explanation

1 .                                            § If the input is 1, unify output with 1

;                                           § Else

{      }?:1f                               § Output a list of all inputs which satisfy
§ the predicate in brackets with the input
§ of the main predicate (ie the input number)
§ as output

,1:.e?                                    § True if the input is an integer between 1
§ and . (the output)

cbZ                            § Concatenate everything into a single number,
§ remove the first digit (1) and call it Z

lL,ZoOlM,                   § L is the length of Z, M is the length of O
§ O being Z sorted (which removes the leading
§ 0s)

10^(L-M)=:Oc.      § Concatenate 10^(L-M) at the beginning of O
§ and unify it with the output


## Smalltalk, 76 bytes

As usual in Smalltalk, conceptually very terse, but textually very verbose...

f:l^'1',((2to:l)fold:[:p :q|p asString,q asString])asByteArray sort asString


Add this as a class method for String and call like this, e.g. for 20, String f: 20

seq $1|sed 's/./&\n/g'|sort|tr -d \\n|sed 's/$$0*$$1/1\1/'  Try it online. • I had a similar approach, but a different handling of the "1 at beginning" part (52 bytes). With yours, you can shave off : sort (1 digit long, no need for -n). – Olivier Dulac Jan 8 '16 at 14:02 # Bash, 35 34 bytes printf %d 1seq 2$1|fold -1|sort


This builds on the answers of @DigitalTrauma and @OlivierDulac. Try it online with Ideone.

• seq 2 $1 prints all integers from 2 to the one specified on the command line. • fold -1 wraps all lines with width 1, i.e., places each character on its own line. -1 seems to be an undocumented feature. • sort sorts the characters by their numeric value. • printf %d 1...​ prepends a 1 to the first line and prints each line as an integer (%d), with no separation. This takes advantage of Bash's curious printf implementation, which repeats the format string over and over, until all arguments are consumed. • +1, I like that one better than ours :) – Olivier Dulac Jan 14 '16 at 18:32 # JavaScript ES6, 49 46 Bytes Thanks to edc65 for 2 bytes F= x=>1+[...(g=_=>x>1?x--+g:_)].sort().join ;console.log(F(15)) • I got something similar but 1 byte shorter. Try using a recursive function with no argument, just decreasing the original parameter x – edc65 Mar 19 '18 at 8:19 # Julia, 33 bytes n->"1"*join(sort([join(2:n)...]))  This is a lambda function that accepts an integer and returns a string. To call it, assign it to a variable. We construct the range 2:n, which will be empty for n < 2, join it into a string, splat the string into an array of characters, sort them, join it into a string, and prepend 1. # APL, 21 bytes {' '~⍨⍕1,x[⍋x←⍕1↓⍳⍵]}  This is an unnamed monadic function that accepts an integer on the right and returns a string. To call it, assign it to a variable. Explanation:  x←⍕1↓⍳⍵]} ⍝ Get the numbers 1:input, drop 1, convert to string x[⍋ ⍝ Sort ⍕1, ⍝ Tack 1 onto the front, flatten to string {' '~⍨ ⍝ Remove the spaces (side effect of ⍕ on an array)  Try it online # Python 2, 60 bytes P=input();print"1"+"".join(sorted(range(2,P+1))[P*2-4:-2])  Indexing for the win. :-) # Python 2, 60 bytes P=input();print"1"+"".join(sorted(range(-P,-1)))[P*3-5:-2]  Just an alternative. # Milky Way 1.6.4, 22 bytes (not competing) ^^'LH=^^JB", "-Q"1";+!  I had to add a sorting opcode for this challenge. ### Explanation ^^  pop the TOS '  read input from the command line LH  push a reversed Pythonic range(TOS+1) =^^J  remove the top 2 items from the TOS B  push the string literal of the TOS ", "-  remove ", " from the TOS Q  sort the TOS "1";+  add "1" to the beginning of the TOS !  output the TOS  ## Seriously, 10 bytes ,u2xεjS'1+  Hex Dump: 2c753278ee6a5327312b  Try It Online Explanation: , Read input u2x Push range from 2..n εj Join into string S Sort '1+ Prepend a "1"  # Bash and GNU tools, 58 52 bytes echo 1$(seq 2 $1|sed -e 's/./&\n/g'|sort|tr -d \\n)  • similar approach than @Digital-trauma , but different way to add the 1, and no need for -n as numbers are 1 digit long. (I saw his answer after I did my first try) – Olivier Dulac Jan 8 '16 at 13:58 ## PowerShell, 61 59 bytes param($a)(("1"+-join([char[]]-join(2..$a)|sort)),1)[$a-eq1]


Takes input param($a) and then uses it to index into an array with [$a-eq1]. If true, we index the second element and output 1. Otherwise, we concatenate "1" with the joined array created by 1) defining a new range 2..$a that has been itself joined together, 2) casting that as a char-array, and 3) sending it through the Sort-Object cmdlet, all of which is then output. Edit1 -- Saved 2 bytes by moving the inner -join operator. # Gogh, 9 7 bytes GJT1-1P  You can run this using: $ ./gogh noi 'GJT1-1P' <input>


G     “ Push a range (1, TOS]       ”
J     “ Join the TOS                ”
T     “ Sort the TOS                ”
1-    “ Remove the first 1          ”
P     “ Prepend the TOS to the STOS ”

• "You may use a language that was made after this contest, as long as it was not made for answering this challenge." I allow such languages to be competitive in my challenges, usually ;) – Conor O'Brien Mar 18 '16 at 13:41
• @CᴏɴᴏʀO'Bʀɪᴇɴ: Oh, awesome! Thanks for pointing that out :) – Zach Gates Mar 18 '16 at 19:17

# Jelly, 8 bytes

RDFṢ1œ|Ḍ


Try it online!

### How it works

RDFṢ1œ|Ḍ  Main link. Argument: n (integer)

R         Range; yield r := [1, ..., n].
D        Convert each k in r into the array of its digits in base 10.
F       Flatten the resulting array of lists.
Ṣ      Sort the resulting flat list of digits.
1œ|   Perform multiset union with 1 as left argument.
This moves a single 1 to the beginning of the list.
Ḍ  Convert the resulting list of base 10 to integer.


# Oracle SQL 11.2, 222 211 bytes

SELECT 1||TRIM(xmlagg(xmlelement(e,c)).EXTRACT('//text()'))FROM(SELECT SUBSTR(s,LEVEL+2,1)c FROM(SELECT MAX(sys_connect_by_path(LEVEL,' '))s FROM DUAL CONNECT BY LEVEL<=:1)CONNECT BY LEVEL<=LENGTH(s)ORDER BY 1);


Un-golfed

SELECT 1||TRIM(xmlagg(xmlelement(e,c)).EXTRACT('//text()'))  -- add leading 1, concatenate each char and remove spaces
FROM   (
SELECT SUBSTR(s,LEVEL+2,1)c                          -- split the string in characters, omiting the first number (1)
FROM   (
SELECT MAX(sys_connect_by_path(LEVEL,' '))s -- create a string by concatenating numbers
FROM   DUAL
CONNECT BY LEVEL<=:1
)
CONNECT BY LEVEL<=LENGTH(s)ORDER BY 1
)


## MATL, 17 bytes

Uses current version (7.0.0) of language/compiler.

49[]i:"@YUh]6L)Sh


EDIT (July 29, 2016): You can try it online with some modifications to conform to changes in the language.

### Example

>> matl
> 49[]i:"@YUh]6L)Sh
>
> 12
101111223456789


### Explanation

49        % push '1'
[]        % push empty array
i:        % input "N" and build vector [1, 2, ... N]
"         % for each number in this vector
@      % push that number
YU     % convert to string
h      % concatenate horizontally
]         % end
6L)       % remove first element
S         % sort
h         % concatenate horizontally


# 05AB1E, 6 bytes

Note: This submission uses features that postdate this challenge and is therefore not competitive

Code:

Lß?J{?


Explanation:

L      # Creates the list [1 .. input]
ß     # Extract the smallest value of the list
?    # Print this value (always 1)
J   # ''.join(list)
{  # Sort the string
? # Print this value
`

Uses ISO 8859-1 encoding

• How does this work? – lirtosiast Jan 21 '16 at 20:43