# Square-free integers

Square-free integers
You are encouraged to solve this task according to the task description, using any language you may know.

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)

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

## ALGOL 68

`BEGIN    # count/show some square free numbers                                           #    # a number is square free if not divisible by any square and so not divisible   #    # by any squared prime                                                          #    # to satisfy the task we need to know the primes up to root 1 000 000 000 145   #    # and the square free numbers up to 1 000 000                                   #    # sieve the primes                                                              #    LONG INT one trillion = LENG 1 000 000 * LENG 1 000 000;    INT prime max = ENTIER SHORTEN long sqrt( one trillion + 145 ) + 1;    [ prime max ]BOOL prime; FOR i TO UPB prime DO prime[ i ] := TRUE OD;    FOR s FROM 2 TO ENTIER sqrt( prime max ) DO        IF prime[ s ] THEN            FOR p FROM s * s BY s TO prime max DO prime[ p ] := FALSE OD        FI    OD;    # sieve the square free integers                                                #    INT sf max = 1 000 000;    [ sf max ]BOOL square free;FOR i TO UPB square free DO square free[ i ] := TRUE OD;    FOR s FROM 2 TO ENTIER sqrt( sf max ) DO        IF prime[ s ] THEN            INT q = s * s;            FOR p FROM q BY q TO sf max DO square free[ p ] := FALSE OD        FI    OD;    # returns TRUE if n is square free, FALSE otherwise                             #    PROC is square free = ( LONG INT n )BOOL:         IF n <= sf max THEN square free[ SHORTEN n ]         ELSE            # n is larger than the sieve - use trial division                       #            INT max factor    = ENTIER SHORTEN long sqrt( n ) + 1;            BOOL square free := TRUE;            FOR f FROM 2 TO max factor WHILE square free DO                IF prime[ f ] THEN                    # have a prime                                                  #                    square free := ( n MOD ( LENG f * LENG f ) /= 0 )                FI            OD;            square free         FI # is square free # ;    # returns the count of square free numbers between m and n (inclusive)          #    PROC count square free = ( INT m, n )INT:         BEGIN            INT count := 0;            FOR i FROM m TO n DO IF square free[ i ] THEN count +:= 1 FI OD;            count         END # count square free # ;     # task requirements                                                             #    # show square free numbers from 1 -> 145                                        #    print( ( "Square free numbers from 1 to 145", newline ) );    INT    count := 0;    FOR i TO 145 DO        IF is square free( i ) THEN            print( ( whole( i, -4 ) ) );            count +:= 1;            IF count MOD 20 = 0 THEN print( ( newline ) ) FI        FI    OD;    print( ( newline ) );    # show square free numbers from 1 trillion -> one trillion + 145                #    print( ( "Square free numbers from 1 000 000 000 000 to 1 000 000 000 145", newline ) );    count := 0;    FOR i FROM 0 TO 145 DO        IF is square free( one trillion + i ) THEN            print( ( whole( one trillion + i, -14 ) ) );            count +:= 1;            IF count MOD 5 = 0 THEN print( ( newline ) ) FI        FI    OD;    print( ( newline ) );    # show counts of square free numbers                                            #    INT sf       100 :=              count square free(       1,       100 );    print( ( "square free numbers between 1 and       100: ", whole( sf       100, -6 ), newline ) );    INT sf     1 000 := sf     100 + count square free(     101,     1 000 );    print( ( "square free numbers between 1 and     1 000: ", whole( sf     1 000, -6 ), newline ) );    INT sf    10 000 := sf   1 000 + count square free(   1 001,    10 000 );    print( ( "square free numbers between 1 and    10 000: ", whole( sf    10 000, -6 ), newline ) );    INT sf   100 000 := sf  10 000 + count square free(  10 001,   100 000 );    print( ( "square free numbers between 1 and   100 000: ", whole( sf   100 000, -6 ), newline ) );    INT sf 1 000 000 := sf 100 000 + count square free( 100 001, 1 000 000 );    print( ( "square free numbers between 1 and 1 000 000: ", whole( sf 1 000 000, -6 ), newline ) )END`
Output:
```Square free numbers 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 numbers from 1 000 000 000 000 to 1 000 000 000 145
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
square free numbers between 1 and       100:     61
square free numbers between 1 and     1 000:    608
square free numbers between 1 and    10 000:   6083
square free numbers between 1 and   100 000:  60794
square free numbers between 1 and 1 000 000: 607926```

## C

Translation of: Go
`#include <stdio.h>#include <stdlib.h>#include <math.h> #define TRUE 1#define FALSE 0#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;    /* allocate enough memory to deal with all examples */    sf = malloc(1000000 * 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");    int a[5] = {100, 1000, 10000, 100000, 1000000};    for (i = 0; i < 5; ++i) {        squareFree(1, a[i], sf, &len);        printf("  from %d to %d = %lld\n", 1, a[i], len);    }    free(sf);    return 0;   }`
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.

`USING: formatting grouping io kernel math math.functionsmath.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 ] [email protected]         ! part 12 6 [a,b] [ 10 swap ^ ] map [ sq-free-count ] each    ! part 2`
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
```

## FreeBASIC

`' version 06-07-2018' compile with: fbc -s console Const As ULongInt trillion = 1000000000000ullConst As ULong max = Sqr(trillion + 145) Dim As UByte list(), sieve()Dim As ULong prime()ReDim list(max), prime(max\12), sieve(max) Dim As ULong a, b, c, i, k, stop_ = Sqr(max) For i = 4 To max Step 2   ' prime sieve remove even numbers except 2    sieve(i) = 1NextFor i = 3 To stop_ Step 2 ' proces odd numbers    If sieve(i) = 0 Then        For a = i * i To max Step i * 2            sieve(a) = 1        Next    End IfNext For i = 2 To max          ' move primes to a list    If sieve(i) = 0 Then        c += 1        prime(c) = i    End IfNext ReDim sieve(145): ReDim Preserve prime(c) For i = 1 To c  ' find all square free integers between 1 and 1000000    a = prime(i) * prime(i)    If a > 1000000 Then Exit For    For k = a To 1000000 Step a        list(k) = 1    NextNext k = 0For i = 1 To 145          ' show all between 1 and 145    If list(i) = 0 Then        Print Using"####"; i;        k +=1        If k Mod 20 = 0 Then Print     End IfNextPrint : Print sieve(0) = 1              ' = trillionFor i = 1 To 5            ' process primes 2, 3, 5, 7, 11    a = prime(i) * prime(i)    b = a - trillion Mod a    For k = b To 145 Step a        sieve(k) = 1    NextNext For i = 6 To c            ' process the rest of the primes    a = prime(i) * prime(i)    k = a - trillion Mod a    If k <= 145 Then sieve(k) = 1Next k = 0For i = 0 To 145    If sieve(i) = 0 Then        Print Using "################"; (trillion + i);        k += 1        If k Mod 5 = 0 Then print    End IfNextPrint : Print a = 1 : b = 100 : k = 0Do Until b > 1000000      ' count them    For i = a To b        If list(i) = 0 Then k += 1    Next    Print "There are "; k; " square free integers between 1 and "; b    a = b : b *= 10Loop ' empty keyboard bufferWhile Inkey <> "" : WendPrint : Print "hit any key to end program"SleepEnd`
Output:
```   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

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

There are 61 square free integers between 1 and 100
There are 608 square free integers between 1 and 1000
There are 6083 square free integers between 1 and 10000
There are 60794 square free integers between 1 and 100000
There are 607926 square free integers between 1 and 1000000```

## 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")    a := [...]uint64{100, 1000, 10000, 100000, 1000000}    for _, n := range a {        fmt.Printf("  from %d to %d = %d\n", 1, n, len(squareFree(1, n)))    }}`
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
```

`import Data.List.Split (chunksOf)import Math.NumberTheory.Primes (factorise)import Text.Printf (printf) -- True iff the argument is a square-free number.isSquareFree :: Integer -> BoolisSquareFree = all ((== 1) . snd) . factorise -- All square-free numbers in the range [lo, hi].squareFrees :: Integer -> Integer -> [Integer]squareFrees lo hi = filter isSquareFree [lo..hi] -- The result of `counts limits values' is the number of values less than or-- equal to each successive limit.  Both limits and values are assumed to be-- in increasing order.counts :: (Ord a, Num b) => [a] -> [a] -> [b]counts = go 0  where go c lims@(l:ls) (v:vs) | v > l     = c : go (c+1) ls vs                                | otherwise = go (c+1) lims vs        go _ [] _  = []        go c ls [] = replicate (length ls) c printSquareFrees :: Int -> Integer -> Integer -> IO ()printSquareFrees cols lo hi =  let ns = squareFrees lo hi      title = printf "Square free numbers from %d to %d\n" lo hi      body = unlines \$ map concat \$ chunksOf cols \$ map (printf " %3d") ns  in putStrLn \$ title ++ body printSquareFreeCounts :: [Integer] -> Integer -> Integer -> IO ()printSquareFreeCounts lims lo hi =  let cs = counts lims \$ squareFrees lo hi :: [Integer]      title = printf "Counts of square-free numbers\n"      body = unlines \$ zipWith (printf "  from 1 to %d: %d") lims cs  in putStrLn \$ title ++ body main :: IO ()main = do  printSquareFrees 20 1 145  printSquareFrees 5 1000000000000 1000000000145  printSquareFreeCounts [100, 1000, 10000, 100000, 1000000] 1 1000000`
Output:
```Square free numbers 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 numbers 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

Counts of square-free numbers
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
```

## J

Solution:

`isSqrFree=: (#@~. = #)@q:   NB. are there no duplicates in the prime factors of a number?filter=: adverb def ' #~ u' NB. filter right arg using verb to leftcountSqrFree=: +/@:isSqrFreethru=: <. + [email protected](+ *)@-~     NB. helper verb`

Required Examples:

`   isSqrFree filter 1 thru 145   NB. returns all results, but not all are displayed1 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...   100 list isSqrFree filter 1000000000000 thru 1000000000145  NB. ensure that all results are displayed1000000000001 1000000000002 1000000000003 1000000000005 1000000000006 1000000000007 10000000000091000000000011 1000000000013 1000000000014 1000000000015 1000000000018 1000000000019 10000000000211000000000022 1000000000023 1000000000027 1000000000029 1000000000030 1000000000031 10000000000331000000000037 1000000000038 1000000000039 1000000000041 1000000000042 1000000000043 10000000000451000000000046 1000000000047 1000000000049 1000000000051 1000000000054 1000000000055 10000000000571000000000058 1000000000059 1000000000061 1000000000063 1000000000065 1000000000066 10000000000671000000000069 1000000000070 1000000000073 1000000000074 1000000000077 1000000000078 10000000000791000000000081 1000000000082 1000000000085 1000000000086 1000000000087 1000000000090 10000000000911000000000093 1000000000094 1000000000095 1000000000097 1000000000099 1000000000101 10000000001021000000000103 1000000000105 1000000000106 1000000000109 1000000000111 1000000000113 10000000001141000000000115 1000000000117 1000000000118 1000000000119 1000000000121 1000000000122 10000000001231000000000126 1000000000127 1000000000129 1000000000130 1000000000133 1000000000135 10000000001371000000000138 1000000000139 1000000000141 1000000000142 1000000000145   countSqrFree 1 thru 10061   countSqrFree 1 thru 1000608   1 [email protected]&> 10 ^ 2 3 4 5 6  NB. count square free ints for 1 to each of 100, 1000, 10000, 10000, 100000 and 100000061 608 6083 60794 607926`

## Java

Translation of: Go
`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());        }    }}`
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
```

## jq

Requires jq 1.5 or higher

For brevity and in order to highlight some points of interest regarding jq, this entry focuses on solving the task in a manner that reflects the specification as closely as possible (no prime sieves or calls to sqrt), with efficiency concerns playing second fiddle.

Once a suitable generator for squares and a test for divisibility have been written, the test for whether a number is square-free can be written in one line:

```   def is_square_free: . as \$n | all( squares; divides(\$n) | not);
```

In words: to verify whether an integer, \$n, is square_free, check that no admissible square divides \$n.

squares

We could define the required `squares` generator using `while`:

```   def squares: . as \$n | 2 | while(.*. <= \$n; .+1) | .*.;
```

(Here `.*.` calculates the square of the input number.)

However, this entails performing an unnecessary multiplication, so the question becomes whether there is a more economical solution that closely reflects the specification of the required generator. Since jq supports tail-recursion optimization for 0-arity filters, the answer is:

```   def squares:
. as \$n
| def s:
(.*.) as \$s
| select(\$s <= \$n)
| \$s, ((1+.)|s);
2|s;
```

The point of interest here is the def-within-a-def.

divides

```   def divides(\$x): (\$x % .) == 0;

```

is_square_free

```   `is_square_free` as defined here intentionally returns true for all numeric inputs less than 4.
def is_square_free: . as \$n | all( squares; divides(\$n) | not) ;
```

The primary task is to examine square-free numbers in an inclusive range, so we define `square_free` to emit a stream of such numbers:

```   def square_free(from; including):
range(from;including+1) | select( is_square_free ) ;
```
```   # Compute SIGMA(s) where s is a stream
def sigma(s): reduce s as \$s (null; .+\$s);
```
```   # Group items in a stream into arrays of length at most \$n.
# For generality, this function uses `nan` as the eos marker.
def nwise(stream; \$n):
foreach (stream, nan) as \$x ([];
if length == \$n then [\$x] else . + [\$x] end;
if (.[-1] | isnan) and length>1 then .[:-1]
elif length == \$n then .
else empty
end);
```
```   def prettify_squares(from; including; width):
"Square-free integers from \(from) to \(including) (inclusive):",
(nwise( square_free(from;including); width) | map(tostring) | join(" ")),
"";
```
```   def prettify_count(\$from; \$including):
"Count from \(\$from) to \(\$including) inclusive: \(sigma( square_free(\$from ; \$including) | 1 ))";
```

```   prettify_squares(1;145; 20),
prettify_squares(1E12; 1E12 + 145; 5),
((1E2, 1E3, 1E4, 1E5, 1E6) | prettify_count(1; .))
```

Output

```Square-free integers from 1 to 145 (inclusive):
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 (inclusive):
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

Count from 1 to 100 inclusive: 61
Count from 1 to 1000 inclusive: 608
Count from 1 to 10000 inclusive: 6083
Count from 1 to 100000 inclusive: 60794
Count from 1 to 1000000 inclusive: 607926```

## Julia

`using Primes const maxrootprime = Int64(floor(sqrt(1000000000145)))const sqprimes = map(x -> x * x, primes(2, maxrootprime))possdivisorsfor(n) = vcat(filter(x -> x <= n / 2, sqprimes), n in sqprimes ? n : [])issquarefree(n) = all(x -> floor(n / x) != n / x, possdivisorsfor(n)) function squarefreebetween(mn, mx)    count = 1    padsize = length(string(mx)) + 2    println("The squarefree numbers between \$mn and \$mx are:")    for n in mn:mx        if issquarefree(n)            print(lpad(string(n), padsize))            count += 1        end        if count * padsize > 80            println()            count = 1        end    end    println()end function squarefreecount(intervals, maxnum)    count = 0    for n in 1:maxnum        for i in 1:length(intervals)            if intervals[i] < n                println("There are \$count square free numbers between 1 and \$(intervals[i]).")                intervals[i] = maxnum + 1            end        end        if issquarefree(n)            count += 1        end    end    println("There are \$count square free numbers between 1 and \$maxnum.")    end squarefreebetween(1, 145)squarefreebetween(1000000000000, 1000000000145)squarefreecount([100, 1000, 10000, 100000], 1000000) `
Output:
```
The squarefree numbers between 1 and 145 are:
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
The squarefree numbers between 1000000000000 and 1000000000145 are:
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
There are 61 square free numbers between 1 and 100.
There are 608 square free numbers between 1 and 1000.
There are 6083 square free numbers between 1 and 10000.
There are 60794 square free numbers between 1 and 100000.
There are 607926 square free numbers between 1 and 1000000.

```

## Kotlin

Translation of: Go
`// 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 = 3    while (true) {        val p2 = p * p        if (p2 > limit) break        for (i in p2..limit step 2L * p) c[i.toInt()] = true        do { p += 2 } while (c[p])    }    for (i in 3..limit step 2)        if (!c[i.toInt()])            primes.add(i)     return primes} fun squareFree(r: LongProgression): List<Long> {    val primes = sieve(sqrt(r.last.toDouble()).toLong())    val results = mutableListOf<Long>()    outer@ for (i in r) {        for (p in primes) {            val p2 = p * p            if (p2 > i) break            if (i % p2 == 0L) continue@outer        }        results.add(i)    }    return results} fun printResults(r: LongProgression, c: Int, f: Int) {    println("Square-free integers from \${r.first} to \${r.last}:")    squareFree(r).chunked(c).forEach {        println()        it.forEach { print("%\${f}d".format(it)) }    }    println('\n')} const val TRILLION = 1000000_000000L fun main(args: Array<String>) {    printResults(1..145L, 20, 4)    printResults(TRILLION..TRILLION + 145L, 5, 14)     println("Number of square-free integers:\n")    longArrayOf(100, 1000, 10000, 100000, 1000000).forEach {        j -> println("  from 1 to \$j = \${squareFree(1..j).size}")    }}`
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
```

## Lua

This is a naive method, runs in about 1 second on LuaJIT.

`function squareFree (n)  for root = 2, math.sqrt(n) do    if n % (root * root) == 0 then return false end  end  return trueend function run (lo, hi, showValues)  io.write("From " .. lo .. " to " .. hi)  io.write(showValues and ":\n" or " = ")  local count = 0  for i = lo, hi do    if squareFree(i) then      if showValues then        io.write(i, "\t")      else        count = count + 1      end    end  end  print(showValues and "\n" or count)end local testCases = {  {1, 145, true},  {1000000000000, 1000000000145, true},  {1, 100},  {1, 1000},  {1, 10000},  {1, 100000},  {1, 1000000}}for _, example in pairs(testCases) do run(unpack(example)) end`
Output:
```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

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

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```

## Mathematica

`squareFree[n_Integer] := DeleteCases[Last /@ FactorInteger[n], 1] === {};findSquareFree[n__] := Select[Range[n], squareFree];findSquareFree[45]findSquareFree[10^9, 10^9 + 145]Length[findSquareFree[10^6]]`
Output:
```{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}

{1000000001, 1000000002, 1000000003, 1000000005, 1000000006,
1000000007, 1000000009, 1000000010, 1000000011, 1000000013,
1000000014, 1000000015, 1000000018, 1000000019, 1000000021,
1000000022, 1000000027, 1000000029, 1000000030, 1000000031,
1000000033, 1000000034, 1000000037, 1000000038, 1000000039,
1000000041, 1000000042, 1000000043, 1000000045, 1000000046,
1000000047, 1000000049, 1000000051, 1000000054, 1000000055,
1000000057, 1000000058, 1000000059, 1000000061, 1000000063,
1000000065, 1000000066, 1000000067, 1000000069, 1000000070,
1000000073, 1000000074, 1000000077, 1000000078, 1000000079,
1000000081, 1000000082, 1000000083, 1000000086, 1000000087,
1000000090, 1000000091, 1000000093, 1000000094, 1000000095,
1000000097, 1000000099, 1000000101, 1000000102, 1000000103,
1000000105, 1000000106, 1000000109, 1000000110, 1000000111,
1000000113, 1000000114, 1000000115, 1000000117, 1000000118,
1000000119, 1000000121, 1000000122, 1000000123, 1000000126,
1000000127, 1000000129, 1000000130, 1000000131, 1000000133,
1000000135, 1000000137, 1000000138, 1000000139, 1000000141, 1000000142}

607926
```

## Perl

Library: ntheory
`use ntheory qw/is_square_free moebius/; sub square_free_count {    my (\$n) = @_;    my \$count = 0;    foreach my \$k (1 .. sqrt(\$n)) {        \$count += moebius(\$k) * int(\$n / \$k**2);    }    return \$count;} print "Square─free numbers between 1 and 145:\n";print join(' ', grep { is_square_free(\$_) } 1 .. 145), "\n"; print "\nSquare-free numbers between 10^12 and 10^12 + 145:\n";print join(' ', grep { is_square_free(\$_) } 1e12 .. 1e12 + 145), "\n"; print "\n";foreach my \$n (2 .. 6) {    my \$c = square_free_count(10**\$n);    print "The number of square-free numbers between 1 and 10^\$n (inclusive) is: \$c\n";}`
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 10^12 and 10^12 + 145:
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 10^2 (inclusive) is: 61
The number of square-free numbers between 1 and 10^3 (inclusive) is: 608
The number of square-free numbers between 1 and 10^4 (inclusive) is: 6083
The number of square-free numbers between 1 and 10^5 (inclusive) is: 60794
The number of square-free numbers between 1 and 10^6 (inclusive) is: 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.

`# Prime factorization routinessub 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 ) {    return 2 unless \$n +& 1;    if (my \$gcd = \$n gcd 6541380665835015) > 1 {        return \$gcd if \$gcd != \$n    }    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;} # Task routinesub is-square-free (Int \$n) { my @v = \$n.&prime-factors.Bag.values; @v.sum/@v <= 1 } # The Task# Parts 1 & 2for 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");} # Part 3for 1e2, 1e3, 1e4, 1e5, 1e6 {    say "\nThe number of square─free numbers between 1 and {\$_} (inclusive) is: ",    +(1 .. .Int).race.grep: *.&is-square-free;}`
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```

## Phix

`function square_free(atom start, finish)    sequence res = {}    if start=1 then res = {1} start = 2 end if    while start<=finish do        sequence pf = prime_factors(start, duplicates:=true)        for i=2 to length(pf) do            if pf[i]=pf[i-1] then                pf = {}                exit            end if        end for        if pf!={} then            res &= start        end if        start += 1    end while    return resend function function format_res(sequence res, string fmt)    for i=1 to length(res) do        res[i] = sprintf(fmt,res[i])    end for    return resend function constant ONE_TRILLION = 1_000_000_000_000 procedure main()    sequence res = square_free(1,145)    printf(1,"There are %d square-free integers from 1 to 145:\n",length(res))    puts(1,join_by(format_res(res,"%4d"),1,20,""))     res = square_free(ONE_TRILLION,ONE_TRILLION+145)    printf(1,"\nThere are %d square-free integers from %,d to %,d:\n",              {length(res),ONE_TRILLION, ONE_TRILLION+145})    puts(1,join_by(format_res(res,"%14d"),1,5,""))     printf(1,"\nNumber of square-free integers:\n");    for i=2 to 6 do        integer lim = power(10,i),                len = length(square_free(1,lim))        printf(1,"  from %,d to %,d = %,d\n", {1,lim,len})    end forend proceduremain()`
Output:
```There are 90 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

There are 89 square-free integers from 1,000,000,000,000 to 1,000,000,000,145:
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 1,000 = 608
from 1 to 10,000 = 6,083
from 1 to 100,000 = 60,794
from 1 to 1,000,000 = 607,926
```

## Python

` import math def SquareFree ( _number ) :	max = (int) (math.sqrt ( _number )) 	for root in range ( 2, max+1 ):					# Create a custom prime sieve		if 0 == _number % ( root * root ):			return False 	return True def ListSquareFrees( _start, _end ):	count = 0	for i in range ( _start, _end+1 ):		if True == SquareFree( i ):			print ( "{}\t".format(i), end="" )			count += 1 	print ( "\n\nTotal count of square-free numbers between {} and {}: {}".format(_start, _end, count)) ListSquareFrees( 1, 100 )ListSquareFrees( 1000000000000, 1000000000145 ) `

Output:

```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

Total count of square-free numbers between 1 and 100: 61
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

Total count of square-free numbers between 1000000000000 and 1000000000145: 89
```

## Racket

`#lang racket (define (not-square-free-set-for-range range-min (range-max (add1 range-min)))  (for*/set ((i2 (sequence-map sqr (in-range 2 (add1 (integer-sqrt range-max)))))             (i2.x (in-range (* i2 (quotient range-min i2))                             (* i2 (add1 (quotient range-max i2)))                             i2))             #:when (and (<= range-min i2.x)                         (< i2.x range-max)))    i2.x)) (define (square-free? n #:table (table (not-square-free-set-for-range n)))  (not (set-member? table n))) (define (count-square-free-numbers #:range-min (range-min 1) range-max)  (- range-max range-min (set-count (not-square-free-set-for-range range-min range-max)))) (define ((print-list-to-width w) l)  (let loop ((l l) (x 0))    (if (null? l)        (unless (zero? x) (newline))        (let* ((str (~a (car l))) (len (string-length str)))          (cond [(<= (+ len x) w) (display str) (write-char #\space) (loop (cdr l) (+ x len 1))]                [(zero? x) (displayln str) (loop (cdr l) 0)]                [else (newline) (loop l 0)])))))  (define print-list-to-80 (print-list-to-width 80)) (module+ main  (print-list-to-80 (for/list ((n (in-range 1 (add1 145))) #:when (square-free? n)) n))   (print-list-to-80 (time (let ((table (not-square-free-set-for-range #e1e12 (add1 (+ #e1e12 145)))))                            (for/list ((n (in-range #e1e12 (add1 (+ #e1e12 145))))                                       #:when (square-free? n #:table table)) n))))  (displayln "Compare time taken without the table (rather with table on the fly):")  (void (time (for/list ((n (in-range #e1e12 (add1 (+ #e1e12 145)))) #:when (square-free? n)) n)))   (count-square-free-numbers 100)  (count-square-free-numbers 1000)  (count-square-free-numbers 10000)  (count-square-free-numbers 100000)  (count-square-free-numbers 1000000))`
Output:
```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
cpu time: 1969 real time: 1967 gc time: 876
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
Compare time taken without the table (rather with table on the fly):
cpu time: 283469 real time: 285225 gc time: 118039
61
608
6083
60794
607926```

## REXX

`/*REXX program displays  square─free numbers  (integers > 1)  up to a specified limit.  */numeric digits 20                                /*be able to handle larger numbers.    */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*/ if \$\==''  then say \$                            /*are there any residuals to display ? */TheNum= 'The number of square─free numbers between 'if HI<0    then say TheNum  LO     " and "     abs(HI)      ' (inclusive)  is: '     countexit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/isSquareFree: procedure; parse arg #;  if #<1  then return 0  /*is the number too small?*/              odd=#//2                           /*ODD=1   if # is odd,   ODD=0 if even.*/                       do k=2+odd  to iSqrt(#)  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                           /* [↑]       //  is REXX's ÷ remainder.*//*──────────────────────────────────────────────────────────────────────────────────────*/iSqrt: procedure; parse arg x;       q= 1                  do while q<=x;     q= q * 4                  end   /*while q<=x*/       r= 0                  do while q>1;      q= q % 4;     _=x - r - q;    r= r % 2                  if _>=0  then do;  x= _;         r= r + q;            end                  end   /*while q>1*/       return r                                  /*R  is the integer square root of  X. */`

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
```
output   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
```

## Ruby

`require "prime" class Integer  def square_free?    prime_division.none?{|pr, exp| exp > 1}  endend puts (1..145).select(&:square_free?).each_slice(20).map{|a| a.join(" ")}puts m = 10**12puts (m..m+145).select(&:square_free?).each_slice(6).map{|a| a.join(" ")}puts markers = [100, 1000, 10_000, 100_000, 1_000_000]count = 0(1..1_000_000).each do |n|  count += 1 if n.square_free?  puts "#{count} square-frees upto #{n}" if markers.include?(n)end  `
Output:
```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

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-frees upto 100
608 square-frees upto 1000
6083 square-frees upto 10000
60794 square-frees upto 100000
607926 square-frees upto 1000000

```

## Sidef

In Sidef, the functions is_square_free(n) and square_free_count(min, max) are built-in. However, we can very easily reimplement them in Sidef code, as fast integer factorization methods are also available in the language.

`func is_square_free(n) {     n.abs!       if (n <  0)    return false if (n == 0)     n.factor_exp + [[1,1]] -> all { .[1] == 1 }} func square_free_count(n) {    1 .. n.isqrt -> sum {|k|        moebius(k) * idiv(n, k*k)    }} func display_results(a, c, f = { _ }) {    a.each_slice(c, {|*s|        say s.map(f).join(' ')    })} var a = range(   1,      145).grep {|n| is_square_free(n) }var b = range(1e12, 1e12+145).grep {|n| is_square_free(n) } say "There are #{a.len} square─free numbers between 1 and 145:"display_results(a, 17, {|n| "%3s" % n }) say "\nThere are #{b.len} square─free numbers between 10^12 and 10^12 + 145:"display_results(b, 5)say '' for (2 .. 6) { |n|    var c = square_free_count(10**n)    say "The number of square─free numbers between 1 and 10^#{n} (inclusive) is: #{c}"}`
Output:
```There are 90 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

There are 89 square─free numbers between 10^12 and 10^12 + 145:
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 10^2 (inclusive) is: 61
The number of square─free numbers between 1 and 10^3 (inclusive) is: 608
The number of square─free numbers between 1 and 10^4 (inclusive) is: 6083
The number of square─free numbers between 1 and 10^5 (inclusive) is: 60794
The number of square─free numbers between 1 and 10^6 (inclusive) is: 607926
```

## zkl

`const Limit=1 + (1e12 + 145).sqrt();	// 1000001 because it fits this taskvar [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-placep:=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());}`
`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(" ") });`
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
```
`n:=100; do(5){    squareFree(1,n)[0]:      println("%,9d square-free integers from 1 to %,d".fmt(_,n));   n*=10;}`
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
```