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

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 2

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 3

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 4

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 5

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 6

Python, 17 bytes, ASCIIThenANSI, A000027

def a(n):return n

Trivial?

Answer 7

Python, 60 bytes, ASCIIThenANSI A000930

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

Still trivial

Answer 8

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 9

Lua, 45 bytes, Katenkyo, A001477

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

Answer 10

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 11

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 12

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 13

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 14

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 15

05AB1E, 1 byte, mnbvc, A004085

Code:

Õ

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

Answer 16

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 17

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 18

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 19

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 20

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 21

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 22

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 23

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 24

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 25

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 26

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 27

Python, 42 bytes, TùxCräftîñg

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

This is the Fibonacci sequence.