Square-free integers: Difference between revisions

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Content added Content deleted
m (Reverted edits by Dijkstra (talk) to last revision by Thundergnat)
(→‎{{header|Perl 6}}: Some re-factoring, slightly better performance, DRY, style twiddles. Some commentary on the algorithm.)
Line 569: Line 569:
=={{header|Perl 6}}==
=={{header|Perl 6}}==
{{works with|Rakudo|2018.06}}
{{works with|Rakudo|2018.06}}
The prime factoring algorithm is not really the best option for finding long runs of sequential square-free numbers. It works, but is probably better suited for testing arbitrary numbers rather than testing every sequential number from 1 to some limit. If you know that that is going to be your use case, there are faster algorithms than this.


<lang perl6># Prime factorization routines
<lang perl6># Prime factorization routines
Line 575: Line 576:
return [] if $n == 1;
return [] if $n == 1;
my $factor = find-factor( $n );
my $factor = find-factor( $n );
sort flat prime-factors( $factor ), prime-factors( $n div $factor );
flat prime-factors( $factor ), prime-factors( $n div $factor );
}
}


Line 595: Line 596:
}
}


# Task routines
# Task routine
multi is-square-free (Int $n) { $n.&prime-factors.Bag.max(*.value).value > 1 ?? False !! True }
multi is-square-free (Int $n) { my @v = $n.&prime-factors.Bag.values; not @v.sum/@v > 1 }
multi is-square-free (1) { True }


# The Task
# The Task
# Parts 1 & 2

for 1, 145, 1e12.Int, 145+1e12.Int -> $start, $end {
# Part 1
say "Square─free numbers between 1 and 145:";
say "\nSquare─free numbers between $start and $end:\n",
say (1..145).map({ $_ if .&is-square-free }).list.fmt("%3d").comb(100).join("\n");
($start .. $end).hyper(:4batch).grep( *.&is-square-free ).list.fmt("%3d").comb(84).join("\n");
}

# Part 2
say "\nSquare─free numbers between 1000000000000 and 1000000000145:";
say (1000000000000..1000000000145).hyper(:4batch).map({ $_ if .&is-square-free }).list.comb(98).join("\n");


# Part 3
# Part 3
for 1e2, 1e3, 1e4, 1e5, 1e6 {
for 1e2, 1e3, 1e4, 1e5, 1e6 {
say "\nThe number of square─free numbers between 1 and {$_} (inclusive) is: ",
say "\nThe number of square─free numbers between 1 and {$_} (inclusive) is: ",
+(1 .. .Int).race.map: { 1 if .&is-square-free };
+(1 .. .Int).race.grep: *.&is-square-free;
}</lang>
}</lang>
{{out}}
{{out}}
<pre>
<pre>Square─free numbers between 1 and 145:
Square─free numbers between 1 and 145:
1 2 3 5 6 7 10 11 13 14 15 17 19 21 22 23 26 29 30 31 33 34 35 37 38
39 41 42 43 46 47 51 53 55 57 58 59 61 62 65 66 67 69 70 71 73 74 77 78 79
1 2 3 5 6 7 10 11 13 14 15 17 19 21 22 23 26 29 30 31 33
82 83 85 86 87 89 91 93 94 95 97 101 102 103 105 106 107 109 110 111 113 114 115 118 119
34 35 37 38 39 41 42 43 46 47 51 53 55 57 58 59 61 62 65 66 67
69 70 71 73 74 77 78 79 82 83 85 86 87 89 91 93 94 95 97 101 102
122 123 127 129 130 131 133 134 137 138 139 141 142 143 145
103 105 106 107 109 110 111 113 114 115 118 119 122 123 127 129 130 131 133 134 137
138 139 141 142 143 145


Square─free numbers between 1000000000000 and 1000000000145:
Square─free numbers between 1000000000000 and 1000000000145:
1000000000001 1000000000002 1000000000003 1000000000005 1000000000006 1000000000007 1000000000009
1000000000001 1000000000002 1000000000003 1000000000005 1000000000006 1000000000007
1000000000011 1000000000013 1000000000014 1000000000015 1000000000018 1000000000019 1000000000021
1000000000009 1000000000011 1000000000013 1000000000014 1000000000015 1000000000018
1000000000022 1000000000023 1000000000027 1000000000029 1000000000030 1000000000031 1000000000033
1000000000019 1000000000021 1000000000022 1000000000023 1000000000027 1000000000029
1000000000037 1000000000038 1000000000039 1000000000041 1000000000042 1000000000043 1000000000045
1000000000030 1000000000031 1000000000033 1000000000037 1000000000038 1000000000039
1000000000041 1000000000042 1000000000043 1000000000045 1000000000046 1000000000047
1000000000046 1000000000047 1000000000049 1000000000051 1000000000054 1000000000055 1000000000057
1000000000049 1000000000051 1000000000054 1000000000055 1000000000057 1000000000058
1000000000058 1000000000059 1000000000061 1000000000063 1000000000065 1000000000066 1000000000067
1000000000059 1000000000061 1000000000063 1000000000065 1000000000066 1000000000067
1000000000069 1000000000070 1000000000073 1000000000074 1000000000077 1000000000078 1000000000079
1000000000069 1000000000070 1000000000073 1000000000074 1000000000077 1000000000078
1000000000081 1000000000082 1000000000085 1000000000086 1000000000087 1000000000090 1000000000091
1000000000079 1000000000081 1000000000082 1000000000085 1000000000086 1000000000087
1000000000093 1000000000094 1000000000095 1000000000097 1000000000099 1000000000101 1000000000102
1000000000090 1000000000091 1000000000093 1000000000094 1000000000095 1000000000097
⚫
1000000000103 1000000000105 1000000000106 1000000000109 1000000000111 1000000000113 1000000000114
1000000000099 1000000000101 1000000000102 1000000000103 1000000000105 1000000000106
⚫
1000000000115 1000000000117 1000000000118 1000000000119 1000000000121 1000000000122 1000000000123
⚫
1000000000109 1000000000111 1000000000113 1000000000114 1000000000115 1000000000117
⚫
1000000000126 1000000000127 1000000000129 1000000000130 1000000000133 1000000000135 1000000000137
⚫
1000000000118 1000000000119 1000000000121 1000000000122 1000000000123 1000000000126
⚫
1000000000127 1000000000129 1000000000130 1000000000133 1000000000135 1000000000137
1000000000138 1000000000139 1000000000141 1000000000142 1000000000145
1000000000138 1000000000139 1000000000141 1000000000142 1000000000145


Revision as of 21:25, 5 July 2018

Square-free integers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Write a function to test if a number is   square-free.


A   square-free   is an integer which is divisible by no perfect square other than   1   (unity).

For this task, only positive square-free numbers will be used.


Show here (on this page) all square-free integers (in a horizontal format) that are between:

  •   1   ───►   145     (inclusive)
  •   1 trillion   ───►   1 trillion + 145     (inclusive)


(One trillion = 1,000,000,000,000)


Show here (on this page) the count of square-free integers from:

  •   1   ───►   one hundred     (inclusive)
  •   1   ───►   one thousand     (inclusive)
  •   1   ───►   ten thousand     (inclusive)
  •   1   ───►   one hundred thousand     (inclusive)
  •   1   ───►   one million     (inclusive)


See also



C

Translation of: Go

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  2. include <math.h>
  1. define TRUE 1
  2. define FALSE 0
  3. define TRILLION 1000000000000

typedef unsigned char bool; typedef unsigned long long uint64;

void sieve(uint64 limit, uint64 *primes, uint64 *length) { 

   uint64 i, count, p, p2;
   bool *c = calloc(limit + 1, sizeof(bool));  /* composite = TRUE */
   primes[0] = 2;
   count  = 1;
   /* no need to process even numbers > 2 */
   p = 3;
   for (;;) {
       p2 = p * p;
       if (p2 > limit) break;
       for (i = p2; i <= limit; i += 2 * p) c[i] = TRUE;
       for (;;) {
           p += 2;
           if (!c[p]) break;
       }
   }
   for (i = 3; i <= limit; i += 2) {
       if (!c[i]) primes[count++] = i;
   }
   *length = count;
   free(c);

}

void squareFree(uint64 from, uint64 to, uint64 *results, uint64 *len) { 

   uint64 i, j, p, p2, np, count = 0, limit = (uint64)sqrt((double)to);
   uint64 *primes = malloc((limit + 1) * sizeof(uint64));
   bool add;
   sieve(limit, primes, &np);
   for (i = from; i <= to; ++i) {
       add = TRUE;
       for (j = 0; j < np; ++j) {
           p = primes[j];
           p2 = p * p;
           if (p2 > i) break;
           if (i % p2 == 0) {
               add = FALSE;
               break;
           }
       }
       if (add) results[count++] = i;
   }
   *len = count;
   free(primes);

}

int main() { 

   uint64 i, *sf, len, from = 1, to;
   to = 145;
   sf = malloc((to - from + 1) * sizeof(uint64));
   printf("Square-free integers from 1 to 145:\n");
   squareFree(1, 145, sf, &len);
   for (i = 0; i < len; ++i) {
       if (i > 0 && i % 20 == 0) {
           printf("\n");
       }
       printf("%4lld", sf[i]);
   }
   printf("\n\nSquare-free integers from %ld to %ld:\n", TRILLION, TRILLION + 145);
   squareFree(TRILLION, TRILLION + 145, sf, &len);
   for (i = 0; i < len; ++i) {
       if (i > 0 && i % 5 == 0) {
           printf("\n");
       }
       printf("%14lld", sf[i]);
   }

   printf("\n\nNumber of square-free integers:\n");

   squareFree(1, 100, sf, &len);
   printf("  from %d to %d = %lld\n", 1, 100, len);

   /* need to allocate a lot more memory to process remaining cases */
   sf = realloc(sf, 1000000 * sizeof(uint64));
   squareFree(1, 1000, sf, &len);
   printf("  from %d to %d = %lld\n", 1, 1000, len);
   squareFree(1, 10000, sf, &len);
   printf("  from %d to %d = %lld\n", 1, 10000, len);
   squareFree(1, 100000, sf, &len);
   printf("  from %d to %d = %lld\n", 1, 100000, len);
   squareFree(1, 1000000, sf, &len);
   printf("  from %d to %d = %lld\n", 1, 1000000, len);
   free(sf);
   return 0;

}</lang>

Output:
Square-free integers from 1 to 145:
   1   2   3   5   6   7  10  11  13  14  15  17  19  21  22  23  26  29  30  31
  33  34  35  37  38  39  41  42  43  46  47  51  53  55  57  58  59  61  62  65
  66  67  69  70  71  73  74  77  78  79  82  83  85  86  87  89  91  93  94  95
  97 101 102 103 105 106 107 109 110 111 113 114 115 118 119 122 123 127 129 130
 131 133 134 137 138 139 141 142 143 145

Square-free integers from 1000000000000 to 1000000000145:
 1000000000001 1000000000002 1000000000003 1000000000005 1000000000006
 1000000000007 1000000000009 1000000000011 1000000000013 1000000000014
 1000000000015 1000000000018 1000000000019 1000000000021 1000000000022
 1000000000023 1000000000027 1000000000029 1000000000030 1000000000031
 1000000000033 1000000000037 1000000000038 1000000000039 1000000000041
 1000000000042 1000000000043 1000000000045 1000000000046 1000000000047
 1000000000049 1000000000051 1000000000054 1000000000055 1000000000057
 1000000000058 1000000000059 1000000000061 1000000000063 1000000000065
 1000000000066 1000000000067 1000000000069 1000000000070 1000000000073
 1000000000074 1000000000077 1000000000078 1000000000079 1000000000081
 1000000000082 1000000000085 1000000000086 1000000000087 1000000000090
 1000000000091 1000000000093 1000000000094 1000000000095 1000000000097
 1000000000099 1000000000101 1000000000102 1000000000103 1000000000105
 1000000000106 1000000000109 1000000000111 1000000000113 1000000000114
 1000000000115 1000000000117 1000000000118 1000000000119 1000000000121
 1000000000122 1000000000123 1000000000126 1000000000127 1000000000129
 1000000000130 1000000000133 1000000000135 1000000000137 1000000000138
 1000000000139 1000000000141 1000000000142 1000000000145

Number of square-free integers:
  from 1 to 100 = 61
  from 1 to 1000 = 608
  from 1 to 10000 = 6083
  from 1 to 100000 = 60794
  from 1 to 1000000 = 607926

Factor

The sq-free? word merits some explanation. Per the Wikipedia entry on square-free integers, A positive integer n is square-free if and only if in the prime factorization of n, no prime factor occurs with an exponent larger than one.

For instance, the prime factorization of 12 is 2 * 2 * 3, or in other words, 22 * 3. The 2 repeats, so we know 12 isn't square-free. <lang factor>USING: formatting grouping io kernel math math.functions math.primes.factors math.ranges sequences sets ; IN: rosetta-code.square-free

sq-free? ( n -- ? ) factors all-unique? ;

! Word wrap for numbers.

numbers-per-line ( m -- n ) log10 >integer 2 + 80 swap /i ;
sq-free-show ( from to -- )
   2dup "Square-free integers from %d to %d:\n" printf
   [ [a,b] [ sq-free? ] filter ] [ numbers-per-line group ] bi
   [ [ "%3d " printf ] each nl ] each nl ;
sq-free-count ( limit -- )
   dup [1,b] [ sq-free? ] count swap
   "%6d square-free integers from 1 to %d\n" printf ;

1 145 10 12 ^ dup 145 + [ sq-free-show ] 2bi@  ! part 1 2 6 [a,b] [ 10 swap ^ ] map [ sq-free-count ] each  ! part 2</lang>

Output:
Square-free integers from 1 to 145:
  1   2   3   5   6   7  10  11  13  14  15  17  19  21  22  23  26  29  30  31 
 33  34  35  37  38  39  41  42  43  46  47  51  53  55  57  58  59  61  62  65 
 66  67  69  70  71  73  74  77  78  79  82  83  85  86  87  89  91  93  94  95 
 97 101 102 103 105 106 107 109 110 111 113 114 115 118 119 122 123 127 129 130 
131 133 134 137 138 139 141 142 143 145 

Square-free integers from 1000000000000 to 1000000000145:
1000000000001 1000000000002 1000000000003 1000000000005 1000000000006 
1000000000007 1000000000009 1000000000011 1000000000013 1000000000014 
1000000000015 1000000000018 1000000000019 1000000000021 1000000000022 
1000000000023 1000000000027 1000000000029 1000000000030 1000000000031 
1000000000033 1000000000037 1000000000038 1000000000039 1000000000041 
1000000000042 1000000000043 1000000000045 1000000000046 1000000000047 
1000000000049 1000000000051 1000000000054 1000000000055 1000000000057 
1000000000058 1000000000059 1000000000061 1000000000063 1000000000065 
1000000000066 1000000000067 1000000000069 1000000000070 1000000000073 
1000000000074 1000000000077 1000000000078 1000000000079 1000000000081 
1000000000082 1000000000085 1000000000086 1000000000087 1000000000090 
1000000000091 1000000000093 1000000000094 1000000000095 1000000000097 
1000000000099 1000000000101 1000000000102 1000000000103 1000000000105 
1000000000106 1000000000109 1000000000111 1000000000113 1000000000114 
1000000000115 1000000000117 1000000000118 1000000000119 1000000000121 
1000000000122 1000000000123 1000000000126 1000000000127 1000000000129 
1000000000130 1000000000133 1000000000135 1000000000137 1000000000138 
1000000000139 1000000000141 1000000000142 1000000000145 

    61 square-free integers from 1 to 100
   608 square-free integers from 1 to 1000
  6083 square-free integers from 1 to 10000
 60794 square-free integers from 1 to 100000
607926 square-free integers from 1 to 1000000

Go

<lang go>package main

import ( 

   "fmt"
   "math"

)

func sieve(limit uint64) []uint64 { 

   primes := []uint64{2}
   c := make([]bool, limit+1) // composite = true
   // no need to process even numbers > 2
   p := uint64(3)
   for {
       p2 := p * p
       if p2 > limit {
           break
       }
       for i := p2; i <= limit; i += 2 * p {
           c[i] = true
       }
       for {
           p += 2
           if !c[p] {
               break
           }
       }
   }
   for i := uint64(3); i <= limit; i += 2 {
       if !c[i] {
           primes = append(primes, i)
       }
   }
   return primes

}

func squareFree(from, to uint64) (results []uint64) { 

   limit := uint64(math.Sqrt(float64(to)))
   primes := sieve(limit)

outer: 

   for i := from; i <= to; i++ {
       for _, p := range primes {
           p2 := p * p
           if p2 > i {
               break
           }
           if i%p2 == 0 {
               continue outer
           }
       }
       results = append(results, i)
   }
   return

}

const trillion uint64 = 1000000000000

func main() { 

   fmt.Println("Square-free integers from 1 to 145:")
   sf := squareFree(1, 145)
   for i := 0; i < len(sf); i++ {
       if i > 0 && i%20 == 0 {
           fmt.Println()
       }
       fmt.Printf("%4d", sf[i])
   }

   fmt.Printf("\n\nSquare-free integers from %d to %d:\n", trillion, trillion+145)
   sf = squareFree(trillion, trillion+145)
   for i := 0; i < len(sf); i++ {
       if i > 0 && i%5 == 0 {
           fmt.Println()
       }
       fmt.Printf("%14d", sf[i])
   }

   fmt.Println("\n\nNumber of square-free integers:\n")

   fmt.Printf("  from %d to %d = %d\n", 1, 100, len(squareFree(1, 100)))
   fmt.Printf("  from %d to %d = %d\n", 1, 1000, len(squareFree(1, 1000)))
   fmt.Printf("  from %d to %d = %d\n", 1, 10000, len(squareFree(1, 10000)))
   fmt.Printf("  from %d to %d = %d\n", 1, 100000, len(squareFree(1, 100000)))
   fmt.Printf("  from %d to %d = %d\n", 1, 1000000, len(squareFree(1, 1000000)))

}</lang>

Output:
Square-free integers from 1 to 145:
   1   2   3   5   6   7  10  11  13  14  15  17  19  21  22  23  26  29  30  31
  33  34  35  37  38  39  41  42  43  46  47  51  53  55  57  58  59  61  62  65
  66  67  69  70  71  73  74  77  78  79  82  83  85  86  87  89  91  93  94  95
  97 101 102 103 105 106 107 109 110 111 113 114 115 118 119 122 123 127 129 130
 131 133 134 137 138 139 141 142 143 145

Square-free integers from 1000000000000 to 1000000000145:
 1000000000001 1000000000002 1000000000003 1000000000005 1000000000006
 1000000000007 1000000000009 1000000000011 1000000000013 1000000000014
 1000000000015 1000000000018 1000000000019 1000000000021 1000000000022
 1000000000023 1000000000027 1000000000029 1000000000030 1000000000031
 1000000000033 1000000000037 1000000000038 1000000000039 1000000000041
 1000000000042 1000000000043 1000000000045 1000000000046 1000000000047
 1000000000049 1000000000051 1000000000054 1000000000055 1000000000057
 1000000000058 1000000000059 1000000000061 1000000000063 1000000000065
 1000000000066 1000000000067 1000000000069 1000000000070 1000000000073
 1000000000074 1000000000077 1000000000078 1000000000079 1000000000081
 1000000000082 1000000000085 1000000000086 1000000000087 1000000000090
 1000000000091 1000000000093 1000000000094 1000000000095 1000000000097
 1000000000099 1000000000101 1000000000102 1000000000103 1000000000105
 1000000000106 1000000000109 1000000000111 1000000000113 1000000000114
 1000000000115 1000000000117 1000000000118 1000000000119 1000000000121
 1000000000122 1000000000123 1000000000126 1000000000127 1000000000129
 1000000000130 1000000000133 1000000000135 1000000000137 1000000000138
 1000000000139 1000000000141 1000000000142 1000000000145

Number of square-free integers:

  from 1 to 100 = 61
  from 1 to 1000 = 608
  from 1 to 10000 = 6083
  from 1 to 100000 = 60794
  from 1 to 1000000 = 607926

Java

Translation of: Go

<lang java>import java.util.ArrayList; import java.util.List;

public class SquareFree { 

   private static List<Long> sieve(long limit) {
       List<Long> primes = new ArrayList<Long>();
       primes.add(2L);
       boolean[] c = new boolean[(int)limit + 1]; // composite = true
       // no need to process even numbers > 2
       long p = 3;
       for (;;) {
           long p2 = p * p;
           if (p2 > limit) break;
           for (long i = p2; i <= limit; i += 2 * p) c[(int)i] = true;
           for (;;) {
               p += 2;
               if (!c[(int)p]) break;
           }
       }
       for (long i = 3; i <= limit; i += 2) {
           if (!c[(int)i]) primes.add(i);
       }
       return primes;
   }

   private static List<Long> squareFree(long from, long to) {
       long limit = (long)Math.sqrt((double)to);
       List<Long> primes = sieve(limit);
       List<Long> results = new ArrayList<Long>();

       outer: for (long i = from; i <= to; i++) {
           for (long p : primes) {
               long p2 = p * p;
               if (p2 > i) break;
               if (i % p2 == 0) continue outer;
           }
           results.add(i);
       }
       return results;
   }

   private final static long TRILLION = 1000000000000L;

   public static void main(String[] args) {
       System.out.println("Square-free integers from 1 to 145:");
       List<Long> sf = squareFree(1, 145);
       for (int i = 0; i < sf.size(); i++) {
           if (i > 0 && i % 20 == 0) {
               System.out.println();
           }
           System.out.printf("%4d", sf.get(i));
       }

       System.out.print("\n\nSquare-free integers");
       System.out.printf(" from %d to %d:\n", TRILLION, TRILLION + 145);
       sf = squareFree(TRILLION, TRILLION + 145);
       for (int i = 0; i < sf.size(); i++) {
           if (i > 0 && i % 5 == 0) System.out.println();
           System.out.printf("%14d", sf.get(i));
       }

       System.out.println("\n\nNumber of square-free integers:\n");
       long[] tos = {100, 1000, 10000, 100000, 1000000};
       for (long to : tos) {
           System.out.printf("  from %d to %d = %d\n", 1, to, squareFree(1, to).size());
       }
   }

}</lang>

Output:
Square-free integers from 1 to 145:
   1   2   3   5   6   7  10  11  13  14  15  17  19  21  22  23  26  29  30  31
  33  34  35  37  38  39  41  42  43  46  47  51  53  55  57  58  59  61  62  65
  66  67  69  70  71  73  74  77  78  79  82  83  85  86  87  89  91  93  94  95
  97 101 102 103 105 106 107 109 110 111 113 114 115 118 119 122 123 127 129 130
 131 133 134 137 138 139 141 142 143 145

Square-free integers from 1000000000000 to 1000000000145:
 1000000000001 1000000000002 1000000000003 1000000000005 1000000000006
 1000000000007 1000000000009 1000000000011 1000000000013 1000000000014
 1000000000015 1000000000018 1000000000019 1000000000021 1000000000022
 1000000000023 1000000000027 1000000000029 1000000000030 1000000000031
 1000000000033 1000000000037 1000000000038 1000000000039 1000000000041
 1000000000042 1000000000043 1000000000045 1000000000046 1000000000047
 1000000000049 1000000000051 1000000000054 1000000000055 1000000000057
 1000000000058 1000000000059 1000000000061 1000000000063 1000000000065
 1000000000066 1000000000067 1000000000069 1000000000070 1000000000073
 1000000000074 1000000000077 1000000000078 1000000000079 1000000000081
 1000000000082 1000000000085 1000000000086 1000000000087 1000000000090
 1000000000091 1000000000093 1000000000094 1000000000095 1000000000097
 1000000000099 1000000000101 1000000000102 1000000000103 1000000000105
 1000000000106 1000000000109 1000000000111 1000000000113 1000000000114
 1000000000115 1000000000117 1000000000118 1000000000119 1000000000121
 1000000000122 1000000000123 1000000000126 1000000000127 1000000000129
 1000000000130 1000000000133 1000000000135 1000000000137 1000000000138
 1000000000139 1000000000141 1000000000142 1000000000145

Number of square-free integers:

  from 1 to 100 = 61
  from 1 to 1000 = 608
  from 1 to 10000 = 6083
  from 1 to 100000 = 60794
  from 1 to 1000000 = 607926

Kotlin

Translation of: Go

<lang scala>// Version 1.2.50

import kotlin.math.sqrt

fun sieve(limit: Long): List<Long> { 

   val primes = mutableListOf(2L)
   val c = BooleanArray(limit.toInt() + 1) // composite = true
   // no need to process even numbers > 2
   var p = 3L
   while (true) {
       val p2 = p * p
       if (p2 > limit) break
       for (i in p2..limit step 2 * p) c[i.toInt()] = true
       while (true) {
           p += 2
           if (!c[p.toInt()]) break
       }
   }
   for (i in 3L..limit step 2) {
       if (!c[i.toInt()]) {
           primes.add(i)
       }
   }
   return primes

}

fun squareFree(from: Long, to: Long): List<Long> { 

   val limit = sqrt(to.toDouble()).toLong()
   val primes = sieve(limit)
   val results = mutableListOf<Long>()
   outer@ for (i in from..to) {
       for (p in primes) {
           val p2 = p * p
           if (p2 > i) break
           if (i % p2 == 0L) continue@outer
       }
       results.add(i)
   }
   return results

}

const val TRILLION = 1000000000000L

fun main(args: Array<String>) { 

   println("Square-free integers from 1 to 145:")
   var sf = squareFree(1, 145)
   for (i in 0 until sf.size) {
       if (i > 0 && i % 20 == 0) println()
       System.out.printf("%4d", sf[i])
   }

   println("\n\nSquare-free integers from $TRILLION to ${TRILLION + 145}:")
   sf = squareFree(TRILLION, TRILLION + 145)
   for (i in 0 until sf.size) {
       if (i > 0 && i % 5 == 0) println()
       System.out.printf("%14d", sf[i])
   }

   println("\n\nNumber of square-free integers:\n")

   System.out.printf("  from %d to %d = %d\n", 1, 100, squareFree(1, 100).size)
   System.out.printf("  from %d to %d = %d\n", 1, 1000, squareFree(1, 1000).size)
   System.out.printf("  from %d to %d = %d\n", 1, 10000, squareFree(1, 10000).size)
   System.out.printf("  from %d to %d = %d\n", 1, 100000, squareFree(1, 100000).size)
   System.out.printf("  from %d to %d = %d\n", 1, 1000000, squareFree(1, 1000000).size)

}</lang>

Output:
Square-free integers from 1 to 145:
   1   2   3   5   6   7  10  11  13  14  15  17  19  21  22  23  26  29  30  31
  33  34  35  37  38  39  41  42  43  46  47  51  53  55  57  58  59  61  62  65
  66  67  69  70  71  73  74  77  78  79  82  83  85  86  87  89  91  93  94  95
  97 101 102 103 105 106 107 109 110 111 113 114 115 118 119 122 123 127 129 130
 131 133 134 137 138 139 141 142 143 145

Square-free integers from 1000000000000 to 1000000000145:
 1000000000001 1000000000002 1000000000003 1000000000005 1000000000006
 1000000000007 1000000000009 1000000000011 1000000000013 1000000000014
 1000000000015 1000000000018 1000000000019 1000000000021 1000000000022
 1000000000023 1000000000027 1000000000029 1000000000030 1000000000031
 1000000000033 1000000000037 1000000000038 1000000000039 1000000000041
 1000000000042 1000000000043 1000000000045 1000000000046 1000000000047
 1000000000049 1000000000051 1000000000054 1000000000055 1000000000057
 1000000000058 1000000000059 1000000000061 1000000000063 1000000000065
 1000000000066 1000000000067 1000000000069 1000000000070 1000000000073
 1000000000074 1000000000077 1000000000078 1000000000079 1000000000081
 1000000000082 1000000000085 1000000000086 1000000000087 1000000000090
 1000000000091 1000000000093 1000000000094 1000000000095 1000000000097
 1000000000099 1000000000101 1000000000102 1000000000103 1000000000105
 1000000000106 1000000000109 1000000000111 1000000000113 1000000000114
 1000000000115 1000000000117 1000000000118 1000000000119 1000000000121
 1000000000122 1000000000123 1000000000126 1000000000127 1000000000129
 1000000000130 1000000000133 1000000000135 1000000000137 1000000000138
 1000000000139 1000000000141 1000000000142 1000000000145

Number of square-free integers:

  from 1 to 100 = 61
  from 1 to 1000 = 608
  from 1 to 10000 = 6083
  from 1 to 100000 = 60794
  from 1 to 1000000 = 607926

Perl 6

Works with: Rakudo version 2018.06

The prime factoring algorithm is not really the best option for finding long runs of sequential square-free numbers. It works, but is probably better suited for testing arbitrary numbers rather than testing every sequential number from 1 to some limit. If you know that that is going to be your use case, there are faster algorithms than this.

<lang perl6># Prime factorization routines sub prime-factors ( Int $n where * > 0 ) { 

   return $n if $n.is-prime;
   return [] if $n == 1;
   my $factor = find-factor( $n );
   flat prime-factors( $factor ), prime-factors( $n div $factor );

}

sub find-factor ( Int $n, $constant = 1 ) { 

   my $x      = 2;
   my $rho    = 1;
   my $factor = 1;
   while $factor == 1 {
       $rho *= 2;
       my $fixed = $x;
       for ^$rho {
           $x = ( $x * $x + $constant ) % $n;
           $factor = ( $x - $fixed ) gcd $n;
           last if 1 < $factor;
       }
   }
   $factor = find-factor( $n, $constant + 1 ) if $n == $factor;
   $factor;

}

  1. Task routine

multi is-square-free (Int $n) { my @v = $n.&prime-factors.Bag.values; not @v.sum/@v > 1 }

  1. The Task
  2. Parts 1 & 2

for 1, 145, 1e12.Int, 145+1e12.Int -> $start, $end { 

   say "\nSquare─free numbers between $start and $end:\n",
   ($start .. $end).hyper(:4batch).grep( *.&is-square-free ).list.fmt("%3d").comb(84).join("\n");

}

  1. Part 3

for 1e2, 1e3, 1e4, 1e5, 1e6 { 

   say "\nThe number of square─free numbers between 1 and {$_} (inclusive) is: ",
   +(1 .. .Int).race.grep: *.&is-square-free;

}</lang>

Output:
Square─free numbers between 1 and 145:
  1   2   3   5   6   7  10  11  13  14  15  17  19  21  22  23  26  29  30  31  33 
 34  35  37  38  39  41  42  43  46  47  51  53  55  57  58  59  61  62  65  66  67 
 69  70  71  73  74  77  78  79  82  83  85  86  87  89  91  93  94  95  97 101 102 
103 105 106 107 109 110 111 113 114 115 118 119 122 123 127 129 130 131 133 134 137 
138 139 141 142 143 145

Square─free numbers between 1000000000000 and 1000000000145:
1000000000001 1000000000002 1000000000003 1000000000005 1000000000006 1000000000007 
1000000000009 1000000000011 1000000000013 1000000000014 1000000000015 1000000000018 
1000000000019 1000000000021 1000000000022 1000000000023 1000000000027 1000000000029 
1000000000030 1000000000031 1000000000033 1000000000037 1000000000038 1000000000039 
1000000000041 1000000000042 1000000000043 1000000000045 1000000000046 1000000000047 
1000000000049 1000000000051 1000000000054 1000000000055 1000000000057 1000000000058 
1000000000059 1000000000061 1000000000063 1000000000065 1000000000066 1000000000067 
1000000000069 1000000000070 1000000000073 1000000000074 1000000000077 1000000000078 
1000000000079 1000000000081 1000000000082 1000000000085 1000000000086 1000000000087 
1000000000090 1000000000091 1000000000093 1000000000094 1000000000095 1000000000097 
1000000000099 1000000000101 1000000000102 1000000000103 1000000000105 1000000000106 
1000000000109 1000000000111 1000000000113 1000000000114 1000000000115 1000000000117 
1000000000118 1000000000119 1000000000121 1000000000122 1000000000123 1000000000126 
1000000000127 1000000000129 1000000000130 1000000000133 1000000000135 1000000000137 
1000000000138 1000000000139 1000000000141 1000000000142 1000000000145

The number of square─free numbers between 1 and 100 (inclusive) is: 61

The number of square─free numbers between 1 and 1000 (inclusive) is: 608

The number of square─free numbers between 1 and 10000 (inclusive) is: 6083

The number of square─free numbers between 1 and 100000 (inclusive) is: 60794

The number of square─free numbers between 1 and 1000000 (inclusive) is: 607926

REXX

<lang rexx>/*REXX program displays square─free numbers (integers > 1) up to a specified limit. */ numeric digits 20 parse arg LO HI . /*obtain optional arguments from the CL*/ if LO== | LO=="," then LO= 1 /*Not specified? Then use the default.*/ if HI== | HI=="," then HI=145 /* " " " " " " */ sw= linesize() - 1 /*use one less than a full line. */ count = 0 /*count of square─free numbers found. */ $= /*variable that holds a line of numbers*/ 

  do j=LO  to abs(HI)                           /*process all integers between LO & HI.*/
  if \isSquareFree(j)   then iterate            /*Not square─free?   Then skip this #. */
  count= count + 1                              /*bump the count of square─free numbers*/
  if HI<0  then iterate                         /*Only counting 'em? Then look for more*/
  if length($ || j)<sw  then $= strip($ j)      /*append the number to the output list.*/
                        else do;  say $;  $=j;  end        /*display a line of numbers.*/
  end   /*j*/

TheNum= 'The number of square─free numbers between ' if HI<0 then say TheNum LO " and " abs(HI) ' (inclusive) is: ' count if $\== then say $ /*are there any residuals to display ? */ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ isSquareFree: procedure; parse arg #; if #<1 then return 0 /*is the number too small?*/ 

             limit= iSqrt(#)                    /*obtain the integer square root of #.?*/
             odd=#//2                           /*ODD=1   if # is odd,   ODD=0 if even.*/
                      do k=2+odd  to limit  by 1+odd    /*use all numbers, or just odds*/
                      if # // k**2 == 0  then return 0  /*Is # divisible by a square?  */
                      end   /*k*/                       /* [↑]  Yes? Then ^ square─free*/
             return 1

/*──────────────────────────────────────────────────────────────────────────────────────*/ iSqrt: procedure; parse arg x; r=0; q=1; do while q<=x; q=q*4; end 

                 do while q>1; q=q%4; _=x-r-q; r=r%2; if _>=0 then do;x=_;r=r+q; end; end
      return r                                  /*R  is the integer square root of  X. */</lang>

This REXX program makes use of   linesize   REXX program (or BIF)   which is used to determine the screen width (or linesize) of the terminal (console);   not all REXXes have this BIF.

The   LINESIZE.REX   REXX program is included here   ──►   LINESIZE.REX.


output   when using the default input:

(Shown at three-quarter size.) 

1 2 3 5 6 7 10 11 13 14 15 17 19 21 22 23 26 29 30 31 33 34 35 37 38 39 41 42 43 46 47 51 53 55 57 58 59 61 62 65 66 67 69 70 71 73 74 77 78 79 82 83
85 86 87 89 91 93 94 95 97 101 102 103 105 106 107 109 110 111 113 114 115 118 119 122 123 127 129 130 131 133 134 137 138 139 141 142 143 145
output   when using the input of:     1000000000000   1000000000145

(Shown at three-quarter size.) 

1000000000001 1000000000002 1000000000003 1000000000005 1000000000006 1000000000007 1000000000009 1000000000011 1000000000013 1000000000014
1000000000015 1000000000018 1000000000019 1000000000021 1000000000022 1000000000023 1000000000027 1000000000029 1000000000030 1000000000031
1000000000033 1000000000037 1000000000038 1000000000039 1000000000041 1000000000042 1000000000043 1000000000045 1000000000046 1000000000047
1000000000049 1000000000051 1000000000054 1000000000055 1000000000057 1000000000058 1000000000059 1000000000061 1000000000063 1000000000065
1000000000066 1000000000067 1000000000069 1000000000070 1000000000073 1000000000074 1000000000077 1000000000078 1000000000079 1000000000081
1000000000082 1000000000085 1000000000086 1000000000087 1000000000090 1000000000091 1000000000093 1000000000094 1000000000095 1000000000097
1000000000099 1000000000101 1000000000102 1000000000103 1000000000105 1000000000106 1000000000109 1000000000111 1000000000113 1000000000114
1000000000115 1000000000117 1000000000118 1000000000119 1000000000121 1000000000122 1000000000123 1000000000126 1000000000127 1000000000129
1000000000130 1000000000133 1000000000135 1000000000137 1000000000138 1000000000139 1000000000141 1000000000142 1000000000145

{{out|output|text=  when using the (separate runs) inputs of:     1   -100   (and others) 

The number of square─free numbers between  1  and  100  (inclusive)  is:  61

The number of square─free numbers between  1  and  1000  (inclusive)  is:  608

The number of square─free numbers between  1  and  10000  (inclusive)  is:  6083

The number of square─free numbers between  1  and  100000  (inclusive)  is:  60794

The number of square─free numbers between  1  and  1000000  (inclusive)  is:  607926

zkl

<lang zkl>const Limit=1 + (1e12 + 145).sqrt(); // 1000001 because it fits this task var [const] 

  BI=Import.lib("zklBigNum"),    // GNU Multiple Precision Arithmetic Library
  primes=List.createLong(Limit); // one big allocate (vs lots of allocs)

// GMP provide nice way to generate primes, nextPrime is in-place p:=BI(0); while(p<Limit){ primes.append(p.nextPrime().toInt()); } // 78,499 primes

fcn squareFree(start,end,save=False){ //-->(cnt,list|n) 

  sink := Sink(if(save) List else Void);  // Sink(Void) is one item sink
  cnt, numPrimes := 0, (end - start).toFloat().sqrt().toInt() - 1;
  foreach n in ([start..end]){
     foreach j in ([0..numPrimes]){
        p,p2 := primes[j], p*p;

if(p2>n) break; if(n%p2==0) continue(2); // -->foreach n 

     }
     sink.write(n); cnt+=1
  }
  return(cnt,sink.close());

}</lang> <lang zkl>println("Square-free integers from 1 to 145:"); squareFree(1,145,True)[1].pump(Console.println, 

  T(Void.Read,14,False),fcn{ vm.arglist.apply("%4d ".fmt).concat() });

println("\nSquare-free integers from 1000000000000 to 1000000000145:"); squareFree(1000000000000,1000000000145,True)[1].pump(Console.println, 

  T(Void.Read,4,False),fcn{ vm.arglist.concat(" ") });</lang>
Output:
Square-free integers from 1 to 145:
   1    2    3    5    6    7   10   11   13   14   15   17   19   21   22 
  23   26   29   30   31   33   34   35   37   38   39   41   42   43   46 
  47   51   53   55   57   58   59   61   62   65   66   67   69   70   71 
  73   74   77   78   79   82   83   85   86   87   89   91   93   94   95 
  97  101  102  103  105  106  107  109  110  111  113  114  115  118  119 
 122  123  127  129  130  131  133  134  137  138  139  141  142  143  145 

Square-free integers from 1000000000000 to 1000000000145:
1000000000001 1000000000002 1000000000003 1000000000005 1000000000006
1000000000007 1000000000009 1000000000011 1000000000013 1000000000014
1000000000015 1000000000018 1000000000019 1000000000021 1000000000022
1000000000023 1000000000027 1000000000029 1000000000030 1000000000031
1000000000033 1000000000037 1000000000038 1000000000039 1000000000041
1000000000042 1000000000043 1000000000045 1000000000046 1000000000047
1000000000049 1000000000051 1000000000054 1000000000055 1000000000057
1000000000058 1000000000059 1000000000061 1000000000063 1000000000065
1000000000066 1000000000067 1000000000069 1000000000070 1000000000073
1000000000074 1000000000077 1000000000078 1000000000079 1000000000081
1000000000082 1000000000085 1000000000086 1000000000087 1000000000090
1000000000091 1000000000093 1000000000094 1000000000095 1000000000097
1000000000099 1000000000101 1000000000102 1000000000103 1000000000105
1000000000106 1000000000109 1000000000111 1000000000113 1000000000114
1000000000115 1000000000117 1000000000118 1000000000119 1000000000121
1000000000122 1000000000123 1000000000126 1000000000127 1000000000129
1000000000130 1000000000133 1000000000135 1000000000137 1000000000138
1000000000139 1000000000141 1000000000142 1000000000145

<lang zkl>n:=100; do(5){ 

  squareFree(1,n)[0]:
     println("%,9d square-free integers from 1 to %,d".fmt(_,n));
  n*=10;

}</lang>

Output:
       61 square-free integers from 1 to 100
      608 square-free integers from 1 to 1,000
    6,083 square-free integers from 1 to 10,000
   60,794 square-free integers from 1 to 100,000
  607,926 square-free integers from 1 to 1,000,000
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