[Question]
  [
This is a Cops and Robbers challenge. This is the cop's thread. The [robber's thread is here](https://codegolf.stackexchange.com/q/188143/31716).
As a cop, you must pick any sequence from the [OEIS](https://oeis.org/), and write a program **p** that prints the first integer from that sequence. You must also find some string **s**. If you insert **s** somewhere into **p**, this program must print the second integer from the sequence. If you insert **s + s** into the same location in **p**, this program must print the third integer from the sequence. **s + s + s** in the same location will print the fourth, and so on and so forth. Here's an example:
> 
> # Python 3, sequence [A000027](https://oeis.org/A000027)
> 
> 
> 
> ```
> print(1)
> 
> ```
> 
> The hidden string is **two bytes**.
> 
> 
> 
The string is `+1`, because the program `print(1+1)` will print the second integer in A000027, the program `print(1+1+1)` will print the third integer, etc.
Cops must reveal the sequence, the original program **p**, and the length of the hidden string **s**. Robbers crack a submission by finding any string *up to* that length and the location to insert it to create the sequence. The string does not need to match the intended solution to be a valid crack, nor does the location it's inserted at.
## Rules
* Your solution must work for any number in the sequence, or at least until a reasonable limit where it fails do to memory restrictions, integer/stack overflow, etc.
* The winning robber is the user who cracks the most submissions, with the tiebreaker being who reached that number of cracks first.
* The winning cop is the cop with the shortest string **s** that isn't cracked. Tiebreaker is the shortest **p**. If there are no uncracked submissions, the cop who had a solution uncracked for longest wins.
* To be declared safe, your solution must stay uncracked for 1 week and then have the hidden string (and location to insert it) revealed.
* **s** may not be nested, it must concatenated end to end. For example, if **s** was `10`, each iteration would go `10, 1010, 101010, 10101010...` rather than `10, 1100, 111000, 11110000...`
* It is acceptable to start on the second term of the sequence rather than the first.
* If your sequence has a finite number of terms, going past the last term is allowed to result in undefined behavior.
* All cryptographic solutions (for example, checking the hash of the substring) are banned.
* If **s** contains any non-ASCII characters, you must also specify the encoding being used.
  
[Answer]
# [MATL](https://github.com/lmendo/MATL), sequence [A005206](http://oeis.org/A005206). [Cracked](https://codegolf.stackexchange.com/questions/188143/robbers-the-hidden-oeis-substring/188407#188407) by [SamYonnou](https://codegolf.stackexchange.com/users/12324/samyonnou)
```
voOdoO
```
[Try it online!](https://tio.run/##y00syfn/vyzfPyXf//9/AA "MATL – Try It Online")
The hidden string has **8 bytes**.
[Answer]
# [Python 2](https://docs.python.org/2/), sequence [A138147](https://oeis.org/A138147) ([cracked](https://codegolf.stackexchange.com/a/188212/20260))
```
print 10
```
[Try it online!](https://tio.run/##K6gsycjPM/r/v6AoM69EwdDg/38A "Python 2 – Try It Online")
The hidden string is **7 bytes**. The sequence goes:
```
10, 1100, 111000, 11110000, 1111100000, ...
```
[Answer]
## [Keg](https://esolangs.org/wiki/Keg), sequence [A000045](https://oeis.org/A000045)
```
0.
```
The hidden string is ≤*6 bytes*(in order to conform the updated cracking rules)
[Answer]
# [Cracked](https://codegolf.stackexchange.com/a/188156/56656)
# ~~[Brain-Flak](https://github.com/Flakheads/BrainHack), sequence [A000290](http://oeis.org/A000290) (Square numbers)~~
```
((()))({}<>)
```
[Try it online!](https://tio.run/##SypKzMzLSEzO/v9fQ0NDU1NTo7rWxk7z/38A "Brain-Flak (BrainHack) – Try It Online")
The hidden string is **6 bytes**.
### Fun fact:
I "discovered" this property while playing [this brain-flak based game](https://gitlab.com/e-neighborhood-watch/cal_final). The hidden string was a randomly generated item that I discovered was very useful.
[Answer]
# [Pyret](https://code.pyret.org/), sequence [A083420](https://oeis.org/A083420), [Cracked](https://codegolf.stackexchange.com/questions/188143/robbers-the-hidden-oeis-substring/188249#188249)
```
fold({(b,e):(2 * b) + 1},1,[list: ])
```
The hidden string has **4 bytes** or fewer.
[Answer]
# Java 8+, sequence [A010686](https://oeis.org/A010686) ([cracked by xnor](https://codegolf.stackexchange.com/a/188268/84303))
Lambda function.
```
()->System.out.println(1);
```
The hidden string is ≤ 5 bytes
[Answer]
# [V](https://github.com/DJMcMayhem/V), sequence [A000290](http://oeis.org/A000290). Cracked by [Cows quack](https://codegolf.stackexchange.com/a/188408/41805)
```
é*Ø.
```
[Try it online!](https://tio.run/##K/v///BKrcMz9P7/BwA "V – Try It Online")
The hidden string is **5 bytes**.
[Answer]
## [Python 3](https://www.python.org), sequence [A096582](https://oeis.org/A096582)
This is really trivial, as I haven't tried Cops and Robbers challenges before.
```
print(100)
```
The hidden string is *3 bytes.*
[Answer]
# [Python 3](https://docs.python.org/3/), sequence [A014092](http://oeis.org/A014092) - ([cracked](https://codegolf.stackexchange.com/questions/188143/robbers-the-hidden-oeis-substring/188311#188311))
```
from sympy import isprime, primerange
from itertools import count
r=1
print(r)
```
[Try it online!](https://tio.run/##Nco7CoAwDADQvafIqOAizh5GpGrAfEjjkNNHKTi95Wn4JbxkHiYELUgDkFTMAZsaUp2gYxuftfSFXs1F7vbPXR72Yutcvso@2Jj5Ag "Python 3 – Try It Online")
The hidden sequence is **82 bytes**.
My intended code (which doesn't rely on the Goldbach Conjecture) was:
```
i=(n for n in count(2)if all(not isprime(n-x) for x in primerange(1,n)))
r=next(i)
#
```
**Cracked** by [NieDzejkob](https://codegolf.stackexchange.com/users/55934/niedzejkob), who uses the [Goldbach Conjecture](https://en.wikipedia.org/wiki/Goldbach%27s_conjecture) to solve it in a [magical 42](https://hitchhikers.fandom.com/wiki/42) characters. Great job!
[Answer]
# [Forth (gforth)](http://www.complang.tuwien.ac.at/forth/gforth/Docs-html/), [A000042](http://oeis.org/A000042) - ([cracked](https://codegolf.stackexchange.com/a/188346/55934))
```
1 .
```
[Try it online!](https://tio.run/##S8svKsnQTU8DUf//Gyro/f8PAA "Forth (gforth) – Try It Online")
The hidden sequence is **5 bytes**, and it can easily handle hundreds of terms.
A one-byte solution that breaks due to integer overflow is also possible. In fact, I'd say it is embarrassingly trivial. While the challenge text, under some interpretations, might allow you to call that a crack, I urge you not to.
[Answer]
## [Desktop Calculator](https://www.gnu.org/software/bc/manual/dc-1.05/html_mono/dc.html), sequence [A006125](https://oeis.org/A006125)
```
1n
```
The hidden string *≤ 12 bytes*.
[Answer]
# [Brain-Flak](https://github.com/Flakheads/BrainHack), sequence [A000984](https://oeis.org/A000984) (central binomial coefficients)
```
(())({{}}{})
```
[Try it online!](https://tio.run/##SypKzMzLSEzO/v9fQ0NTU6O6ura2ulbz/38A "Brain-Flak (BrainHack) – Try It Online")
The hidden string has **36 bytes** or fewer.
[Answer]
# [Python 3](https://docs.python.org/3/), [A268575](http://oeis.org/A268575)
```
from itertools import product
S,F,D=lambda*x:tuple(map(sum,zip(*x))),lambda f,s:(v for x in s for v in f(x)),lambda s:{(c-48>>4,c&15)for c in map(ord,s)}
W=D("6@AQUVW")
print(len(W))
```
[Try it online!](https://tio.run/##Rcy7CsIwGEDhvU8ROsifEgexihRaFIq7iHauaYOB3EjSEhWfPRId3M7wcczD37Vax8isloj70XqthUNcGm09MlYPE/XZmRxJW4te3oa@CJWfjBhB9gbcJMmTGygCxpj8AGLEVTAjpi0KiCvkvjmnZBD@zlUvoMty1zQloYvVBidGE0trbQfi8Dvr6hby7f5wuly7HGfGcuVBjAo6jGP8AA "Python 3 – Try It Online")
The hidden sequence is **102 bytes.**
[Answer]
# [AsciiDots](https://github.com/aaronduino/asciidots), 36 bytes, [A019523](http://oeis.org/A019523)
```
/>1#*$_#\
*^-{+})./
|/\/\/\
\/\/\/\&
```
[Try it online!](https://tio.run/##SyxOzsxMyS8p/v9f385QWUslXjmGSytOt1q7VlNPn6tGPwYEuWIgtNr//wA "AsciiDots – Try It Online")
The hidden string has **12 bytes**.
[Answer]
# [Haskell](https://www.haskell.org/), sequence [A083318](http://oeis.org/A083318) ([cracked](https://codegolf.stackexchange.com/a/188546/20260))
```
f=length[2]
```
[Try it online!](https://tio.run/##y0gszk7Nyfn/P802JzUvvSQj2ij2f25iZp6CrUJBUWZeiYKKQtr/f8lpOYnpxf91kwsKAA "Haskell – Try It Online\"n/P802JzUvvSQj2jD2f25iZp6CrUJBUWZeiYKKQtr/f8lpOYnpxf91kwsKAA \"Haskell – Try It Online")
The hidden string is **5 bytes**. The sequence goes:
```
1, 3, 5, 9, 17, 33, 65, 129, 257, 513, ...
```
[Answer]
# [Brain-Flak](https://github.com/Flakheads/BrainHack), sequence [A000578](http://oeis.org/A000578) (Cube numbers)
```
((())<>)
```
[Try it online!](https://tio.run/##SypKzMzLSEzO/v9fQ0NDU9PGTvP/fwA "Brain-Flak (BrainHack) – Try It Online")
The hidden string is **16 bytes**
[Answer]
# [Cracked](https://codegolf.stackexchange.com/a/188245/73884)
# ~~[cQuents](https://github.com/stestoltz/cQuents), sequence [A003617](https://oeis.org/A003617)~~
```
=1#2:pZ
```
[Try it online!](https://tio.run/##Sy4sTc0rKf7/39ZQ2ciqIOr/fwA "cQuents – Try It Online")
The hidden string is **1 byte**.
[Answer]
# [Unefunge-98 (PyFunge)](https://pythonhosted.org/PyFunge/), sequence [A000108](http://oeis.org/A000108)
```
1.@
```
[Try it online!](https://tio.run/##K81LTSvNS0/VtbTQLagEM///N9Rz@P8fAA "Unefunge-98 (PyFunge) – Try It Online")
The hidden sequence is **19 bytes**.
[Answer]
# [MATL](https://github.com/lmendo/MATL), sequence [A000796](https://oeis.org/A000796). [Cracked](https://codegolf.stackexchange.com/questions/188143/robbers-the-hidden-oeis-substring/188314#188314) by [SamYonnou](https://codegolf.stackexchange.com/users/12324/samyonnou)
```
'pi'td1_&:_1)Y$J)
```
[Try it online!](https://tio.run/##y00syfn/X70gU70kxTBezSreUDNSxUvz/38A "MATL – Try It Online")
The hidden string has **3 bytes**.
[Answer]
# [Python 3](https://docs.python.org/3/), [A008574](https://oeis.org/A008574), [cracked by xnor](https://codegolf.stackexchange.com/a/188375/44718)
```
print(1)
```
[Try it online!](https://tio.run/##K6gsycjPM/7/v6AoM69Ew1Dz/38A "Python 3 – Try It Online")
Insert 4 bytes to complete A008574. A008574: a(0)=1, thereafter a(n) = 4n.
[Answer]
# [Haskell](https://haskell.org), sequence [A014675](https://oeis.org/A014675), [cracked by nimi](https://codegolf.stackexchange.com/a/188353/73884)
```
main=print$uncurry(!!)([2],0)
```
[Try it online!](https://tio.run/##y0gszk7Nyfn/PzcxM8@2oCgzr0SlNC@5tKioUkNRUVMj2ihWx0Dz/38A)
The hidden sequence is **35 bytes**.
Here's my intended solution:
> 
>  
> ```
> 
>  main=print$uncurry(!!)$((l:n)->(l>>=flip take[2,1],n+1))([2],0)
>                        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
> ```
> 
> 
> 
> 
[Answer]
# [VDM-SL](https://raw.githubusercontent.com/overturetool/documentation/master/documentation/VDM10LangMan/VDM10_lang_man.pdf), sequence [A000312](https://oeis.org/A000312)
```
let m={1|->{0}}in hd reverse[x**x|x in set m(1)&x<card m(1)]
```
The hidden string has **33 bytes** or fewer
[Answer]
# Haskell, [A000045 (Fibonacci)](https://oeis.org/A000045) -- [Cracked](https://codegolf.stackexchange.com/a/188505/73884)
```
f = head [0, 1]
```
I've got a solution with a whopping 23 bytes. I don't expect this to be safe for long, but it was super fun to come up with.
Solution:
> 
>  I thought Haskell would be a fun language to try this challenge with -- the natural thing is to do to end up adding a function call every time, but if the sequence can't be written recursively in terms of the last term only, you run into some trickiness with Haskell's strictness and function application.  
> 
>  Khuldraeseth na'Barya found a super clever way to do this with an applicative functor. I did something much less brilliant, using where-hacking:  
> 
>   
> 
> `f = head [b,a+b]where[a,b]=[0,1]
>  ^^^^^^^^^^^^^^^^^^`  
> 
>  (This is actually 18 bytes. My less-golfed 23 byte version, where I'd totally forgot about pattern matching, used `[last a,sum a]where a=` instead.)
> 
> 
> 
[Answer]
# [Java 8+](http://jdk.java.net/), 1044 bytes, sequence [A008008](https://oeis.org/A008008) (Safe)
```
class c{long[]u={1,4,11,21,35,52,74,102,136,172,212,257,306,354,400,445,488,529,563,587,595,592,584,575,558,530,482,421,354,292,232,164,85,0,-85,-164,-232,-292,-354,-421,-482,-530,-558,-575,-584,-592,-595,-587,-563,-529,-488,-445,-400,-354,-306,-257,-212,-172,-136,-102,-74,-52,-35,-21,-11,-4,-1},v={0,0,0,0,0,0,0,0,0,0,0,0,-1,0,-1,0,-1,0,0,0,0,0,0,0,-1,0,1,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,-1,0,0,0,0,0,0,0,-1,0,-1,0,-1,0,0,0,0,0,0,0,0,0,0,0,1},w={1,0,0,-1,5};long d=1,e=1;void f(long a,long b){long[]U=u,V=v,W,X;while(a-->0){U=g(U);w=h(v,w);}W=h(v,U);while(b-->0){V=g(V);v=h(v,v);}X=h(V,u);if(w[0]!=v[0]){int i,j,k=0;u=new long[i=(i=W.length)>(j=X.length)?i:j];for(;k<i;k++)u[k]=(k<i?W[k]:0)-(k<j?X[k]:0);d*=e++;}}long[]g(long[]y){int s=y.length,i=1;long[]Y=new long[s-1];for(;i<s;){Y[i-1]=y[i]*i++;}return Y;}long[]h(long[]x,long[]y){int q=x.length,r=y.length,i=0,j;long[]z=new long[q+r-1];for(;i<q;i++)if(x[i]!=0)for(j=0;j<r;)z[i+j]+=x[i]*y[j++];return z;}c(){f(3,0);System.out.println(u[0]/d);}public static void main(String[]args){new c();}}
```
[Try it online!](https://tio.run/##bVLbUqNAEP0VfGOk2@WaRMfRj7DMpSgeMCHJkEiUW4wU357tBlxjrUXNTN/mnNM9pHEdY7ranc/LfVwUxrLZH7JNGFWqccAHxwHXAS@AwIUxubYLjjcCZ@xSnFYwBs8eUYEPvm2D7wfgTyZUfQvByINgMobglm7fumT7EIzJDijvUe3EBb8D98GlvOsR9siHSQA2IO3IHnIYOY9ciHwD@SoyBjIYMioyPDIPMiEyM7IEZC3IopDVIcvsoVg4cgfIrSD3hNwccpc4ZriOlfMUZF46WqhVY8PvHzo/tv9Sl@vru4z9fut3wH/3WzjyYw3lQSv5BY2VciBRjqwPemWszS4WQ3e8iOGRn1UFU1XDDObyuNX7xIwRH2zRPKuN@SzkUW3NGo5CtrPO4lBX9tKXTalsKmTdJWsqm5M1hUpIvTaPoR1dqZp20eisNDSksFO2rFSWHI1OgFamVrObfZJtyq14MFM1/3Ie9V0ayfUhN@XuXsudZYkq3EXKJO9xRtadLZCc9HHeO3J1rRLLkm3b97Yx@/PUsxfqNECDpqn0ucW3lAKdgU7fF1I0i1BTRJ1CHV1rhs2TssozYyEH/O2A/wE/eN7VxxdPfklpQzqQfn6Tvlv5Be27JCJBk/sg0itlC46nNLH0PpfiM9RWGlmKk9enMLWsSA6aPmW7NEWzNj2gOTydijJ5vTlU5c1bTpL2mVnRI/xZ0fu8VS97vTSKMi7p6H6N11hn5lNJlaQtzjeFaFgfAdIoz@e/ "Java (JDK) – Try It Online")
Can be solved using a hidden string of size **12**. Can definitely be golfed more, but there is no way this is actually winning. I just wanted to contribute out of respect for the number 8008.
Note: before anyone complains that the sequence is hard-coded, I've tested this up to the first term that diverges from the hard-coding (13th term = 307) and it gets it correctly albeit slowly. This is also why it's using `long` instead of `int`, otherwise it overflows before that term.
**Update (Jul 12 2019)**: updated to be a bit more performant. Computes the 13th term in 30 seconds on my computer now instead of 5 minutes.
**Update (Jul 17 2019)**: fixed bugs in for loop bounds for the `g` function, and array length bounds in the bottom of the `f` function. These bugs should have eventually caused problems, but not early enough to get caught by just checking the output. In either case, since the presence of these bugs 5 days into the game might have confused some people enough into being unable to solve this puzzle, I am totally fine with extending the "safe" deadline until July 24th for this submission.
**Update (Jul 18 2019)**: After some testing I have confirmed that overflows start after the 4th term in the sequence and start affecting the validity of the output after the 19th term. Also in the program as it is written here, each consecutive term takes roughly 5 times longer than the previous to compute. The 15th term takes about 14 minutes on my computer. So actually computing the 19th term using the program as written would take over 6 days.
Also, here is the code with sane spacing/indentation so it is a bit easier to read if people don't have an IDE with auto-formatting on hand.
```
class c {
  long[] u = {1, 4, 11, 21, 35, 52, 74, 102, 136, 172, 212, 257, 306, 354, 400, 445, 488, 529, 563, 587, 595, 592, 584,
      575, 558, 530, 482, 421, 354, 292, 232, 164, 85, 0, -85, -164, -232, -292, -354, -421, -482, -530, -558, -575,
      -584, -592, -595, -587, -563, -529, -488, -445, -400, -354, -306, -257, -212, -172, -136, -102, -74, -52, -35,
      -21, -11, -4, -1},
      v = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0,
          0, 1, 0, 1, 0, 1, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, -1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1},
      w = {1, 0, 0, -1, 5};
  long d = 1, e = 1;
  void f(long a, long b) {
    long[] U = u, V = v, W, X;
    while (a-- > 0) {
      U = g(U);
      w = h(v, w);
    }
    W = h(v, U);
    while (b-- > 0) {
      V = g(V);
      v = h(v, v);
    }
    X = h(V, u);
    if (w[0] != v[0]) {
      int i, j, k = 0;
      u = new long[i = (i = W.length) > (j = X.length) ? i : j];
      for (; k < i; k++)
        u[k] = (k < i ? W[k] : 0) - (k < j ? X[k] : 0);
      d *= e++;
    }
  }
  long[] g(long[] y) {
    int s = y.length, i = 1;
    long[] Y = new long[s - 1];
    for (; i < s;) {
      Y[i - 1] = y[i] * i++;
    }
    return Y;
  }
  long[] h(long[] x, long[] y) {
    int q = x.length, r = y.length, i = 0, j;
    long[] z = new long[q + r - 1];
    for (; i < q; i++)
      if (x[i] != 0)
        for (j = 0; j < r;)
          z[i + j] += x[i] * y[j++];
    return z;
  }
  c() {
    f(3, 0);
    System.out.println(u[0] / d);
  }
  public static void main(String[] args) {
    new c();
  }
}
```
# Solution
> 
>  `f(1,v[0]=1);` right before the `System.out.println`  
> 
>  The program works by computing the n'th Taylor expansion coefficient at 0. Where the original function is a quotient of polynomials, represented by `u` and `v` which I got from [here](https://oeis.org/A008000/a008000_1.pdf), except that in the linked document the denominator is not multiplied out, and nowhere do they say that you have to compute the Taylor series, I stumbled on that by accident and then confirmed via another source.  
> 
>  The calculation is done via repeated application of the quotient rule for derivatives.  
> 
>  The incorrect first term of `v`, the entire array `w` and a few other things like the function `f` having any arguments are thrown in to mess with people.
> 
> 
> 
[Answer]
# [Brachylog](https://github.com/JCumin/Brachylog), 7 bytes ([Brachylog SBCS](https://github.com/JCumin/Brachylog/wiki/Code-page)), [A114018](https://oeis.org/A114018) ([Cracked](https://codegolf.stackexchange.com/a/188485/85334))
```
≜ṗ↔ṗb&w
```
[Crack it online!](https://tio.run/##SypKTM6ozMlPN/r//1HnnIc7pz9qmwIkk9TK//8HAA "Brachylog – Try It Online")
The string has 2 or fewer bytes.
> 
>  Fatalize's solution, `ẹb`, is the original string which I had intended. Note that `ẹk` also works, for the same reasons. In addition to the issue with `9001` beheading to `001`=`1`, it actually turns out that `b` on a number just won't fail, because all single digit numbers behead to `0`, including `0` itself.
> 
> 
> 
[Answer]
# C# (.NET Core), A003678, 29727 bytes (Safe)
```
using System;using System.Linq;using System.CodeDom.Compiler;class P{static void Main(){Int32 z=0;\u0049nt32 T(\u0049nt32 i){i--;var \u0064="";for(;i>0;\u0069/=5)d=i%5+d;return d.Aggre\u0067a\u0074e(0,(a,b)=>a*5+b%48*2%5)+1;}System.D\u0069agn\u006fs\u0074ics.Pro\u0063ess.\u0053tart(CodeDo\u006dProvi\u0064er.\u0043reateP\u0072ovider("CSharp").Co\u006dpi\u006ceAssembly\u0046romSource(new 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```
The hidden sequence is 4 bytes or less.
> 
>  The way the program works is by decoding the long string into a program. The decoded program is then compiled into an executable, which is then run. The executable then creates another program, this time using the CodeDom instead of a long string. Finally, the last program outputs the result.
>  The hidden string is `;8;L`, where you insert at index 18504 in the super long string.
> 
> 
> 
> 
> 
> 
[Answer]
# [1+](https://esolangs.org/wiki/1%2B), sequence [A000079](http://oeis.org/A000079)
```
1:
```
Trivial. The hidden string is 2 bytes. How to multiply by 2 when pushing 2 is three bytes (`11+`)?
[Answer]
# [Prolog (SWI)](http://www.swi-prolog.org), 28 bytes, [A011557](https://oeis.org/A011557), safe
```
+ 1.
?- +X,X<2,write(X),X>2.
```
[Try it online!](https://tio.run/##KyjKz8lP1y0uz/z/X1vBUI/LXldBO0InwsZIp7wosyRVI0JTJ8LOSO//fwA "Prolog (SWI) – Try It Online")
(I'm not really sure what counts as a full program for Prolog, but this works as a program on TIO.)
The hidden string is 5 bytes or less.
I'm a bit surprised this survived a week... The hidden string is
> 
>  
> ```
> 
> 
>  + 0.
> ```
> 
>  , inserted after `+ 1.` (note the leading newline). [Try it online](https://tio.run/##KyjKz8lP1y0uz/z/X1vBUI9LW8EAStjrKmhH6ETYGOmUF2WWpGpEaOpE2Bnp/f8PAA "Prolog (SWI) – Try It Online"). Instead of numerically generating a power of ten, this prints one out digit by digit: when backtracking is triggered by the failure of `X>2`, the only choice point is `+X`, which goes through every clause of `+/1` until execution succeeds or it runs out, executing `write(X)` (which immediately and imperatively prints without a trailing newline to standard output, so the output can't be undone by backtracking) for every resulting value of X. `X<2` is just there to prevent the 1-byte solution `0`.
> 
> 
> 
[Answer]
# [1+](https://esolangs.org/wiki/1%2B), sequence [A058891](https://oeis.org/A058891)
```
1:
```
Hopefully better, although still pretty much straight forward. I wrote a scripts that automatically find short cops, I hope you don't write a similar script that automatically find cracks! The hidden string is 4 bytes.
[Answer]
# Pure [Bash](https://www.gnu.org/software/bash/), 13 bytes, sequence [A010888](https://oeis.org/A010888), [cracked (17 bytes)](https://codegolf.stackexchange.com/a/235788/78410).
I think this is an easy problem.
```
x=
echo ${#x}
```
I've hidden an 18 bytes of string.
[Try it online!](https://tio.run/##S0oszvj/v8KWKzU5I19BpVq5ovb/fwA "Bash – Try It Online")
## Intended solution
Insert the following string between first and second lines:
```
x=x${x#xxxxxxxxx}
```
]