Is it a Mersenne Prime?

Question

A number is a Mersenne Prime if it is both prime and can be written in the form 2ⁿ-1, where n is a positive integer.

Your task is to, given any positive integer, determine whether or not it is a Mersenne prime. You may submit either a function which returns a truthy/falsy value, or a full program which performs IO.

Rules:

• As this is code-golf, you should aim to do this in the shortest byte count possible. Builtins are allowed.
• Standard golfing loopholes apply - you cannot read the Mersenne primes from external files, or hardcode them into your program.
• Your program should work for values within your language's standard integer size.

---

Test Cases

For reference, a list of (known) Mersenne Primes can be found here. Some handy test cases are:

2 -> False
1 -> False 
20 -> False
51 -> False
63 -> False
3 -> True
31 -> True
8191 -> True

---

_{Merry Christmas, everybody! Have a great holiday, whatever you celebrate :)}

• flawr: If I could I'd vote this as a dupe of the isprime challenge, as it doesn't really add anything new.
• FlipTack: @flawr They are very similar - but for this challenge, there is less likely to be a builtin and there are lots of interesting approaches to determining whether a number is representable as 2^n-1
• SuperJedi224: I believe the definition of a Mersenne number also mandates that n be prime (a condition that has also been proven necessary, but not sufficient, for (2^n)-1 to be prime.)
• FlipTack: @SuperJedi224 n is always prime, but knowing that changes nothing, the definition is still correct.
• FlipTack: @TheBitByte Yes - if you're implementing some probability-based algorithm which doesn't work 100% of the time, you can still post it, but it wouldn't be competing :)

Answer 1

05AB1E, 5 bytes

A positive number in the form 2ⁿ - 1 in binary only consists of 1's.

Code:

b`1pP

Explanation:

b` # Push each digit of the binary representation of the number onto the stack
 1p # Check if the input is prime
 P # Take the product of all these digits

Uses the CP-1252 encoding. Try it online! or Verify all test cases.

• FlipTack: I wondered how long until someone used that trick :)
• Kelly Lowder: ¹ takes 2 bytes, so this is 6.
• Adnan: @KellyLowder In UTF-8, yes. However, 05AB1E uses the CP-1252 encoding rather than the UTF-8 encoding.
• Command Master: p*bW seems to work for 4 bytes

Answer 2

Jelly, 5 bytes

&‘<ÆP

Try it online!

How it works

&‘<ÆP Main link. Argument: x
 ‘ Yield x+1.
& Take the bitwise AND of x and x+1.
 This yields 0 iff x is a Mersenne number, i.e., iff x+1 is a power of 2.
 ÆP Yield 1 if x is a prime, 0 if not.
 < Compare the results to both sides,
 This yields 1 iff x is both a Mersenne number and a prime.

• Kelly Lowder: Same issue as Adnan's answer. See mothereff.in/byte-counter
• Dennis: @KellyLowder That byte counter uses UTF-8. Both Jelly and 05AB1E use single byte character sets.

Answer 3

Python, 45 bytes

lambda n:-~n&n<all(n%i for i in range(2,n))<n

Try it online!

How it works

The three terms of the chained comparison

-~n&n<all(n%i for i in range(2,n))<n

do the following:

• -~n&n computes the bitwise AND of n + 1 and n. Since n consists solely of 1 bits if it is a Mersenne number, the bitwise AND will return 0 if (and only if) this is the case.

• all(n%i for i in range(2,n)) returns True if and only if n mod i is non-zero for all values of i in [2, ..., n - 1], i.e., if and only if n has no positive divisors apart from 1 and n.

  In other words, all returns True if and only if n is a composite number, i.e., n is either 1 or a prime.

• n is self-explanatory.

The chained comparison returns True if and only if the individual comparisons do the same.

• Since all returns either True/1 or False/0, -~n&n<all(n%i for i in range(2,n)) can only return True if -~n&n yields 0 (i.e., if n is a Mersenne number) and all returns True (i.e., if n either 1 or a prime).

• The comparison all(n%i for i in range(2,n))<n holds whenever n> 1, but since all returns True if n = 1, it does not hold in this case.

• ABcDexter: Wow, that's amazing :)

Answer 4

Brachylog, 7 bytes

#p+~^h2

Try it online!

A Brachylog program is basically a sequence of constraints which form a chain: the first constraint is between the input and an anonymous unknown (let's call it A for the purpose of this discussion), the second constraint is between that anonymous unknown and a second anonymous unknown (which we'll call B), and so on. As such, the program breaks down like this:

#p Input = A, and is prime
+ B = A + 1
~^ B = X to the power Y, C = the list [X, Y]
h D = the head of list C (= X)
2 D = 2

The only way all these constraints can be satisfied simultaneously is if B is a power of 2, i.e. the input is a power of 2 minus 1, and the input is also prime. (Brachylog uses a constraint solver internally, so the program won't be as inefficient as the evaluation order looks; it'll be aware that C is of the form [2, Y] before it tries to express B as the exponentiation of two numbers.)

Interestingly, #p+~^ almost works, because Mersenne-like primes can only use 2 as the base in non-degenerate cases (proof), but a) it fails for non-Mersenne primes B-1 as they can be expressed as B1, and b) the existing Brachylog interpreter seems to be confused (going into an infinite, or at least long-duration, loop) by a program that's that poorly constrained. So 7 bytes seems unlikely to be beaten in Brachylog.

• Fatalize: I'm impressed! As for the infinite loop problem, this is due to the overloading of predicates. Looking back I think I should not have implemented any overloading for predicates. This also causes problems in things like findall.

Answer 5

Mathematica 26 Bytes

PerfectNumberQ[# (#+1)/2]&

See this proof

Works so long as there are no odd perfect numbers, and none are known to exist.

• Jonathan Frech: So your answer is not proven to be valid?
• Jonathan Frech: I do not think that space is necessary.
• Jeppe Stig Nielsen: @JonathanFrech The formula n(n+1)/2 produces (even) perfect numbers whenever n is a Mersenne prime (Euclid). It appears to be unknown whether an odd perfect number can have the form n(n+1)/2, i.e. be a triangular number. All even perfect numbers are triangular where this n is a Mersenne prime (Euler).
• Jonathan Frech: @JeppeStigNielsen The question is if it is valid to use an unknown fact to base one's solution upon.

Answer 6

Mathematica, (削除) 29 (削除ここまで) 26 bytes

Edit: Saved 3 bytes thanks to Martin Ender

(削除) PrimeQ@#&&IntegerQ@Log2[#+1]& (削除ここまで)

PrimeQ@#&&1>BitAnd[#,#+1]&

I suspect this would be faster since the first 42 exponents are hard-coded:

MersennePrimeExponentQ@Log2[#+1]&

• Martin Ender: PrimeQ@#&&1>BitAnd[#,#+1]&

Answer 7

Perl 6, 29 bytes

{.base(2)~~/^1*$/&&.is-prime}

Try it

Expanded:

{ # bare block lambda with implicit parameter 「$_」
 .base(2) # is its binary representation ( implicit method call on 「$_」 )
 ~~
 /^ 1* $/ # made entirely of 「1」s
 && # and
 .is-prime # is it prime
}

since Perl 6 has arbitrarily large Ints, it doesn't pad the front of .base(2) with 0s.

Answer 8

Python, (削除) 83 (削除ここまで) (削除) 82 (削除ここまで) (削除) 79 (削除ここまで) (削除) 76 (削除ここまで) 73 bytes

def f(m):
 s,n=(m!=3)*4,m>>2
 while-~m&m<n:s,n=(s*s-2)%m,n>>1
 return s<1

Python 2, 71 bytes

def f(m):
 s,n=(m!=3)*4,m/4
 while-~m&m<n:s,n=(s*s-2)%m,n/2
 return s<1

This function implements the Lucas–Lehmer primality test, so while it isn't as short as some of the other Python offerings it's much faster at handling huge inputs.

---

Here's some test code that runs on Python 2 or Python 3.

from __future__ import print_function
def primes(n):
 """ Return a list of primes < n """
 # From http://stackoverflow.com/a/3035188/4014959
 sieve = [True] * (n//2)
 for i in range(3, int(n**0.5) + 1, 2):
 if sieve[i//2]:
 sieve[i*i//2::i] = [False] * ((n - i*i - 1) // (2*i) + 1)
 return [2] + [2*i + 1 for i in range(1, n//2) if sieve[i]]
def lucas_lehmer_old(p):
 m = (1 << p) - 1
 s = 4
 for i in range(p - 2):
 s = (s * s - 2) % m
 return s == 0 and m or 0
# much faster
def lucas_lehmer(p):
 m = (1 << p) - 1
 s = 4
 for i in range(p - 2):
 s = s * s - 2
 while s > m:
 s = (s & m) + (s >> p)
 return s == 0 or s == m and m or 0
def f(m):
 s,n=(m!=3)*4,m>>2
 while-~m&m<n:s,n=(s*s-2)%m,n>>1
 return s<1
# Make a list of some Mersenne primes
a = [3]
for p in primes(608):
 m = lucas_lehmer(p)
 if m:
 print(p, m)
 a.append(m)
print()
# Test that `f` works on all the numbers in `a`
print(all(map(f, a))) 
# Test `f` on numbers that may not be Mersenne primes
for i in range(1, 525000):
 u = f(i)
 v = i in a
 if u or v:
 print(i, u, v)
 if u != v:
 print('Error:', i, u, v)

output

3 7
5 31
7 127
13 8191
17 131071
19 524287
31 2147483647
61 2305843009213693951
89 618970019642690137449562111
107 162259276829213363391578010288127
127 170141183460469231731687303715884105727
521 6864797660130609714981900799081393217269435300143305409394463459185543183397656052122559640661454554977296311391480858037121987999716643812574028291115057151
607 531137992816767098689588206552468627329593117727031923199444138200403559860852242739162502265229285668889329486246501015346579337652707239409519978766587351943831270835393219031728127
True
3 True True
7 True True
31 True True
127 True True
8191 True True
131071 True True
524287 True True

FWIW, here's a slightly more efficient version of f that doesn't re-test m on every loop:

def f(m):
 s,n=m!=3and 4,m>>2
 if-~m&m<1:
 while n:
 s=(s*s-2)%m
 n>>=1
 return s<1

• FlipTack: You can write the while loop all on one line (no need for a newline and indent)
• PM 2Ring: @FlipTack D'oh! Thankyou! I really don't know why I missed that... And I just noticed I can shave off a couple more bytes by reverting to Python 2.

Answer 9

R, (削除) 41 (削除ここまで) 40 bytes

matlab::isprime(x<-scan())&!log2(x+1)%%1

Oddly enough the builtin in R mersenne takes n as argument, not 2^n-1.

This takes x from STDIN, checks if it is prime using the matlab package and checks if the 2-log of x+1 is a whole number by taking mod 1 and checking for 'not zero-ness'.

Also, if you use the mersenne builtin, it ends up being slightly shorter, but feels like cheating:

numbers::mersenne(log2(scan()+1))

Saved 1 byte thanks to @Billywob

• Billywob: Posted a similar answer but I deleted it now. May i suggest matlab::isprime to save one byte. Also you have to use <- for in-function assignment.
• JAD: @billywob Just noticed aswell that matlab::isprime was 1 byte shorter. (got a 1 second peak at your solution).
• Billywob: You can also use log2(x+1) instead log(x+1,2).

Answer 10

Pyke, 10 bytes

_PQb2+}1円q

Try it here!

_P - is_prime(input)
 + - ^ + V
 Qb2 - base_2(input)
 } - uniquify(^)
 1円q - ^ == "1"

Answer 11

Actually, 9 bytes

;├╔'1=@p*

Try it online!

Explanation:

Since every number of the form 2ⁿ-1 has all 1's in its binary representation, a Mersenne prime can be identified as a prime number with that quality.

;├╔'1=@p*
 ├╔'1= only unique binary digit is 1
 * and
; @p is prime

Answer 12

Jelly, 5 bytes

Alternate approach to @Dennis' existing 5-byte Jelly answer:

B;ÆPP

Try it online!

How it works:

B Returns the binary representation of the input as a list [1, 0, 1, 1, ...]
 ; And attach to this list 
 ÆP a 1 if the input is a prime, 0 otherwise
 P Calculates the product of this list of 1's and 0's

Since a Mersenne Prime is one less than a power of 2, its binary representation is excusively 1's. The output therefor is 1 for Mersenne primes, and 0 in all other cases .

Answer 13

Ceylon, 66 bytes

Boolean m(Integer c)=>c>2&&c.and(c+1)<1&&!(2:c-2).any((d)=>c%d<1);

Formatted (and commented):

// Check whether a (positive integer) number is a mersenne prime number.
//
// Question: http://codegolf.stackexchange.com/q/104508/2338
// My Answer: http://codegolf.stackexchange.com/a/104805/2338
Boolean m(Integer c) =>
 // check whether c+1 is a power of two
 c.and(c+1)<1 &&
 // the standard primality check by trial division
 !(2 : c-2).any((d) => c%d < 1) &&
 // we need to exclude 1, which is unfortunately
 // matched by both criteria above, but is no prime.
 c>1;

---

With cheating (hardcoding the results in the range of Ceylon's Integer), we can get a byte shorter (65):

Boolean h(Integer c) =>
 c.and(c+1)<1 && #20000000800a20ac.and(c+1)>0;

(It looks like the syntax highlighter misunderstands Ceylon's hex numerals as start-of-comment.)

If an anonymous function is okay, this one is 49 bytes:

[2,3,5,7,13,17,19,31,61].map((p)=>2^p-1).contains

Answer 14

Pushy, 7 bytes

oBoIpP#

Try it online!

This takes advantage of the fact that mersenne numbers have only ones in their binary representation:

oB \ Pop input, push its binary digits.
 oI \ Re-push the input
 p \ Test its primality (0/1)
 P# \ Print the product of the stack

The stack product will only be 1 if the number has no zeroes in its binary representation, and its primality is True.

Answer 15

Wolfram Language (Mathematica), 23 bytes

PrimeQ[BitAnd[#,#+2]#]&

Try it online!

1 is handled correctly because PrimeQ[BitAnd[1,1+2]*1] == PrimeQ@1 == False. Otherwise, for BitAnd[#,#+2]# to be prime, we need that # is prime and BitAnd[#,#+2] == 1, which happens when # is a Mersenne number.

• Deadcode: Nicely done! As someone who's never used Mathematica, however, your TIO code was confusing at first. Then I realized you're comparing your function against ngenisis's previous tied record holder. I think it'd be better just to show the function's output and maybe have a second link comparing it to the other solution(s).

Answer 16

Regex (ECMAScript), 29 bytes

^(?!(xx+|(x(x))+)(1円3円)+$)xxx

Try it online!

Inspired by Grimy in chat

The regex asserts that the input is greater than 3, and that it is neither of the form: (xx+)1円+ or ((xx)+)(1円x)+.

The first matches composite numbers.
The second matches a number that is 1 less than a multiple of some odd number greater than 2.

The first will not match prime numbers, or 0 or 1.
The second will not match numbers of the form \2ドル^n-1\$, or numbers that are 1 less than an odd prime.

Since 2 is the only prime that is 1 less than an odd prime, the negative lookahead, together with the assertion that the input is greater than 3, will match only mersenne primes.

Answer 17

Regex (ECMAScript), (削除) 42 (削除ここまで) 31 bytes

^(?!(xx+)1円+$)(x(x*)(?=3円$))+x$

^
(?!(xx+)1円+$) # Assert that N is prime or 0 or 1.
(x(x*)(?=3円$))+x$ # Assert that N is a power of 2 minus 1 and is >= 3.
 # The >=3 part of this prevents the match of 0 and 1.

Try it online!

Edit: Down to 31 bytes thanks to Neil.

The basic "is a power of 2 minus 1" test is ^(x(x*)(?=2円$))*$. This works by looping the operation "subtract 1, then divide evenly by 2" until it can be done no further, then asserting that the result is zero. This can be modified to match only numbers ≥1 by changing the last * to a +, forcing the loop to iterate at least once. Inserting an x before the last $ further modifies it to match only numbers ≥3 by asserting that the final result after looping at least once is 1.

The related "is a power of 2" test is ^((x+)(?=2円$))*x$. There is also a shorthand for matching powers of 2 minus 2, discovered by Grimmy: ^((x+)(?=2円$)x)*$. All three of these regexes are of the same length.

Alternative 31 byte version, by Grimmy:

^(?!(xx+)1円+$|((xx)+)(2円x)*$)xx

Try it online!

# Match Mersenne primes in the domain ^x*$
^ # N = input number
(?! # "(?!p|q)" is equivalent to "(?!p)(?!q)"; evaluate the
 # logical AND of the following negative lookaheads:
 (xx+)1円+$ # Assert that N is prime or 0 or 1
|
 ((xx)+)(2円x)*$ # Assert that N is a power of 2 minus 1; this is based
 # on "(?!(x(xx)+)1円*$)" which matches powers of 2.
)
xx # Assert that N >= 2, to prevent the unwanted match of
 # 0 and 1 by both of the negative lookahead statements.

• Neil: Save 11 bytes by checking directly for a number 1 less than a power of 2: Try it online!
• Deadcode: @Neil Thank you very much! I wish I'd thought of that, but then, this is exactly the kind of thing I wanted to happen!
• Neil: Actually thinking about it would x(x+)(?=3円$) be slightly more efficient?
• Deadcode: Yep, you're absolutely right.

Answer 18

Vyxal, 6 bytes

æ[› ̈2c

Try it Online!

If N is prime, check if n+1 is a power of 2.

Vyxal, 5 bytes

æ0bJΠ

Try it Online!

Port of the Jelly 5 byte answer.

Answer 19

Ruby, 47 bytes

->b{!("%b"%(b/2)=~/0/||(2...b).find{|a|b%a<1})}

Answer 20

Julia, 26 bytes

n->isprime(n)&&ispow2(n+1)

Answer 21

Python, 65 bytes

f=lambda n,i=3:(n^i)-all(n%i for i in range(2,n))<0 or f(n,-~i|i)

Outputs via Exit Code. Recursion Error for False. No error for True.

How it works

Since 2^n-1 in binary is made entirely from 1's, the next 2^n-1 number can be generated by number|number+1.

This function uses this by recursively going through each 2^n-1number checking to see if it's a prime number and eqaul to the input. If the number is not a mersenne prime, python will eventually throw an error as the maximum recursion depth would have been exceeded.

• Jonathan Frech: If I am not mistaken, <0 ~> 0>.

Answer 22

Pyth, 8 bytes

&.AjQ2P_

Verify all the test cases.

Pyth, 8 bytes

<.&QhQP_

Verify all the test cases.

---

How?

Code Breakdown #1

&.AjQ2P_ Full program with implicit input.
 P_ Is Prime?
 jQ2 Convert the input to binary as a list of digits.
 .A All the elements are truthy (i.e. all are 1).
& Logical AND.
 Output implicitly.

How does that work?

A number of the form 2ⁿ - 1 always contains 1 only when written in binary. Hence, we test if all its binary digits are 1 and if it is prime.

Code Breakdown #2

<.&QhQP_ Full program with implicit input.
 P_ Is Prime?
 hQ Input + 1.
 .&Q Bitwise AND between the input and ^.
< Is smaller than? I.e. The bitwise AND results in 0 and the primality test results in 1.
 Output implicitly.

How does that work?

This tests if the input + 1 is a power of two (i.e. if it is a Mersenne number), and then performs the primality test. In Python, bool is a subclass of int, so truthy is treated as 1 and falsy is treated as 0. To avoid checking explicitly that one is 0 and the other is 1, we compare their values using < (since we only have 1 such case).

Answer 23

Java 8, (削除) 53 (削除ここまで) (削除) 52 (削除ここまで) 49 bytes

n->{int i=1;for(;n%++i>0;);return(n&n+1|i^n)==0;}

Bug-fixed and golfed by 4 bytes thanks to @Nevay.

Explanation:

Try it here.

n->{ // Method with integer parameter and boolean return-type
 int i=1; // Temp integer `i`, starting at 1
 for(;n%++i>0;); // Loop and increase `i` as long as `n` is divisible by `i`
 return(n&n+1|i^n) // Then return if `n` bitwise-AND `n+1` bitwise-OR `i` bitwise-XOR `n`
 ==0; // is exactly 0
} // End of method

• Nevay: The current solution returns true for every prime >2, not only for Mersenne primes, 56 bytes: n->{for(int i=2;i<n;n&=-n%i++>>-1);return(n&n+1)<1&n>2;}
• Nevay: 52 bytes: n->{int i=1;for(;++i<n&n%i>0;);return(n&n+1|i^n)<1;}
• Kevin Cruijssen: @Nevay Thanks.. And not sure why the test cases didn't include any primes that aren't mersenne primes.. Added them myself, and you were indeed right.
• Nevay: 49 bytes: n->{int i=1;for(;n%++i>0;);return(n&n+1|i^n)==0;}

Answer 24

Python 3, 68 bytes

a=int(input());print(a&-~a<1and a>1and all(a%b for b in range(2,a)))

Try it here

Python 2, 63 bytes

a=input();print(a&-~a<1)and a>1and all(a%b for b in range(2,a))

Try it here

---

Thanks for suggestion Jonathan

---

Open to any suggestions for reducing the bytecount.

• Jonathan Frech: 1 and ~> 1and.

Answer 25

Japt, 6 bytes

×ばつ&Uj

Run it or Run all test cases

• Shaggy: Very nice :) You could replace logical © with bitwise & for consistent output, if you want.

Answer 26

Prolog (SWI), 48 bytes

Y+X:-X=<Y;X mod Y>0,Y+1+X.
\X:-X>1,2+X,X+1/\X<1.

Try it online!

-7 thanks to Jo King

Answer 27

Pari/GP, 29 bytes

A port of @lirtosiast's Mathematica answer in Pari/GP.

---

Golfed version. Try it online!

g(n)=isprime(bitand(n,n+2)*n)

Answer 28

Python, 93 Bytes

def f(a):
 for b in range(a):
 if(a+1==2**b and not[i for i in range(2,a)if a%i<1]):return 1

This code would work in both Python 2 and Python 3 so I have not specified a version.

Answer 29

Racket 76 bytes

(define(g m)(for/or((i m))(= m(-(expt 2 i)1))))(if(and(prime? n)(g n))#t #f)

Ungolfed:

(require math)
(define(f n)
 (define (ispowerminus1 m)
 (for/or ((i m))
 (= m (-(expt 2 i)1))))
 (if (and (prime? n)
 (ispowerminus1 n))
 #t #f))

Testing:

(f 1)
(f 2)
(f 20)
(f 51)
(f 63)
(f 3)
(f 31)
(f 8191)

Output:

#f
#f
#f
#f
#f
#t
#t
#t

Answer 30

PHP, 53 bytes

for($i=$n=$argv[1];--$i&&$n%$i;);echo!($i-1|$n+1&$n);

takes command line argument; prints 1 for Mersenne prime, empty string else. Run with -r.

breakdown

for($i=$n=$argv[1];--$i&&$n%$i;); // loop $i down from $n-1 until $i divides $n
 // If $n is prime, loop ends with $i=1. ($n=1 -> $i=0)
echo!($i-1|$n+1&$n); // If $i!=1, $n is not prime. If ($n+1&$n)>0, $n is not Mersenne.
 // If either $i-1 or $n+1&$n is truthy, the negation will be false.