Primes with prime bit-counts

Question

Task

Find all the non-negative integers up to and including a given non-zero positive integer n, that are prime and the count of 1's and 0's in their binary representation (having no leading zeroes) are prime too.

Here are the first five such primes,

17, 19, 37, 41, 79

10001, 10011, 100101, 101001, 1001111

Clarifications and rules

• Default I/O methods are accepted.
• The answer can be a program or a function.
• If there are no such primes then output garbage or nothing.
• Standard loopholes are forbidden.
• 2 3 5 7 did not make it to the list because in their binary representation number of occurrences of 0's and 1's are not prime. Consider 7whose binary representation is 111, here 0 occurs zero times and zero is not prime.

• Built-ins are allowed.

• The shortest code in bytes wins!

Test cases

10
[]

100
[17, 19, 37, 41, 79]

150
[17, 19, 37, 41, 79, 103, 107, 109, 131, 137]

/* Configuration */

var QUESTION_ID = 107050; // Obtain this from the url
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var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk";
var OVERRIDE_USER = 47650; // This should be the user ID of the challenge author.

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getAnswers();

var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;

var OVERRIDE_REG = /^Override\s*header:\s*/i;

function getAuthorName(a) {
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function process() {
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        language: match[1],
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  langs.sort(function (a, b) {
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body { text-align: left !important}

#answer-list {
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#language-list {
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table thead {
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table td {
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<table style="display: none">
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    <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr>
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Answer 1

Jelly, 14 13 bytes

RBċþd`ÆPPTfÆR

Try it online!

How it works

RBċþd`ÆPPTfÆR  Main link. Argument: n

R              Range; yield [1, ..., n].
 B             Binary; convert each integer to base 2.
    d`         Divmod self; yield [n : n, n % n], i.e., [1, 0].
  ċþ           Count table; count the occurrences of 1 and 0 in each binary
               representation, yielding a pair of lists of length n.
      ÆP       Test each count for primality.
        P      Product; multiply the corresponding Booleans.
         T     Truth; get all indices of 1, yielding the list of integers in
               [1, ..., n] that have a prime amount of 0's and 1's.
           ÆR  Prime range; yield all primes in [1, ..., n].
          f    Filter; intersect the lists.

Answer 2

Python 2, 106 102 100 bytes

k=m=1;p={0}
exec"p^={m%k*k,0};c=bin(k).count\nif{k,c('1'),c('0')-1}<p:print k\nm*=k*k;k+=1;"*input()

Try it online!

Background

To identify primes, we use a corollary of Wilson's theorem:

How it works

We start by initializing k and m as 1 and p as the set {0}. Note that m = 1 = 0!² = (k - 1)!². Immediately afterwards, the following code is executed n times, where n is the integer read from standard input.

p^={m%k*k,0};c=bin(k).count
if{k,c('1'),c('0')-1}<p:print k
m*=k*k;k+=1

By the corollary, m % k will be 1 if k is prime and 0 otherwise. Thus, {m%k*k,0} will return the set {k, 0} if k is prime and the set {0} otherwise.

If (and only if) k is prime, since p cannot contain k at this point, the in-place symmetric difference p^={m%k*k,0} will add k to the set p. Also, p will contain 0 after the update if and only if it does not contain 0 before, so 0 ∊ p if and only if k is even.

On the same line, we define a function c via c=bin(k).count, which will count the occurrences of its argument in k's binary representation.

The second line produces the actual output. {k,c('1'),c('0')-1} returns the set that consists of k itself, the number of set bits in k, and the number of unset bits in k. Since the output of bin(k) begins with 0b, we have to decrement c('0') to account for the leading 0.

If all of them are prime, they will all belong to p, which by now contains all prime numbers up to k (and potentially 0). If k is a Mersenne number (i.e., if it has only set bits), c('0')-1 will yield 0. Since Mersenne numbers are odd, p will not contain 0, so the condition will fail.

After (potentially) printing k, we multiply m by k². Since m = (k-1)!² before the update, m = k!² after it. After incrementing k, the relation m = (k-1)!² holds again and we're ready for the next iteration.

Answer 3

Mathematica, 80 68 54 bytes

Select[p=PrimeQ;Range@#,p@#&&And@@p/@#~DigitCount~2&]&

Explanation

Select[p=PrimeQ;Range@#,p@#&&And@@p/@#~DigitCount~2&]&

       p=PrimeQ                                        (* Store prime check func. in p *)
Select[                                             ]  (* Select *)
                Range@#                                (* from a list {1..n} *)
                        p@#                            (* numbers that are prime *)
                           &&                          (* And *)
                                     #~DigitCount~2    (* whose digit counts in binary *)
                             And@@p/@                  (* are prime as well *)

Answer 4

JavaScript (ES6), 123 118 115 111 104 96 bytes

Saved 4 bytes thanks to @Arnauld

G=n=>n?G(n>>1,++a[n%2]):a.some(n=>(P=x=>n%--x?P(x):x)(n)-1)
F=n=>F(n-1,G(n,a=[0,0,n])||alert(n))

A combination of three typical recursive functions. Alerts the sequence in reverse order and terminates on a "too much recursion" error.

Test snippet

(modified to output to the page)

alert = s => O.textContent += s + "\n"

G=n=>n?G(n>>1,++a[n%2]):a.some(n=>(P=x=>n%--x?P(x):x)(n)-1)
F=n=>F(n-1,G(n,a=[0,0,n])||alert(n))

F(1000)

<pre id=O></pre>

The main function can return an array for 104 bytes:

G=n=>n?G(n>>1,++a[n%2]):a.some(n=>(P=x=>n%--x?P(x):x)(n)-1)
F=n=>n?F(n-1).concat(G(n,a=[0,0,n])?[]:n):[]

It can also be non-recursive at the cost of another byte:

G=n=>n?G(n>>1,++a[n%2]):a.some(n=>(P=x=>n%--x?P(x):x)(n)-1)
n=>[for(_ of Array(n))if(!G(--n,a=[0,0,n]))n]

---

Here's the one I started with: (Saved 6 bytes thanks to @Arnauld)

P=(n,x=n)=>n>1&--x<2||n%x&&P(n,x)
G=n=>n?G(n>>1,o+=n%2,t++):P(o)&P(t-o)
F=n=>n?F(n-1).concat(P(n)&G(n,o=t=0)?n:[]):[]

I tried golfing this further and managed to do it in 104 bytes—then realized I had already found that solution (it's at the bottom of the answer). Don't you hate it when that happens? :P

A non-recursive attempt at the main function (again, same byte count):

n=>[for(i of Array(n))if(P(--n)&G(n,o=t=0))n]

---

This one takes the easy route of counting how many 0's and 1's are in the binary representation:

F=n=>n?F(n-1).concat([n,(G=x=>n.toString(2).split(x).length-1)(0),G(1)].some(n=>(P=x=>n%--x?P(x):x)(n)-1)?[]:n):[]

Same thing with an array comprehension:

n=>[for(_ of Array(n))if(![--n,(G=x=>n.toString(2).split(x).length-1)(0),G(1)].some(n=>(P=x=>n%--x?P(x):x)(n)-1))n]

This one takes a slightly harder route to do the same thing:

F=n=>n?F(n-1).concat([n,(G=(x,w=n)=>w&&G(x,w>>1)+(w%2==x))(0),G(1)].some(n=>(P=x=>n%--x?P(x):x)(n)-1)?[]:n):[]

And this one takes yet another related route that's as short as the original:

F=n=>n?F(n-1).concat([n,o=(G=x=>x&&x%2+G(n>>++t))(n,t=0),t-o].some(n=>(P=x=>n%--x?P(x):x)(n)-1)?[]:n):[]

Again, you can golf 8 bytes by making it alert the sequence in reverse order:

F=n=>F(n-1,[n,o=(G=x=>x&&x%2+G(n>>++t))(n,t=0),t-o].some(n=>(P=x=>n%--x?P(x):x)(n)-1)||alert(n))

Answer 5

Jelly, 17 16 bytes

BĠL€µ;LÆPẠ
ÆRÇÐf

Try it online!

How?

BĠL€µ;LÆPẠ - Link 1, hasPrimeBits: n  e.g. 7         or 13
B          - convert to binary        e.g. [1,1,1]  or [1,1,0,1]
 Ġ         - group indices            e.g. [1,2,3]  or [3,[1,2,4]]
  L€       - length of €ach           e.g. 3          or [1,3]
    µ      - monadic chain separation
     ;L    - concatenate length       e.g. [3,1]      or [1,3,2]
           -     this caters for the edge case of Mersenne primes (like 7), which have
           -     no zero bits, the length will be 1, which is not prime.
       ÆP  - isPrime?                 e.g. [1,0]      or [0,1,1]
         Ạ - All?                     e.g. 0          or 0

ÆRÇÐf - Main link: N
ÆR    - prime range (primes up to and including N)
   Ðf - filter keep those satisfying
  Ç   - last link (1) as a monad

Answer 6

05AB1E, 14 bytes

ƒNpNbSD_‚OpP&–

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Explanation

ƒ               # for N in [0 ... input]
 Np             # push N is prime
   Nb           # push N converted to binary
     S          # split into individual digits
      D_        # push an inverted copy
        ‚       # pair the 2 binary lists
         O      # sum them
          p     # elementwise check for primality
           P    # product of list
            &   # and with the primality of N
             –  # if true, print N

Answer 7

05AB1E, 17 15 bytes

LDpÏvybSD_)OpP—

Uses the CP-1252 encoding. Try it online!

Answer 8

Perl, 101 bytes

99 bytes of code + -nl flags.

$r=q/^1?$|^(11+)\1+$/;for$@(2..$_){$_=sprintf"%b",$@;print$@if(1x$@)!~$r&y/01/1/dr!~$r&s/0//gr!~$r}

To run it:

perl -lne '$r=q/^1?$|^(11+)\1+$/;for$@(2..$_){$_=sprintf"%b",$@;print$@if(1x$@)!~$r&y/01/1/dr!~$r&s/0//gr!~$r}' <<< 150

Some short explanations
$r holds the classical prime checking regex (q/^1?$|^(11+)\1+$/).
For all numbers between 2 and the input,
(1x$@)!~$r checks if the number is prime,
y/01/1/dr!~$r checks if the number of 0 in the binary representation is prime,
s/0//gr!~$r checks if the number 1 in the binary representation is prime.
(if the 3 conditions are met, print$@ prints it).

Answer 9

MATL, 16 15 bytes

:"@tB!t~hshZp?@

Try it online!

:         % Input n (implicit). Push range [1 2 ... n]
"         % For each k in [1 2 ... n]
  @       %   Push k
  tB!     %   Duplicate. Binary expansion as an m×1 vector
  t~      %   Duplicate. Negate
  hs      %   Concatenate horizontally into an m×2 matrix. Sum of each column.
          %   This gives a 1×2 vector. However, for k==1 this gives the sum of
          %   the original 1×2 matrix (m==1). But fortunately it doesn't matter
          %   because 1 is not a prime anyway
  h       %   Concatenate horizontally into a 1×3 row vector
  Zp      %   Isprime function applied on each of those three numbers
  ?       %   If all gave true
    @     %     Push k
          %   End (implicit)
          % End (implicit)
          % Display (implicit)

Answer 10

Python, 129 125 123 bytes

If only zero were prime...

p=lambda n:all(n%m for m in range(2,n))*n>1
lambda n:[i for i in range(n+1)if p(i)*all(p(bin(i)[2:].count(x))for x in'01')]

Try it online

Function p is the prime-checking function, which has >0 at the end so that it works for zero as well, which would otherwise return -1. The anonymous lambda is the lambda that checks all the conditions required.

---

Here's a slightly different method using a set comprehension. The result will be a set. (124 bytes):

p=lambda n:all(n%m for m in range(2,n))*n>1
lambda n:{i*p(i)*all(p(bin(i)[2:].count(x))for x in'01')for i in range(n+1)}-{0}

Answer 11

Perl 6, 65 bytes

{grep {all($^a,|map {+$a.base(2).comb(~$_)},0,1).is-prime},0..$_}

Try it

Expanded:

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

  grep      # find all of the values where

  {         # lambda with placeholder parameter ｢$a｣

    all(    # all junction ( treat multiple values as a single value )

      $^a,  # declare parameter to block

      |\    # slip the following list into the outer list

      # get the count of 0's and 1's in the binary representation
      map

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

        +           # turn to numeric ( count the values in the list )
        $a          # the outer parameter
        .base(2)    # in base 2
        .comb(~$_)  # find the characters that are the same as
      },0,1         # 0 or 1

    # ask if $a, count of 0's, count of 1's are all prime
    ).is-prime

  },

  0 .. $_ # Range from 0 to the input inclusive

}

Answer 12

Octave, 73 bytes

@(n)(p=primes(n))(all(isprime([s=sum(dec2bin(p)'-48);fix(log2(p))+1-s])))

Try It Online!

@(n)
    (p=primes(n))                           generate prime numbers
        (all(                               from them select those that
            isprime(                        are prime both
                [s=sum(dec2bin(p)'-48);     sum of 1s
                fix(log2(p))+1-s])))        and sum of 0s

Answer 13

CJam, 26 27 bytes

ri){mp},{2b_1-,mp\0-,mp&},p

Straightforward algorithm, will golf further.

EDIT: Forgot n was inclusive.

Try it online!

For fun, this is very similar and also 27 bytes:

ri){_mp\2b_1-,mp\0-,mp&&},p

Explanation

ri                          e# Take an int as input
  ){mp},                    e# Take the range up and including to the input, 
                            e#   including only prime numbers
       {                    e# For each prime...
        2b_                 e# Convert it to binary and copy it
           1-,mp            e# Take the top binary number, remove 1's, check if the length 
                            e#   is prime
                \           e# Swap top elements
                 0-,mp      e# Do the same but remove 0's
                      &     e# And
                       },   e# Filter out primes for which the above block was false
                         p  e# Print nicely

Answer 14

Python, 172 170 168 159 154 133 130 bytes

p=lambda n:[i for i in range(1,n)if n%i==0]==[1]
lambda n:[i for i in range(n+1)if p(i)*all(p(bin(i)[2:].count(x))for x in'10')]

---

Usage: Call the anonymous lambda function
Method p checks whether a number is prime

---

Saved 2+2+9=13 bytes thanks to Gurupad Mamadapur
Saved 5 bytes thanks to mbomb007

Answer 15

Jelly, 14 bytes

BċÆPðÐf
ÆRç0ç1

Try it online!

Unfortunately, I could only tie with @Dennis, but the algorithm seems to be somewhat different so I'm posting this anyway.

Explanation

Helper function (deletes list elements that don't have a prime number of occurrences of a given bit):

BċÆPðÐf
     Ðf   Filter {the first argument}, looking for truthy returns from
    ð     the preceding code, which takes two arguments:
B         Convert {the element being checked} to binary
 ċ        Count the number of occurrences of {the second argument}
  ÆP      Return true if prime, false if not prime

Main program:

ÆRç0ç1
ÆR        Generate all primes from 2 to the input
  ç0      Call the subroutine to require a prime 0 count
    ç1    Call the subroutine to require a prime 1 count

Answer 16

Bash + GNU utilities, 129 126 123 114 111 109 bytes

seq '-fp()([ `factor|wc -w` = 2 ]);g()(dc -e2o${n}n|tr -cd $1|wc -c|p);n=%.f;p<<<$n&&g 0&&g 1&&echo $n' $1|sh

Try it online!

Noticed the comment that the output doesn't have to be in increasing order -- shaved off 3 bytes by counting down instead of counting up.

Replaced grep with wc to save 3 additional bytes.

Eliminated a variable (9 more bytes off).

Changed the way factor is used -- 3 more bytes.

Made it golfier with seq (2 bytes).

Answer 17

Haxe, 169 171 bytes

function f(n,?p)return[for(j in(p=[for(i in 0...++n)i<2?0:i]))if(j>0){var x=j*2,y=0,z=0,k=j;while(k<n)p[k+=j]=0;while((x>>=1)>0)x&1>0?y++:z++;p[y]<1||p[z]<1?continue:j;}];

Nothing crazy. Essentially a modified sieve of Eratosthenes that evaluates the primality of the 0 / 1 count in the same iteration as crossing out higher non-primes. With some whitespace:

function f(n, ?p)
  return [ for (j in (p = [ for(i in 0...++n) i < 2 ? 0 : i ]))
    if (j > 0){
      var x = j * 2, y = 0, z = 0, k = j;
      while (k < n)
        p[k += j] = 0;
      while((x >>= 1) > 0)
        x & 1 > 0 ? y++ : z++;
      p[y] < 1 || p[z] < 1 ? continue : j;
    }
  ];

But today I learned I could put continue in a ternary operator and Haxe doesn't even mind.

Edit: +2 because n is an inclusive upper bound!

Answer 18

Pyke, 21 bytes

S#j_PI\0[jb2R/_P)I\1[

Try it here!

S#j                   - for j in filter(range(1, input+1))
   _PI                -  if is_prime(j):
        [jb2R/_P)     -   function_[(a) = V
         jb2          -      base_2(j)
            R/        -     ^.count(a)
      \0         I    -   if function_[("0"):
                  \1[ -    function_[("1")

Answer 19

Pyth, 18 bytes

f.APM_M+T/LSjT2U2S

Try it online here.

Answer 20

Groovy, 120 bytes

p={x->(2..<x).every{x%it!=0}&x>1|x==2}
{x->(1..x).findAll({y->p(y)&'10'.every{p(Integer.toBinaryString(y).count(it))}})}

This is an unnamed closure.

Try it here!

Answer 21

JavaScript (ES6), 110 bytes

B=n=>n?B(n>>1,v[n%2]++):v.every(n=>(P=i=>n%--i?P(i):1==i)(n))
F=(n,a=[])=>n?F(n-1,B(n,v=[0,0,n])?[n,...a]:a):a

B=n=>n?B(n>>1,v,v[n%2]++):v.every(n=>(P=i=>n%--i?P(i):1==i)(n))
F=(n,a=[])=>n?F(n-1,B(n,v=[0,0,n])?[n,...a]:a):a

const update = () => {
  console.clear();
  console.log(F(input.value))
}

input.oninput = update;
update();

<input id="input" type="number" min="1" value="100" />

Answer 22

Python 3, 97 bytes

def f(n,k=2,m=1,p={0}):f(n-1,k+1,m*k*k,p^{m%k*k,{*map(bin(m%k%n*k)[2:].count,'01')}<p==print(k)})

This is the same algorithm as in my Python 2 answer, but the implementation is quite different. The function prints to STDOUT and terminates with an error.

Try it online!

Answer 23

MATLAB, 50 bytes

@(n)all(isprime([n,sum(de2bi(n)),sum(~de2bi(n))]))

Answer 24

Haxe, 158 147 bytes

function p(n,x=1)return n%++x<1||n<2?x==n:p(n,x);function f(n){var q=n,t=0,o=0;while(q>0){t++;o+=q%2;q>>=1;}p(n)&&p(o)&&p(t-o)?trace(n):0;f(n-1);};

Outputs to STDOUT and terminates on a "too much recursion" error; call with e.g. f(1000);. You can use a while loop with no error for 156 bytes:

function p(n,x=1)return n%++x<1||n<2?x==n:p(n,x);function f(n){while(n>0){var q=n,t=0,o=0;while(q>0){t++;o+=q%2;q>>=1;}p(n)&&p(o)&&p(t-o)?trace(n):0;n--;}};

Using a range turned out three bytes longer:

function p(n,x=1)return n%++x<1||n<2?x==n:p(n,x);function f(n){for(i in 0...n+1){var q=i,t=0,o=0;while(q>0){t++;o+=q%2;q>>=1;}p(i)&&p(o)&&p(t-o)?trace(i):0;}};

Test it online!

Answer 25

Rust, 147 bytes

fn f(x:u32){let p=|n|n>1&&(2..n).all(|x|n%x!=0);for n in 9..x+1{if p(n)&&p(n.count_ones())&&p(n.count_zeros()-n.leading_zeros()){print!("{} ",n)}}}

playground link

The count_ones, count_zeros, and leading_zeros methods came in handy.

Formatted version

fn f(x: u32) {
    // primality test closure using trial division
    let p = |n| n > 1 && (2..n).all(|x| n % x != 0);

    for n in 9..x + 1 {
        if p(n) && p(n.count_ones()) && p(n.count_zeros() - n.leading_zeros()) {
            print!("{} ", n)
        }
    }
}

Answer 26

Perl 6, 50 bytes

{grep {($_&+.base(2).comb(~0&~1)).is-prime},0..$_}

Variation of b2gills' answer, using the & junction operator to great effect.

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How it works

{                                                }  # Lambda
                                            0..$_   # Range from 0 to the lambda argument.
 grep {                                   },        # Filter values satisfying:
       ($_                      ).is-prime          #  a) The Number itself is prime.
       (   +.base(2).comb(~0   )).is-prime          #  b) The count of 0's is prime.
       (   +.base(2).comb(   ~1)).is-prime          #  c) The count of 1's is prime.
          &                 &                       # Junction magic! :)