# Is it a Mersenne Prime?

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 , 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 :)

• If I could I'd vote this as a dupe of the isprime challenge, as it doesn't really add anything new. Dec 25, 2016 at 14:32
• @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 Dec 25, 2016 at 14:39
• 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.) Dec 25, 2016 at 15:04
• @SuperJedi224 n is always prime, but knowing that changes nothing, the definition is still correct. Dec 25, 2016 at 15:13
• @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 :) Dec 28, 2016 at 16:06

# 05AB1E, 5 bytes

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

Code:

b¹pP


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.

• I wondered how long until someone used that trick :) Dec 25, 2016 at 15:14
• ¹ takes 2 bytes, so this is 6. Jan 9, 2017 at 22:19
• @KellyLowder In UTF-8, yes. However, 05AB1E uses the CP-1252 encoding rather than the UTF-8 encoding. Jan 9, 2017 at 23:19
• p*bW seems to work for 4 bytes Sep 5 at 16:17

# 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.

• Same issue as Adnan's answer. See mothereff.in/byte-counter Jan 9, 2017 at 22:20
• @KellyLowder That byte counter uses UTF-8. Both Jelly and 05AB1E use single byte character sets. Jan 9, 2017 at 22:38

# 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.

• Wow, that's amazing :) Dec 26, 2016 at 17:52

# 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.

• 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. Jan 2, 2017 at 15:47

## 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.

• So your answer is not proven to be valid? Sep 15, 2017 at 4:19
• I do not think that space is necessary. Sep 15, 2017 at 4:22
• @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). Nov 20, 2018 at 1:33
• @JeppeStigNielsen The question is if it is valid to use an unknown fact to base one's solution upon. Nov 20, 2018 at 12:44

## Mathematica, 29 26 bytes

Edit: Saved 3 bytes thanks to Martin Ender

PrimeQ@#&&IntegerQ@Log2[#+1]&

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


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

MersennePrimeExponentQ@Log2[#+1]&

• PrimeQ@#&&1>BitAnd[#,#+1]& Dec 25, 2016 at 23:57

# 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.

## Python, 83827976 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

• You can write the while loop all on one line (no need for a newline and indent) Feb 2, 2017 at 19:16
• @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. Feb 3, 2017 at 7:12

# R, 41 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

• 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. Dec 25, 2016 at 14:48
• @billywob Just noticed aswell that matlab::isprime was 1 byte shorter. (got a 1 second peak at your solution).
Dec 25, 2016 at 14:50
• You can also use log2(x+1) instead log(x+1,2). Dec 25, 2016 at 15:02

## Pyke, 10 bytes

_PQb2+}\1q


Try it here!

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


# 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


## 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 .

# 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

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


# 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.

• 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). Jan 28, 2019 at 6:55

^(?!(xx+|(x(x))+)(\1\3)+$)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+ or ((xx)+)(\1x)+. 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^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. # Regex (ECMAScript), 42 31 bytes ^(?!(xx+)\1+$)(x(x*)(?=\3$))+x$

^
(?!(xx+)\1+$) # Assert that N is prime or 0 or 1. (x(x*)(?=\3$))+x$# Assert that N is a power of 2 minus 1 and is >= 3. # The >=3 part of this prevents the match of 0 and 1.  Try it online! Edit: Down to 31 bytes thanks to Neil. The basic "is a power of 2 minus 1" test is ^(x(x*)(?=\2$))*$. 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$))*x$. There is also a shorthand for matching powers of 2 minus 2, discovered by Grimmy: ^((x+)(?=\2$)x)*$. All three of these regexes are of the same length.

Alternative 31 byte version, by Grimmy:

^(?!(xx+)\1+$|((xx)+)(\2x)*$)xx

Try it online!

# Match Mersenne primes in the domain ^x*$^ # N = input number (?! # "(?!p|q)" is equivalent to "(?!p)(?!q)"; evaluate the # logical AND of the following negative lookaheads: (xx+)\1+$       # Assert that N is prime or 0 or 1
|
((xx)+)(\2x)*$# Assert that N is a power of 2 minus 1; this is based # on "(?!(x(xx)+)\1*$)" 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.

• Save 11 bytes by checking directly for a number 1 less than a power of 2: Try it online!
– Neil
Jan 23, 2019 at 9:27
• @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! Jan 23, 2019 at 9:37
• Actually thinking about it would x(x+)(?=\3$) be slightly more efficient? – Neil Jan 23, 2019 at 11:53 • Yep, you're absolutely right. Jan 23, 2019 at 12:01 ## Ruby, 47 bytes ->b{!("%b"%(b/2)=~/0/||(2...b).find{|a|b%a<1})}  # Julia, 26 bytes n->isprime(n)&&ispow2(n+1)  # 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. • If I am not mistaken, <0 ~> 0>. Jan 23, 2019 at 10:22 # 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. # 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). # Java 8, 5352 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  • 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;} Aug 31, 2017 at 13:11 • 52 bytes: n->{int i=1;for(;++i<n&n%i>0;);return(n&n+1|i^n)<1;} Aug 31, 2017 at 13:17 • @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. Aug 31, 2017 at 13:36 • 49 bytes: n->{int i=1;for(;n%++i>0;);return(n&n+1|i^n)==0;} Aug 31, 2017 at 13:58 # 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. • 1 and ~> 1and. Jan 23, 2019 at 10:23 # Japt, 6 bytes ¢¬×&Uj  • Very nice :) You could replace logical © with bitwise & for consistent output, if you want. Jan 26, 2019 at 23:35 # 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 # 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. ## 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  # 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.


# C, 94 bytes

g(n,i){return--i?g(2*n,i):n;}n,r;f(x){for(n=r=1;++n<x;)r=x%n?x^g(2,n)-1?r:r|2:r&2;return r>2;}


Returns 1 if the number is a Mersenne Prime, 0 otherwise.

• Suggest ~x+g(2,n) instead of x^g(2,n)-1 Jan 23, 2019 at 9:56

## Scala, 59 Bytes

def f(t:BigInt)=t.isProbablePrime(t.bitLength*9)&(1+t)%2==0


This function requires the input to be a BigInt. You can easily convert a string "162259276829213363391578010288127" (2**107-1 is a Mersenne prime) into BigInt by doing BigInt("162259276829213363391578010288127"). It might go wrong as the name of isProbablePrime() method suggests. But the probability is not more than 0.5^(t.bigLength)*9.

Standalone script version is 72 bytes long.

val t=BigInt(args(0));print(t.isProbablePrime(t.bitLength*9)&(1+t)%2==0)


Assume we save it as "t.scala", then then program can be run as

>scala t.scala 162259276829213363391578010288127
>true

• You can remove the Probable from isProbablePrime if Scala has an isPrime function. Feb 5, 2019 at 14:51