Число - это простое число Мерсенна, если оно простое и может быть записано в виде 2 ⁿ -1 , где n - положительное целое число.

Ваша задача состоит в том, чтобы с учетом любого положительного целого числа определить, является ли оно простым числом Мерсенна. Вы можете отправить либо функцию, которая возвращает истинное / ложное значение, либо полную программу, которая выполняет ввод-вывод.

Правила:

• Так как это код-гольф , вы должны стремиться сделать это как можно быстрее. Встроенные разрешены.
• Применяются стандартные лазейки для игры в гольф - вы не можете прочитать простые числа Мерсенна из внешних файлов или жестко закодировать их в своей программе.
• Ваша программа должна работать для значений в пределах стандартного целочисленного размера вашего языка.

Тестовые случаи

Для справки, список (известных) простых чисел Мерсенна можно найти здесь . Некоторые удобные тестовые примеры:

2  -> False
1  -> False 
20 -> False
51 -> False
63 -> False

3    -> True
31   -> True
8191 -> True

_{Всех с Рождеством! Хорошего вам праздника, что бы вы ни отмечали :)}

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.

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.

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.

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.

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.

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]&

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, 83 82 79 76 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

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

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
// 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

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.

ECMAScript regex, 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 Grimy: ^((x+)(?=\2$)x)*$. All three of these regexes are of the same length.

Alternative 31 byte version, by Grimy:

^(?!(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.

Regex (ECMAScript), 29 bytes

^(?!(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.

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.

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, 53 52 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

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.

Japt, 6 bytes

¢¬×&Uj

Run it or Run all test cases

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.

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

Perl 5, 53 bytes

52 bytes of code + 1 for -p

$f=0|sqrt;1while$_%$f--;$_=!$f*(sprintf'%b',$_)!~/0/

Try it online!