I absolutely hate the digit 1. So, I need your help to convert numbers to their "proper forms".

Numbers in proper form never have two 1s in a row. 101 is okay, but 110 is hideous.

To convert, simply skip all the improper numbers and count normally. For instance...

1 -> 1
2 -> 2
...
10 -> 10
11 -> 12
12 -> 13
...
108 -> 109
109 -> 120
110 -> 121
111 -> 122


and so on.

Your program should take an integer and output it in proper form. This is , so shortest code in bytes wins.

• Is there an upper bound on input? – lirtosiast Oct 23 '15 at 18:48
• I don't get the 109 -> 120 conversion... – kirbyfan64sos Oct 23 '15 at 19:52
• @kirbyfan64sos Since 108 maps to 109, 109 would map to the next number, which is 110. But that one has two 1s in a row, so it goes to the next number until it reaches one that does not. Which is 120, since all of 110-119 are excluded. – Reto Koradi Oct 23 '15 at 19:59
• @Corey Ogburn It's not about binairy. See it as how a list of numbers would be when you count up to the given number with the no-11 rule for each number in the list – LukStorms Oct 23 '15 at 20:26
• @leymannx The number on the left represents the number in the series. So the first value in the series is 1, the second value in the series is 2, yadda yadda(lobster bisque), the tenth value in the series is 10, and the eleventh value in the series is 12, because we skipped 11 as talex finds it an abomination unto the lord. This idea continues on, hence why the 108th value in the series is 109, and the 110th value in the series is 120, as we skip everything from 110 to 119. Hope I clarified well enough. – ahall Oct 24 '15 at 23:35

## Burlesque, 10 bytes

{11~[n!}FO

Older versions:

ro{11~[n!}f[

ro{Sh"11"~=n!}f[

• done. It's been around since 2012, there's also an article on RosettaCode and the esowiki as well. – mroman Oct 23 '15 at 21:00

# Bash + GNU utils, 36

seq $1$1|grep -v 11|sed -n "$1{p;q}"  • Surely 1$1 suffices, rather than $1$1? – Neil Oct 24 '15 at 10:39

# Perl 5, 34 Bytes

Looping a counter and changing the occasional double-one.

map{$i++;$i=~s/11/12/}1..pop;say$i  ### Test $ perl -M5.012 -e 'map{$i++;$i=~s/11/12/}1..pop;say$i' 111$ 122


# JavaScript, 53 bytes

n=>[...Array(n*2).keys()].filter(a=>!/11/.test(a))[n]


Alternate (using comprehensions, same length):

n=>[for(i of Array(n*2).keys())if(!/11/.test(i))i][n]


# Pyth, 13 11 bytes

e.f!}hTZQ


Saved 2 bytes thanks to @FryAmTheEggman.

Live demo and test cases.

## 13-byte version

e.f!}"11"+ZkQ

• Context: Pyth, Pyth, and Pyth. – Peter Mortensen Oct 25 '15 at 16:21
• @PeterMortensen The third one is unrelated to the Pyth used here - it's from 8 years before this Pyth was created, actually. Pyth is just a popular name for Python inspired languages. – isaacg Oct 26 '15 at 9:18

# Jelly, 8 bytes

$- Group the previous 2 links into a monad f(k): 11 - Yield 11 w - Yield 1 if 11 is a sublist of k's digit else 0 ¬ - Map 1 to 0 and 0 to 1 1 # - Count up k = 1, 2, 3, ... until n integers return true under f(k) Ṫ - Return the last k  # Jelly, 8 bytes ²wÐḟ11ị@  Try it online! An alternate approach, still only 8 bytes though. Relies on the assumption that the output is always less than the square of the input ## How it works ²wÐḟ11ị@ - Main link. Takes n on the left ² - n² Ðḟ - Filter-false the range [1, 2, ..., n²] on: 11 - 11 w - Is a sublist of the digits ị@ - Take the n'th value in this list  • The assumption in your second answer is not correct, see A322054. f(A322054(n)) = 10^n and f(A322054(n))/10^n converges to 0. – ovs Nov 12 '20 at 15:41 • @ovs How about $n^2$ as an upper bound? Also, yep, I'm an idiot, fixed – caird coinheringaahing Nov 12 '20 at 15:45 • $n^2$ seems to work. FYI your original assumption would have failed at around $12\cdot10^{73}$. – ovs Nov 12 '20 at 15:52 ## Python 2, 50 lambda n:[i for i in range(n*2)if'11'not ini][n]  An anonymous function that lists the numbers not containing 11 in order, and takes the nth one. The off-by-one error of zero-indexing cancels with the inclusion of 0 in the list. In theory, this will fail for sufficiently high numbers where f(n)>2*n, but this shouldn't happen until n is at least 10**50. 51 bytes: n=input();i=0 while n:i+=1;n-='11'not ini print i  Counts up numbers i until the quota of n numbers without 11 is met. A function is the same length because of the off-by-one corrections needed. f=lambda n,i=0:n+1and f(n-('11'not ini),i+1)or~-i  # JavaScript (ES6) 43 bytes As an anonymous function n=>eval('for(i=0;/11/.test(i)||n--;i++);i')  Note: the very simplest way would be 44: n=>{for(i=0;/11/.test(i)||n--;i++);return i}  Test running the snippet below. f=n=>eval('for(i=0;/11/.test(i)||n--;i++);i') var o='',v=0,r,c; for(r=0;r<30;r++) { o += '<tr>'; for(c=0;c<10;c++) { o += '<td>' +v + '⇨' + f(v) + '</td>'; v++; } o += '</tr>' } O.innerHTML=o <table id=O style='font-size:80%'></table> • This fails for entering 108, it returns 119 – Samathingamajig Nov 12 '20 at 16:33 • @Samathingamajig you're right, I stand corrected after 5 years, better late than never. Should work now – edc65 Nov 13 '20 at 13:49 # Python 3 74 Still need a bit of golfing. n=int(input()) c=0 for x in ' '*n: c+=1 while'11'in str(c):c+=1 print(c)  It's pretty brute force right now. # Haskell, 51 bytes ([x|x<-[0..],notElem('1','1')$zip=<<tail$show x]!!)  Usage example: ([x|x<-[0..],notElem('1','1')$zip=<<tail$show x]!!) 110 -> 121. How it works: [x|x<-[0..] ] -- take all x starting with 0 , -- where ('1','1') -- the pair of two chars '1' notElem -- is not part of zip=<<tail -- the list of pairs of neighbor elements of show x -- the string representation of x !! -- take nth element, where n is the parameter  # Brachylog, 15 bytes ∧{ℕ₁≜¬s11&}ᶠ↖?t  Try it online! ∧{ℕ₁≜¬s11&}ᶠ↖?t ∧{ … }ᶠ↖?t get the N'th output from … ℕ₁≜ try a number, starting with 1 ¬s11 doesn't contain 11 & output that number  # 05AB1E, 7 bytes µND11å≠  Explanation: µ # Loop until the counter_variable is equal to the (implicit) input-integer: # (the counter_variable is 0 by default) ND # Push the 0-based loop index twice 11å # Pop one index, and check if it contains 11 as substring ≠ # Invert this boolean (1→0; 0→1) # (implicitly pop the top of the stack, and increase the counter_variable by # 1 if it's truthy) # (after which the top of the stack - the duplicated index - is output # implicitly as result)  • 05AB1E's µ is like a much better version of Jelly's # :/ – caird coinheringaahing Nov 12 '20 at 14:34 • @cairdcoinheringaahing Yeah, it's pretty useful for challenges like this. Especially that it by default increases the counter_variable at the end of every iteration if the top is truthy. – Kevin Cruijssen Nov 12 '20 at 14:42 # MUMPS, 37 bytes t(i) f j=1:1 s:j'[11 k=k+1 q:k=i q j  Pretty straightforward. The only "interesting" thing here is the construct j'[11 - '[ is the "does not contain" operator, so that "abc"'["ab" is false and "abc"'["cd" is true. Despite both operands of j'[11 being numbers, MUMPS remains unperturbed. It will happily autocoerce both operands to strings and move on with its life. Hooray! (Incidentally, if you're okay with the program never terminating, we can shorten this to 35 bytes: t2(i) f j=1:1 s:j'[11 k=k+1 w:k=i j) # Perl 5-p, 25 bytes $\++while$\=~/11/||$_--}{


Try it online!

# Japt, 9 bytes

@!XsèB}iU


Try it

# Raku, 24 bytes

{grep(?1^/11/,^∞)[$_]}  Try it online! grep(..., ^∞) filters the numbers from zero to infinity according to the given test object, and [$_] indexes into that sequence with the function argument, returning the number at that index.

/11/ matches any number whose decimal representation contains two consecutive one digits. The sense of the match can be inverted by writing True ^ /11/, where ^ creates an exclusive-or junction matcher that succeeds only if exactly one of its operands succeeds. True will always succeed, so the overall junction can only succeed if the /11/ fails. Finally, True can be expressed more concisely as ?1, that is, 1 coerced to a Boolean.

none(/11/) also works as a matcher, but is longer.

# GolfScript, 2924 18 bytes

~0\{{).11?)}do}*


Try it online!

This is my first code golf, so this is probably the least effecient way of doing things

-5 bytes by not being dumb and using the increment symbol as well as just keeping the input on the stack and decrementing instead of wasting space saving it as a variable and comparing it, probably more things I could save on

-6 bytes the loop is fixed size and I don't need the input after the loop so I can just use repeat

## Perl 5, 47 bytes

@_[$_]=++$i!~/11/?$i:redo for 1..<>;print$_[-1]


while(preg_match('/11/',$i)){$i++;}print$i;  # Husk, 11 10 bytes Edit: -1 byte thanks to Razetime !fö¬€ḋ3QdN  Try it online! Finds index of input among natural numbers filtered to exclude those with digit-sublists that include [1,1]. • -1 – Razetime Nov 13 '20 at 17:28 • Very clever! Thanks! – Dominic van Essen Nov 13 '20 at 23:01 # Julia 1.0, 44 bytes x->[i for i=1:9x if !occursin("11","$i")][x]


Try it online!