Find the program that prints this integer sequence (Robbers' thread)

Question

This is the robbers' thread. The cops' thread goes here.

In the cops thread, the task was to write a program/function that takes a positive (or non-negative) integer and outputs/returns another number (not necessarily integer). The robbers task is to unscramble the code the cops used to produce this output.

The cracked code doesn't have to be identical, as long as it has the same length and any revealed characters are in the correct positions. The language must also be the same (version numbers can be different). The output must of course be identical.

No-ops can be used in robber's solution.

The winner of the robbers thread will be the user who has cracked the most submissions by May 7th 2016. If there's a tie, the user who has cracked submissions with the longest combined code will win.

The submission should be formatted like this:

Language, nn characters (including link to answer), Cop's username

Code:

function a(n)
    if n<2 then
        return n
    else
        return a(n-1) + a(n-2)
    end
end

Output

a(0) returns 0
a(3) returns 2

Optional explanation and comments.

Answer 1

MATL, 5 bytes, Luis Mendo

H5-*|

This code calculates abs((2-5)*input) which is just a(n)=3*n for positive numbers, which is http://oeis.org/A008585

Answer 2

Hexagony, 7 bytes, Adnan, A005843

?{2'*!@

or

 ? {
2 ' *
 ! @

Try it online!

Simply doubles the input (and assumes positive input). The code is (for once) simply executed in reading order. The code uses three memory edges A, B, C with the memory pointer starting out as shown:

?    Read integer from STDIN into edge A.
{    Move memory pointer forwards to edge B.
2    Set edge B to 2.
'    Move memory pointers backwards to edge C.
*    Multiply edges A and B and store result in C.
!    Print result to STDOUT.
@    Terminate program.

Answer 3

J, 7 bytes, Cᴏɴᴏʀ O'Bʀɪᴇɴ

Code

2+*:@p:

Output

   f =: 2+*:@p:
   f 0
6
   f 2
27

Try it with J.js.

How it works

Sequence A061725 is defined as a(n) := p_n² + 2, where p_n is the (n + 1)^th prime number.

2+*:@p:  Monadic verb. Argument: n

    @    Atop; combine the verbs to the right and to the left, applying one after
         the other.
     p:  Compute the (n+1)th prime number.
  *:     Square it.
2+       Add 2 to the result.

Answer 4

05AB1E, 5 bytes, Adnan, A001788

Læ€OO

Try it online! This uses an alternative definition given on the page. Explanation:

Læ€OO
L     range;      [1..n]
 æ    powerset;   [[], [1], ..., [1..n]]
  €O  mapped sum; [0, 1, ..., T(n)]
    O sum;        [a(n)]

Answer 5

JavaScript, 10 bytes, user81655, A033999

I think I got it. Yeah. This one was really hard. I like the submission because it relies heavily on precedences.

---

It's the sequence A033999:

a(n) = (-1)^n.

Source

t=>~t.z**t

Explanation

If you split this code according to the JavaScript operator precedences you get:

1. . (precedence 18) gets evaluated first and t.z will return undefined.
2. ~ (precedence 15) tries to cast undefined, resulting in 0, and returns -1 after bitwise not.
3. ** (precedence 14) will return -1 ^ t, where t is odd or even, resulting in -1 or 1.

Demo

console.log(
    (t=>~t.z**t)(0),
    (t=>~t.z**t)(1),
);

Try before buy

---

I will award a 100 rep bounty on this cool Cop submission.

Answer 6

Element, 7 bytes, PhiNotPi, A000042

_'[,1`}

Notes: I was misled by the } for soooooo long. So it also matches [.

Try it online!

---

How it works:

_'[,1`}
_        main_stack.push(input());
 '       control_stack.push(main_stack.pop());
  [      Object temp = control_stack.pop();
         for(int i=0;i<temp;i++){
   ,         Object a = main_stack.pop(); //is actually zero
             main_stack.push(a.toChars()[0]);
             main_stack.push(a);
    1        main_stack.push(1);
     `       System.out.println(main_stack.pop());
      }  }

Answer 7

05AB1E, 4 bytes, Paul Picard, A001317

Code:

$Fx^

Try it online!

Explanation:

$      # Pushes 1 and input
 F     # Pops x, creates a for-loop in range(0, x)
  x    # Pops x, pushes x and 2x
   ^   # Bitwise XOR on the last two elements
       # Implicit, ends the for-loop
       # Implicit, nothing has printed so the last element is printed automatically

The sequence basically is a binary Sierpinski triangle:

f(0)=      1                    =1
f(1)=     1 1                   =3
f(2)=    1 0 1                  =5
f(3)=   1 1 1 1                 =15
f(4)=  1 0 0 0 1                =17

And translates to the formula a(n) = a(n - 1) XOR (2 × a(n - 1))

Luckily, I remembered this one :)

Answer 8

PHP, 41 bytes, insertusernamehere, A079978

echo/* does n%3=0 */$argv[1]%3<1?1:0    ;

Returns 1 if its argument is a multiple of 3, and 0 otherwise. Not much beyond that.

Answer 9

Jolf, 3 bytes, Easterly Irk, A001477

axx

Consists of a simple cat (ax) followed by a no-op. Not sure what the cop was going for here.

Answer 10

MATL, 9 bytes, beaker, A022844

Code (with a whitespace at the end):

3x2xYP*k 

Try it online!

Found the following three matches with a script I wrote:

Found match: A022844
info: "name": "Floor(n*Pi).",

Found match: A073934
info: "name": "Sum of terms in n-th row of triangle in A073932.",

Found match: A120068
info: "name": "Numbers n such that n-th prime + 1 is squarefree.",

I tried to do the first one, which is basically done with YP*k:

3x2x       # Push 3, delete it, push 2 and delete that too
    YP     # Push pi
      *    # Multiply by implicit input
       k   # Floor function

Answer 11

Java, 479 bytes, Daniel M., A000073

Code:

import java.util.*;
public class A{

    public static int i=0;
    public boolean b;

    static A a = new A();

    public static void main(String[] args){
        int input = Integer.parseInt(args[0]);

        LinkedList<Integer> l = new LinkedList<>();
        l.add(1);
        l.add(0);
        l.add(0);

        for(int ix = 0; ix<=input; ix++)if(ix>2){
            l.add(0,l//d
            .get(1)+l.peekFirst()+     l.get(2));
        }

        System.out.println(input<2?0:l.pop()
              +(A.i        +(/*( 5*/ 0 )));
    }
}

If you miss non-revealed characters, they are replaced with spaces.

Answer 12

Ruby, 38 bytes, histocrat, A008592

->o{(s="+o++o"*5).sum==03333&&eval(s)}

Could be different from the intended solution as I found this by hand.

Answer 13

S.I.L.O.S, betseg, A001844

readIO
a=i
a+1
i*2
i*a
i+1
printInt i

Try it online!

Answer 14

Jolf, 5 characters, Cᴏɴᴏʀ O'Bʀɪᴇɴ, A033536

Code:

!K!8x

Output:

a(2) = 8
a(10) = 4738245926336

Answer 15

Reng v3.3, 36 bytes, Cᴏɴᴏʀ O'Bʀɪᴇɴ, A005449

iv:#+##->>)2%æ~¡#~
#>:3*1+*^##</div>

Output

a(1) = 2
a(3) = 15

Explanation

I completely ignored the prespecified commands, except the ) because I did not have enough space.

The actually useful commands are here:

iv      >>)2%æ~
 >:3*1+*^

Stretched to a straight line:

i:3*1+*)2%æ~

With explanation:

i:3*1+*)2%æ~ stack
i            [1]      takes input
 :           [1,1]    duplicates
  3          [1,1,3]  pushes 3
   *         [1,3]    multiplies
    1        [1,3,1]  pushes 1
     +       [1,4]    adds
      *      [4]      multiplies
       )     [4]      shifts (does nothing)
        2    [4,2]    pushes 2
         %   [2]      divides
          æ  []       prints
           ~ []       halts

The formula is a(n) = n(3n+1)/2.

Answer 16

05AB1E, 3 bytes, Adnan, A000292

LLO

Output

a(9) = 165
a(10) = 220

How it works

LLO Stack
L   [1,2,3,4,5,6,7,8,9]                         range
 L  [1,1,2,1,2,3,1,2,3,4,...,1,2,3,4,5,6,7,8,9] range of range
  O sum all of them

The mathematical equivalent is sum(sum(n)), where sum is summation.

Answer 17

Python 2, 87 bytes, Sp3000, A083054

n=input()
_=int(3**.5*n)-3*int(n/3**.5)########################################
print _

Not that hard, actually. Just searched for sequences that met the constraints until I found one that could be generated in the given space.

Answer 18

Jolf, 11 bytes, QPaysTaxes, A000005

aσ0xxdxxxxx

Simple enough: alert the σ0 (number of divisors of) x, then put useless stuff at the end.

Try it online! The test suite button's a bit broke, but still shows proper results.

(You could've golfed it down to two bytes! Just σ0 would've done nicely.)

Answer 19

Jolf, 11 bytes, RikerW, A011551

Code:

c*mf^+91x~P

Explanation:

     +91     # add(9, 1) = 10
    ^   x    # 10 ** input
  mf         # floor function (no-op)
 *       ~P  # multiply by phi
c            # ¯\_(ツ)_/¯

Try it here.

Answer 20

JavaScript (ES6), 119 bytes, Cᴏɴᴏʀ O'Bʀɪᴇɴ, A178501

x=>(n="=>[[["|x|"##r(###f#n###;##")|n?Math.pow("#<1##].c####t.##pl##[####nc#"|10,"y([###(###(#]###)"|x-1|``):0|`#h####`

I'm sure the actual code generates a trickier sequence than this, but with just the two outputs, this OEIS sequence is simple and matches them.

Without all the ignored characters, the algorithm is just x=>x?Math.pow(10,x-1):0.

Answer 21

05AB1E, 5 bytes, Luis Mendo, A051696

Code:

Ðms!¿

Explanation:

Ð      # Triplicate input.
 m     # Power function, which calculates input ** input.
  s    # Swap two top elements of the stack.
   !   # Calculate the factorial of input.
    ¿  # Compute the greatest common divisor of the top two elements.

So, basically this calculates gcd(n!, nⁿ), which is A051696.

Try it online!.

Answer 22

PHP, 18 bytes, insertusernamehere, A023443

Code:

echo$argv[1]+0+~0;

Output:

a(0) = -1
a(1) = 0

Answer 23

Octave (34 bytes) by Stewie Griffin

The sequence is A066911.

@(m)(mod(m,u=1:m  )&isprime(u))*u'

Answer 24

PHP, 137 bytes, insertusernamehere, A000959

Code:

for($s=range(1,303);$i<($t=count($s));$s=array_merge($s))for($j=($k=++$i==1?2:$s[$i-1])-1;$j<$t;$j+=$k )unset($s[$j]);echo$s[$argv[1]-1];

Output:

a(3)  =   7
a(7)  =  21
a(23) =  99

Answer 25

05AB1E, 10 bytes, George Gibson, A003215

Code:

Ds3*s1+*1+

Explanation:

Computes 3*n*(n+1)+1 which is the oeis sequence A003215.

Answer 26

Element, 10 bytes, PhiNotPi, A097547

2_4:/2@^^`

Try it online!

Output

a(3) = 6561
a(4) = 4294967296

Answer 27

Pyke, 6 bytes, muddyfish, A005563

0QhXts

Yay hacks! The 0Qh and s are no-ops. hXt just computes (n + 1) ^ 2 - 1.

Answer 28

J, 8 bytes, Kenny Lau, A057427

Code:

(-%- )\.

Output:

a(0) = 0
a(1..29) = 1

I don't think this is intended. And I don't know why J had this behavior. But it works.

Answer 29

Pyth, 70 bytes, FliiFe, A070650

Code (with obfuscated version below):

DhbI|qb"#"qb"#"R!1Iqb"#";=^Q6+""s ]%Q27  ;.qlY+Q1Ih+""Z##;.q)=Z+Z1;@YQ
DhbI|qb"#"qb"#"R!1Iqb"#"#####+""s####2###;##lY+Q1Ih+""Z#####)=Z+Z1;@YQ (obfuscated)

This basically does:

=^Q6%Q27

It calculates a(n) = n⁶ % 27, which is A070650. Explanation:

=^Q6       # Assign Q to Q ** 6
    %Q27   # Compute Q % 27
           # Implicit output

Try it here

Answer 30

Python, 108, CAD97, A005132

def a(n):
 if n == 0: return 0
 f=a(n-1)-n
 return f if f>0 and not f in(a(i)for i in range(n))else a(n-1)+n

Obfuscated code :

def a(n):
 ###n####0######n#0
 f=a#######
 return f #f#####a###### f ####a(##f###i#i###a####n##else a#######

Outputs:

>>> a(0)
0
>>> a(4)
2
>>> a(16)
8
>>> a(20)
42

Answer 31

C (71 bytes) by mIllIbyte

The sequence is floor(n/4).

x;f(_){x/=4;
    }
main(){
 scanf("%d",&x);
 f( 6);
 printf("%d", x);
}

Answer 32

Python 3 (58 bytes) by CAD97

The sequence is A062318. There were a lot of unnecessary characters in this one, so I commented some out.

a=lambda n: ~-( -~(n%2)
# if ###### else 
*3**(n//2))#)##)

Answer 33

Ruby, 11 bytes, histocrat, A011765

It only needs 5...

->i{i%4/3 }

I'd like to note though that the sequence is off-by-one compared to OEIS (which specifies that the first term corresponds to A(1)).

Answer 34

Haskell, 51 bytes, Zgarb, A063866

f e=sum[1|1<-sum<$>mapM(flip(:)=<<pure.(0-))[1..e]]

Output: map f [0..6]: [0,1,1,0,0,3,5].

Answer 35

05AB1E, 5 bytes, Adnan, A102669

Code:

TA«-g

Output:

a(0) = 0
a(1) = 0
a(2) = 1
a(3) = 1
a(4) = 1
a(5) = 1
a(6) = 1
a(7) = 1
a(8) = 1
a(9) = 1
a(10) = 0
a(11) = 0

Too bad you have designed the language in this inconvenient way.

Answer 36

Python, 17 bytes, ASCIIThenANSI, A000027

def a(n):return n

Trivial?

Answer 37

Python, 60 bytes, ASCIIThenANSI A000930

def a(n):
 if n<3:
  return 1
 else:
  return a(n-1)+a(n-3)

Still trivial

Answer 38

05AB1E, (4 bytes) by Adnan

The sequence is A126804, and the code to produce it is

D·ŸP

Try it online

I've never even looked at a program in 05AB1E before, it looks like a pretty cool language!

Answer 39

Lua, 45 bytes, Katenkyo, A001477

a=function(n)return n-1<1 and 0or 1+a(n-1)end

Answer 40

Swift, 55 bytes, Nefrin, A103847

The sequence is McCarthy's 91 Function.
It takes the value 91 for any n up to 101 and then continues with 92,93,94 ...

I wonder what it's actually useful for.

func M(n:Int)->Int{
return(n<=100) ?M(M(n+11)):n-10;
}

Answer 41

05AB1E, 28 bytes, Lause, A256861

Code:

DD>D>D>D>****sD<*6+*6n2n5**/

Explanation:

              # implicit n on stack
DD>D>D>D>     # save n,n+1,n+2,n+3,n+4 on the stack
****          # multiply top 5 on stack together
              #     new stack: n,n(n+1)(n+2)(n+3)(n+4)
sD<*          # rotate, duplicate, decrease, multiply
              #     new stack: n(n+1)(n+2)(n+3)(n+4),n(n-1)
6+*           # add 6 to stack, add top 2, multiply top 2
              #     new stack: n(n+1)(n+2)(n+3)(n+4)(n(n-1)*6)
6n2n5**       # add 720 to stack
/             # divide
              # implicitly print n(n+1)(n+2)(n+3)(n+4)(n(n-1)*6)/720

Answer 42

Jolf, 3 bytes, Cᴏɴᴏʀ O'Bʀɪᴇɴ, A091940

+QQ

Test it here.

Since the output is off-by-one, the second formula on OEIS reduces to n + n^4. Where we implement n^4 by squaring the input twice.

Answer 43

05AB1A, 4 bytes, Adnan, A001127

Code:

$FÂ+

Explained:

$          # push 1 and input
 F         # input number of times, do:
  Â        # duplicate and reverse
   +       # add

The sequence starts at 1 and continues by adding the current number to its reverse.

0: 1
1: 1+1 = 2
2: 2+2 = 4
3: 4+4 = 8
4: 8+8 = 16
5: 16+61 = 77
6: 77+77 = 154
7: 154+451= 605
and so on...

Answer 44

05AB1E, 3 bytes, Emigna, A022559

Code:

!ÓO

Explanation:

!    # Compute the factorial of the implicit input.
 Ó   # Compute the exponents of the prime factors.
  O  # Sum that up together.

Resulting into A022559. Try it online!

Answer 45

05AB1E, 1 byte, mnbvc, A004085

Code:

Õ

Sequence is the sum of digits of Euler totient function of n.

Answer 46

LiveCode, 35 bytes, mnbvc, A001477

I/O:

oeis (0) = 0
oeis (2) = 2

Code:

function oeis n
    return n*1
End oeis

Sequence returns the non-negative integers.

Answer 47

Python, 37 bytes, mnbvc, A000290

I/O:

a(2) = 4
a(4) = 16

Code:

def a(n):
    return n*n#__!______n____*_

Sequence return square of input

Answer 48

05AB1E, 3 bytes, Emigna, A002994

Code:

3m¬

Explanation:

3m   # Compute input ** 3
  ¬  # Take the first digit

This gives us the initial digits of the cubes, which is A002994.

Uses CP-1252 encoding. Try it online!.

Answer 49

05AB1E, 8 bytes, Emigna, A088666

Code (with obfuscated version below):

4m1s+rT%
___s_r__

Explanation:

4m         # Compute input ** 4
  1        # Add one to the stack
   s       # Swap the top two elements
    +      # Add the two numbers (result = input ** 4 + 1)
     r     # Reverse the stack
      T    # Add 10
       %   # Modulo

This comes down to the formula a(n) = (n⁴ + 1) % 10, which is A088666. I'm excited to see what the original code was.

Try it online!.

Answer 50

J, 8 bytes, Kenny Lau, A057427

Code:

(>1-*)<.

Output:

a(0) = 0
a(i) = 1 for 0 < i < 30

The sequence is simply signum(n). The program computes floor(n) > 1 - signum(floor(n)), which is signum(n) for non-negative integers n.

Answer 51

Python 3, 50 bytes, CAD97, A004242

from math import    *
a=lambda n:ceil(1000*log(n))

Try it online

Output:

a(1) => 0
a(10) => 2303

Answer 52

Python 2 (72 bytes) by Sp3000

There's an unbelievable amount of sequences starting with 1,2,3,6,9,18 from the 1 element. I was fortunate to find one on the 3rd page where these are the only 6 terms: Divisors of 18.

print[1,#((#y
2,3,6,9,#.####c###6#####16#.
18][#)#131963
~-input()#3
]#)

Answer 53

Pyth, 4 bytes, FliiFe, A000041

Code:

l./Q

Explanation:

 ./Q  # Calculate the paritions of Q (input)
l     # Take the length, which counts the number of partitions.

Try it here.

Answer 54

JavaScript, 30 bytes, Aplet123, A002061

It's the sequence A002061:

Central polygonal numbers: n^2 - n + 1.

n=>Math.pow(n,!![]+!![])-(n-1)

See it in action on JSFiddle.

Answer 55

Python, 123 bytes, ASCIIThenANSI, A089911

Code:

def f(n):
 if n<2:
  return n
 else:
  return f(n-1)+f(n-2)

def g(n):
 if n<2:
  return n
 else:
  return (f(n-1)+f(n-2))%12

The formula for this sequence is g(n) = f(n) % 12, whereas f(n) is the n^th Fibonacci number.

This was actually a really nice puzzle! :)

Answer 56

Python 3, 58 bytes, Mega Man

import math;x=int(input());print(int(x/math.sqrt(5)*x**x))

The sequence is:

(x/√5)*(x**x)

Answer 57

Python, 42 bytes, TùxCräftîñg

def f(x):return x if x<2else f(x-1)+f(x-2)

This is the Fibonacci sequence.