53
\$\begingroup\$

There are clever ways of determining whether a number is a power of 2. That's no longer an interesting problem, so let's determine whether a given integer is an integer power of -2. For example:

-2 => yes: (-2)¹
-1 => no
0 => no
1 => yes: (-2)⁰
2 => no
3 => no
4 => yes: (-2)²

Rules

  • You may write a program or a function and use any of the standard methods of receiving input and providing output.

  • Your input is a single integer, and output must be a truthy value if the integer is an integer power of -2, and a falsy value otherwise. No other output (e.g. warning messages) is permitted.

  • The usual integer overflow rules apply: your solution must be able to work for arbitrarily large integers in a hypothetical (or perhaps real) version of your language in which all integers are unbounded by default, but if your program fails in practice due to the implementation not supporting integers that large, that doesn't invalidate the solution.

  • You may use any programming language, but note that these loopholes are forbidden by default.

Winning condition

This is a contest: the answer which has the fewest bytes (in your chosen encoding) is the winner.

\$\endgroup\$
12
  • 20
    \$\begingroup\$ @KritixiLithos I don't see why it should. There is no integer i such that (-2)^i = 2 \$\endgroup\$
    – Fatalize
    Apr 6, 2017 at 12:10
  • 2
    \$\begingroup\$ Are the exponents positive or -0.5 should be valid since it's 2^(-1). \$\endgroup\$
    – Mr. Xcoder
    Apr 6, 2017 at 12:13
  • 1
    \$\begingroup\$ @Mr.Xcoder, Since inputs are always integer values, a negative exponent won't be required (or possible). \$\endgroup\$ Apr 6, 2017 at 12:16
  • 2
    \$\begingroup\$ @Jason, as many as supported/natural in your language - see the third rule. And it's code-golf because it needs an objective winning criterion to be on-topic here - "a pleasing solution" doesn't cut it (though I do like the Mathematica answer - that surprised me). \$\endgroup\$ Apr 6, 2017 at 14:55
  • 2
    \$\begingroup\$ @TobySpeight I've proposed an alternate version of this challenge here. \$\endgroup\$
    – Jason C
    Apr 6, 2017 at 16:07

57 Answers 57

28
\$\begingroup\$

Mathematica, 22 bytes

EvenQ@Log2@Max[#,-2#]&

Try it online! (Using Mathics instead, where this solution also works.)

I tried to find a solution with bitwise operators for a while, and while one definitely exists, I ended up finding something which is probably simpler:

  • Max[#,-2#] multiplies the input by -2 if it's negative. Multiplying by another factor of -2 doesn't change whether the value is a power of -2 or not. But now all odd powers of -2 have been turned into even powers of -2.
  • But even powers of -2 are also even powers of 2, so we can use a simple Log2@... and check if the result is an integer (to check whether it's a power of 2). This already saves two bytes over Log[4,...] (another way to look at even powers of -2).
  • As an added bonus, checking whether a value is an even integer is shorter than just checking whether it's an integer: we can save three more bytes by using EvenQ instead of IntegerQ.
\$\endgroup\$
2
  • \$\begingroup\$ Does it help to consider that even powers of -2 are integer powers of 4? I like the idea of multiplying by -2 to get everything positive - though disappointed to see no bit-twiddling so far. \$\endgroup\$ Apr 6, 2017 at 13:04
  • 6
    \$\begingroup\$ @TobySpeight Treating them as powers of 2 actually saves 5 bytes. I used powers of 4 at first, but Log[4,...] is longer than Log2@... and IntegerQ is longer than EvenQ. \$\endgroup\$ Apr 6, 2017 at 13:05
18
\$\begingroup\$

Jelly, 5 bytes

æḟ-2=

Try it online!

How it works

æḟ-2=  Main link. Argument: n

æḟ-2   Round n towards 0 to the nearest power of -2.
    =  Test if the result is equal to n.
\$\endgroup\$
13
\$\begingroup\$

Jelly, 6 bytes

b-2S⁼1

Try it online!

This is based on how Jelly converts an integer N to any arbitrary base B, doing so by converting N to an array, in which each integer is a digit d of (N)B, which can have a value 0≤Vd<B. Here, we will 0-index digits from the right, so every digit adds VdBd to form N. Vd<BVdBd<BBd=Bd+1, therefore every possible N has only one unique representation, if we ignore leading 0s in (N)B.

Here, d=input, B=-2. N=Bd=1Bd=VdBd⇔1=VdVd=1, and, since we're not adding any other multiples of powers of B, every other V would be 0. Right now, the array should be a 1 concatenated with d 0s. Since Jelly 1-indexes from the left, we should check whether the array's 1st element is 1, and all other elements are 0.

Hmm... all good, right? No? What's going on? Oh yeah, I have a better idea! First, let's take the sum of all of the integers in the array, treating it as if it was an integer array and not a number in base -2. If it is 1, it means that there is only one 1, and all other integers are 0. Since there can't be leading zeroes, except in the case of 0-2 (where the sum would be 0≠1 anyways), the 1st integer must be non-zero. The only non-zero integer in the array is the 1, so it must be the first one. Therefore, this is the only case that the sum of all of the integers in the array would be 1, because the smallest possible sum of a pair of positive integers is Σ{1,1}=2, since the smallest positive integer is 1. Every integer in a base representation is non-negative, so the only way the sum is 1 is to only have one 1, and all other integers are 0. Therefore, we can just check if the sum of all of the integers in the array is 1.

Here is what the code does:

b-2S⁼1 Main link. Arguments: d
b-2    Convert d to base -2.
   S   Take the sum.
    ⁼1 Check if the sum is equal to 1.
\$\endgroup\$
3
  • 2
    \$\begingroup\$ Phew, that explanation took time to write... \$\endgroup\$ Apr 6, 2017 at 13:34
  • \$\begingroup\$ I'd hate to see what an explanation for a long program would look like then... \$\endgroup\$
    – boboquack
    Apr 8, 2017 at 8:32
  • 1
    \$\begingroup\$ @boboquack Here I am explaining why I use the base conversion stuff. I don't think explanation for long programs would be this long. A post can contain up to 30000 markdown characters, and explanations for longer programs would be more terse anyways. Also, I have read much longer explanations, and they are not that boring. \$\endgroup\$ Apr 8, 2017 at 8:38
13
\$\begingroup\$

Python, 46 bytes

-2 bytes thanks to @ovs.

def g(x):
 while x%-2==0!=x:x/=-2
 return x==1

Function with usage:

g(4) # put your number between the brackets

Try it online!

\$\endgroup\$
14
  • \$\begingroup\$ print g(8) prints False \$\endgroup\$ Apr 6, 2017 at 13:02
  • 2
    \$\begingroup\$ @FelipeNardiBatista shouldn't it? \$\endgroup\$
    – Mr. Xcoder
    Apr 6, 2017 at 13:04
  • 2
    \$\begingroup\$ sorry, my example was a bad one, print g(4) does the same \$\endgroup\$ Apr 6, 2017 at 13:05
  • \$\begingroup\$ Wait, there is a small error, fixing it shortly \$\endgroup\$
    – Mr. Xcoder
    Apr 6, 2017 at 13:06
  • 1
    \$\begingroup\$ I've put a ; instead of a newline... sorry for that. Fixed @FelipeNardiBatista \$\endgroup\$
    – Mr. Xcoder
    Apr 6, 2017 at 13:08
12
\$\begingroup\$

Python 2, 98 50 bytes

lambda x:x*(x&-x==abs(x))*((x<0)^x.bit_length()&1)

Try it online!

\$\endgroup\$
0
11
\$\begingroup\$

Python 2, 35 34 32 bytes

f=lambda n:n==1or n!=n%2<f(n/-2)

Try it online!

\$\endgroup\$
10
\$\begingroup\$

05AB1E, 8 bytes

Y(IÄÝm¹å

Try it online! or as a Test suite

Explanation

Y(         # push -2
  IÄÝ      # push range [0 ... abs(input)]
     m     # element-wise power
      ¹å   # check if input is in the resulting list
\$\endgroup\$
4
  • \$\begingroup\$ Why the downvote? \$\endgroup\$
    – user41805
    Apr 6, 2017 at 15:28
  • \$\begingroup\$ @KritixiLithos: Seems like someone has downvoted all golfing languages. \$\endgroup\$
    – Emigna
    Apr 6, 2017 at 15:29
  • 7
    \$\begingroup\$ Noticed that too. Although I haven't been around PPCG for long, I've learned that creative and interesting solutions in standard languages are far more appreciated than 3-byte solutions in golfing languages. However, there are some people who (unfortunately) downvote very creative solutions in golfing languages, simply because they think everything is built-in, and don't understand how good the algorithms (although written in golfing languges) are. +1 for the incredible solution @Emigna \$\endgroup\$
    – Mr. Xcoder
    Apr 6, 2017 at 15:32
  • \$\begingroup\$ ÄLY(småO for 8. Y(sÄLm¢Z for 8... Nevermind, all 8. \$\endgroup\$ Jul 13, 2017 at 14:04
10
\$\begingroup\$

JavaScript (ES6), 37 28 24 bytes

f=x=>!x|x%2?x==1:f(x/-2)

Saved 4 bytes thanks to Arnauld.

f=x=>!x|x%2?x==1:f(x/-2)

console.log(f(-2));
console.log(f(-1));
console.log(f(0));
console.log(f(1));
console.log(f(2));
console.log(f(3));
console.log(f(4));

\$\endgroup\$
3
  • \$\begingroup\$ Why do I see some errors (before the true/false values) when I click on "Run code snippet"? \$\endgroup\$ Apr 6, 2017 at 13:49
  • \$\begingroup\$ @numbermaniac I'm not sure, maybe you're using a browser that doesn't fully support ES6? \$\endgroup\$
    – Tom
    Apr 6, 2017 at 13:51
  • \$\begingroup\$ Welp, refreshed and tried again, no errors. Not sure what happened the first time. \$\endgroup\$ Apr 6, 2017 at 13:54
10
\$\begingroup\$

Excel, 40 36 bytes

Saved 4 bytes by CallumDA

Excel can certainly do it but correcting errors adds 11 bytes

=IFERROR(-2^IMREAL(IMLOG2(A1)),1)=A1

Input is in cell A1. Output is TRUE or FALSE

If it was allowed to return either FALSE or #NUM! error for false values, it would be only 25 bytes:

=-2^IMREAL(IMLOG2(A1))=A1
\$\endgroup\$
2
  • \$\begingroup\$ Heres a small improvement: =IFERROR(-2^IMREAL(IMLOG2(A1)),1)=A1 \$\endgroup\$
    – CallumDA
    Apr 6, 2017 at 23:49
  • 1
    \$\begingroup\$ @CallumDA Thanks! I tried to figure a way to use the complex number functions but everything I came up with was longer. \$\endgroup\$ Apr 7, 2017 at 3:43
9
\$\begingroup\$

C (gcc), 34 29 bytes

f(n){n=n%2?n==1:f(n?n/-2:2);}

Try it online!

\$\endgroup\$
0
8
\$\begingroup\$

MATL, 9 8 bytes

2_y|:q^m

Try it online! Or verify all test cases.

How it works

Consider input -8 as an example

2_    % Push -2
      % STACK: -2
y     % Implicit input. Duplicate from below
      % STACK: -8, -2, -8
|     % Absolute value
      % STACK: -8, -2, 8
:     % Range
      % STACK: -8, -2, [1 2 3 4 5 6 7 8]
q     % Subtract 1, element-wise
      % STACK: -8, -2, [0 1 2 3 4 5 6 7]
^     % Power, element-wise
      % STACK: -8, [1 -2 4 -8 16 -32 64 -128]
m     % Ismember. Implicit display
      % STACK: 1
\$\endgroup\$
2
  • \$\begingroup\$ If I understand your explanation correctly, then given input n, this creates an array of size n as an intermediate step. Good job that efficiency is not a criterion here! \$\endgroup\$ Apr 6, 2017 at 15:03
  • 2
    \$\begingroup\$ @Toby Of course! This is code golf, who cares about efficiency? :-D \$\endgroup\$
    – Luis Mendo
    Apr 6, 2017 at 15:05
8
\$\begingroup\$

Regex (ECMAScript), 38 37 34 33 bytes

\b(?=(^|x))((x*)\3\3(?=\3$))*x\1$

Try it online!

Takes its input in unary, as an optional - sign followed by a string of x characters, the count of which represents the absolute value of the number. (As such, it is not bijective unary, as zero can be represented in two ways. The regex works with both.)

Sets a flag for whether a negative sign is present. Operating on the absolute value, repeatedly divides by 4 (requiring that there is no remainder), and then asserts that the remaining number is 1 if there was no negative sign, or 2 if there was.

\b                    # Anchor on the border between {start or "-"} and "x"; in the
                      # domain of this problem, it is actually equivalent to
                      # (?<=^(?=x)|-|x$) - so it can anchor on the end instead of the
                      # beginning, but that would result in a non-match.
(?=(^|x))             # \1 = 1 if negative sign is present, 0 if not
(                     # loop the following:
    (x*)\3\3(?=\3$)   # assert tail is divisible by 4; tail = tail / 4
)*                    # iterate as many times as possible (minimum 0)
x                     # tail -= 1
\1                    # tail -= \1
$                     # assert tail==0

Regex (ECMAScript), 36 bytes

^(-(x*)(?=\2$))?((x*)\4\4(?=\4$))*x$

Try it online!

This was the original version I posted; I'm keeping it here, as it's sufficiently different, yet close enough in length, to be interesting.

If it sees a negative sign, it strips it and divides the number by 2 (requiring that there is no remainder). It then asserts that the remaining number is a power of 4.

^
(
    -                 # eat a leading negative sign
    (x*)(?=\2$)       # assert tail is even; tail = tail / 2
)?                    # do the above optionally
# Assert that tail is a power of 4
(                     # loop the following:
    (x*)\4\4(?=\4$)   # assert tail is divisible by 4; tail = tail / 4
)*                    # iterate as many times as possible (minimum 0)
x$                    # assert tail == 1; if this fails to match, the regex engine will
                      # try backtracking the loops, but that cannot result in a match

Regex (PCRE / .NET), 32 bytes

^(-)?((x*)\3\3(?=\3$))*x(?(1)x)$

Try it online! - PCRE2
Try it online! - .NET

Uses essentially the same algorithm as the 33 byte ECMAScript version, but in a more straightforward way, which never matches strings outside the problem's domain.

^
(-)?                  # \1 = negative sign if it is present; unset otherwise
(                     # loop the following:
    (x*)\3\3(?=\3$)   # assert tail is divisible by 4; tail = tail / 4
)*                    # iterate as many times as possible (minimum 0)
x                     # tail = tail - 1
(?(1)x)               # if \1 is set, then tail = tail - 1, else do nothing
$                     # assert tail == 0
\$\endgroup\$
7
\$\begingroup\$

Python, 24 bytes

lambda n:n*n&n*n-1<n%3%2

Try it online!

The bit trick k&k-1==0 checks whether k is a power of 2 (or k==0). Checking this for k=n*n as n*n&n*n-1==0 tells us whether abs(n) is a power of 2.

To further see if n is a power of -2, we need only check that n%3==1. This works because mod 3, the value -2 is equal to 1, so its powers are 1. In contrast, their negations are 2 mod 3, and of course 0 gives 0 mod 3.

We combine the checks n*n&n*n-1==0 and n%3==1 into a single expression. The first can be written with <1 for ==0, since it's never negative. The n%3==1 is equivalent to n%3%2, giving 0 or 1. So, we can combine them as n*n&n*n-1<n%3%2.

\$\endgroup\$
6
\$\begingroup\$

Octave, 28 bytes

@(n)any((-2).^(0:abs(n))==n)

This defines an anonymous function. The approach is similar to that in my MATL answer.

Try it online!

\$\endgroup\$
6
\$\begingroup\$

PHP, 41 Bytes

for(;$argn%-2==0;)$argn/=-2;echo$argn==1;

PHP, 52 Bytes

echo($l=log(abs($argn),2))==($i=$l^0)&&$argn>0^$i%2;

PHP, 64 Bytes

Working with a Regex

echo preg_match("#^".($argn>0?1:"1+0")."(00)*$#",decbin($argn));
\$\endgroup\$
6
\$\begingroup\$

JavaScript (ES6), 21 bytes

A recursive function that returns 0 or true.

f=n=>n==1||n&&f(n/-2)

How it works

This doesn't include any explicit test -- like n being odd or abs(n) being less than one -- to stop the recursion early when the input is not an exact power of -2.

We exit only when n is exactly equal to either 1 or 0.

This does work however because any IEEE-754 float will eventually be rounded to 0 when divided by 2 (or -2) enough times, because of arithmetic underflow.

Test cases

f=n=>n==1||n&&f(n/-2)

console.log(f(-2));
console.log(f(-1));
console.log(f(0));
console.log(f(1));
console.log(f(2));
console.log(f(3));
console.log(f(4));

\$\endgroup\$
6
\$\begingroup\$

HP 28s 36 bytes

WHILE DUP ABS 1 > REPEAT 2 NEG / END 1 ==
\$\endgroup\$
2
  • 3
    \$\begingroup\$ Welcome to Code Golf! Nice first answer! \$\endgroup\$ Jul 20, 2021 at 2:59
  • \$\begingroup\$ @RedwolfPrograms Thanks! \$\endgroup\$ Jul 20, 2021 at 14:46
5
\$\begingroup\$

Python 3, 34 bytes

lambda n:n==(-2)**~-n.bit_length()
\$\endgroup\$
5
\$\begingroup\$

C (gcc), 33 bytes

f(n){return n%2?n==1:n&&f(n/-2);}

Try it online!

\$\endgroup\$
4
\$\begingroup\$

Java 7, 55 bytes

boolean c(int n){return n==0?0>1:n%-2==0?c(n/-2):n==1;}

Explanation:

boolean c(int n){  // Method with integer parameter and boolean return-type
  return n==0 ?    //  If n is zero:
    0>1//false     //   Return false
   : n%-2==0 ?     //  Else-if n mod -2 is zero:
    c(n/-2)        //   Recursive call for the input divided by -2
   :               //  Else:
    n==1;          //   Return if n is one
}                  // End of method

Test code:

Try it here.

class M{
  static boolean c(int n){return n==0?0>1:n%-2==0?c(n/-2):n==1;}

  public static void main(String[] a){
    for(int i = -2; i <= 4; i++){
      System.out.println(i + ": " + c(i));
    }
  }
}

Output:

-2: true
-1: false
0: false
1: true
2: false
3: false
4: true
\$\endgroup\$
3
  • \$\begingroup\$ The non-recursive way is shorter by 5 bytes: boolean c(int n){while(0==n%-2)n/=-2;return 1==n;}. \$\endgroup\$ Apr 7, 2017 at 14:14
  • \$\begingroup\$ @OlivierGrégoire Unfortunately that one doesn't work for n=0 in Java, because 0%-2==0 will be true and 0/-2 is still 0, causing an infinite loop, which is why I added the n==0?0>1 part to my recursive method. \$\endgroup\$ Apr 7, 2017 at 14:40
  • \$\begingroup\$ Nicely spotted! \$\endgroup\$ Apr 7, 2017 at 14:51
4
\$\begingroup\$

Haskell, 24 23 bytes

f 0=0
f 1=1
f n=f(-n/2)

Defines a function f which returns 1 for powers of -2 and 0 otherwise.

A golfed version of my first submission to the other challenge.

\$\endgroup\$
0
4
\$\begingroup\$

Pyth, 7 bytes

!tsjQ_2

Test suite

Explanation:

     _2   # -2
   jQ     # Cast input to that base
          # Iff input is a power of -2, then jQ_2 returns something of the form [1]+[0]*n
          # where n is the power of -2
          # This is also the only situation in which the sum of that is 1
!ts       # so we check for that: is the (s)um equal to 1 (or, equivalently, "!(sum - 1)" )
\$\endgroup\$
4
\$\begingroup\$

05AB1E, 5 4 bytes

2(вO

Try it online! Beats all other answers.

2(вO  # full program
   O  # push sum of...
  в   # list of base-10 values of base...
 (    # negative...
2     # literal....
  в   # digits of...
      # implicit input
      # implicit output
\$\endgroup\$
4
\$\begingroup\$

Risky, 14 bytes

__-+_-2{[?+_0:__0+_0+__?+_0

Try it online!

Explanation:

  • __-+_-: -2, written as -1 + -1 with padding
  • 2: Power
  • {[?+_0: log_2(abs(input)), with +_0 as padding
  • :: Equals
  • __0+_0+__?+_0: Input (?), with padding

So basically, -2 ** log_2(abs(input)) == input

\$\endgroup\$
1
  • \$\begingroup\$ Do you think you could add and explanation for this, please? Thanks! \$\endgroup\$ Jan 29 at 8:04
4
\$\begingroup\$

x86-32 machine code, 20 18 16 bytes

-2 bytes thanks to @xiver77 -2 bytes by removing an unnecessary jump

(This submission was originally 19 bytes, but I adjusted it to 20 since I forgot to add ret).

00000000: f8a9 ffff 7f00 7507 c1f8 1730 e0d1 f8c3  ......u....0....

Try it online!

Expects a single precision float in eax. Output is in the carry flag.

It takes advantage of some fundamental properties of floats, for some power of -2 n:

  • abs(n) is a power of 2, therefore n's mantissa must be 0.
  • n is positive for odd numbered exponents, and negative for even exponents. In other words, sign ^ exponent[0] = 1 where exponent[0] is the least significant bit of the exponent.

NASM source:

[BITS 32]

isnegpow2:
    ; Clear carry
    clc
    ; Check if mantissa is non-zero
    test eax, 0x7fffff
    ; If it is, then abs(eax) is not a
    ; power of 2, so eax cannot be a power
    ; of -2
    jnz .end

    ; Shift away the mantissa
    sar eax, 23
    ; Now ah contains the sign,
    ; and al contains the exponent.
    ; For powers of -2, sign ^ exponent[0] = 1
    xor al, ah
    ; Check lsb by shifting it to the carry flag
    ; If it's 1, then eax is a power of -2, so the
    ; carry will be set
    sar eax, 1
.end:
    ret
\$\endgroup\$
5
  • \$\begingroup\$ I hadn't even considered floating-point input! \$\endgroup\$ Feb 25 at 18:54
  • \$\begingroup\$ Nice observation! Clever, really. \$\endgroup\$
    – xiver77
    Feb 26 at 3:49
  • \$\begingroup\$ How about this? It fits in 17 bytes. BTW, is it acceptable not to have a ret? \$\endgroup\$
    – xiver77
    Feb 26 at 4:07
  • \$\begingroup\$ It should have a ret, you're right. I'll update the post with your improvement tomorrow when I have time to figure out how it works. \$\endgroup\$
    – nununoisy
    Feb 26 at 5:24
  • \$\begingroup\$ Thanks @xiver77, didn't think to set the carry flag using sar. Neat stuff. Managed to rearrange it slightly to golf off another 2 bytes also. \$\endgroup\$
    – nununoisy
    Feb 26 at 21:51
3
\$\begingroup\$

Javascript(ES7), 45 bytes

x=>-1**Math.log2(Math.abs(x))*Math.abs(x)==x
\$\endgroup\$
4
  • \$\begingroup\$ Math.abs(x) is longer than x>0?x:-x, 11 bytes to 8 bytes. You should also be able to do -2**... instead of -1... to remove the second Math.abs(x) \$\endgroup\$ Apr 7, 2017 at 1:38
  • \$\begingroup\$ What's ES7 specific in this? \$\endgroup\$
    – Arjun
    Apr 8, 2017 at 3:36
  • \$\begingroup\$ @DobbyTheFree-Elf, ** is. \$\endgroup\$
    – Qwertiy
    Apr 10, 2017 at 20:11
  • \$\begingroup\$ What interpreter are you using. This code is syntax error on my Firefox. \$\endgroup\$
    – tsh
    Jul 19, 2021 at 11:35
3
\$\begingroup\$

Perl 6, 21 bytes

{$_==(-2)**(.lsb//0)}

Try it

Expanded:

{  # bare block lambda with implicit parameter 「$_」

  $_                  # is the input
  ==                  # equal to
  (-2)**( .lsb // 0 ) # -2 to the power of the least significant bit of the input
}

Note that 0.lsb returns Nil which produces a warning when used as a number, so the defined or operator // is used.
(Think of // as || with a different slant)

A method call with no invocant where a term is expected is implicitly called on $_. (.lsb)

Also works with .msb.

\$\endgroup\$
1
  • \$\begingroup\$ I like this one! \$\endgroup\$
    – tale852150
    Apr 7, 2017 at 14:03
3
\$\begingroup\$

Prolog (SWI), 44 bytes

p(X):-X=1;X\=0,X mod 2=:=0,Z is X/(-2),p(Z).

Online interpreter

\$\endgroup\$
3
\$\begingroup\$

R, 22 bytes

Takes input from stdin, returns TRUE or FALSE accordingly.

scan()%in%(-2)^(0:1e4)

I'm not 100% sure that this is a valid answer, as it only works for integers up to R's size limit, and if the integers were unbounded it wouldn't work. However, the rules state:

The usual integer overflow rules apply: your solution must be able to work for arbitrarily large integers in a hypothetical (or perhaps real) version of your language in which all integers are unbounded by default, but if your program fails in practice due to the implementation not supporting integers that large, that doesn't invalidate the solution.

In a hypothetical version of R which does allow unbounded integers, then we could use the following code, for the same byte count:

scan()%in%(-2)^(0:Inf)

Of course, in real R, the above code just gives Error in 0:Inf : result would be too long a vector.

\$\endgroup\$
3
\$\begingroup\$

Julia 0.5, 20 bytes

!n=n∈(-2).^(0:n^2)

Try it online!

\$\endgroup\$
1
  • \$\begingroup\$ -1 byte \$\endgroup\$
    – MarcMush
    Jul 19, 2021 at 12:46

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