Square-free integers
You are encouraged to solve this task according to the task description, using any language you may know.
- 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
-
- the Wikipedia entry: square-free integer
11l
F SquareFree(_number)
V max = Int(sqrt(_number))
L(root) 2 .. max
I 0 == _number % (Int64(root) ^ 2)
R 0B
R 1B
F ListSquareFrees(Int64 _start, Int64 _end)
V count = 0
L(i) _start .. _end
I SquareFree(i)
print(i"\t", end' ‘’)
I count % 5 == 4
print()
count++
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
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
Arturo
squareFree?: function [n][
mx: to :integer sqrt n
loop 2..mx 'r ->
if zero? n % r ^ 2 -> return false
return true
]
sqFreeUpTo145: [1 2 3] ++ select 1..145 => squareFree?
print "Square-free integers from 1 to 145:"
loop split.every: 20 sqFreeUpTo145 'x ->
print map x 's -> pad to :string s 4
print ""
sqFreeUpToTrillion: select 1000000000000..1000000000145 => squareFree?
print "Square-free integers from 1000000000000 to 1000000000145:"
loop split.every: 5 sqFreeUpToTrillion 'x ->
print map x 's -> pad to :string s 14
print ""
print ["Number of square-free numbers between 1 and 100: " s100: <= 3 + enumerate 1..100 => squareFree?]
print ["Number of square-free numbers between 1 and 1000: " s1000: <= s100 + enumerate 101..1000 => squareFree?]
print ["Number of square-free numbers between 1 and 10000: " s10000: <= s1000 + enumerate 1001..10000 => squareFree?]
print ["Number of square-free numbers between 1 and 100000: " s100000: <= s10000 + enumerate 10001..100000 => squareFree?]
print ["Number of square-free numbers between 1 and 1000000: " s100000 + enumerate 100001..1000000 => squareFree?]
- 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 numbers between 1 and 100: 61 Number of square-free numbers between 1 and 1000: 608 Number of square-free numbers between 1 and 10000: 6083 Number of square-free numbers between 1 and 100000: 60794 Number of square-free numbers between 1 and 1000000: 607926
AWK
# syntax: GAWK -f SQUARE-FREE_INTEGERS.AWK
# converted from LUA
BEGIN {
main(1,145,1)
main(1000000000000,1000000000145,1)
main(1,100,0)
main(1,1000,0)
main(1,10000,0)
main(1,100000,0)
main(1,1000000,0)
exit(0)
}
function main(lo,hi,show_values, count,i,leng) {
printf("%d-%d: ",lo,hi)
leng = length(lo) + length(hi) + 3
for (i=lo; i<=hi; i++) {
if (square_free(i)) {
count++
if (show_values) {
if (leng > 110) {
printf("\n")
leng = 0
}
printf("%d ",i)
leng += length(i) + 1
}
}
}
printf("count=%d\n\n",count)
}
function square_free(n, root) {
for (root=2; root<=sqrt(n); root++) {
if (n % (root * root) == 0) {
return(0)
}
}
return(1)
}
- Output:
1-145: 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 count=90 1000000000000-1000000000145: 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 count=89 1-100: count=61 1-1000: count=608 1-10000: count=6083 1-100000: count=60794 1-1000000: count=607926
C
#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
C++
#include <cstdint>
#include <iostream>
#include <string>
using integer = uint64_t;
bool square_free(integer n) {
if (n % 4 == 0)
return false;
for (integer p = 3; p * p <= n; p += 2) {
integer count = 0;
for (; n % p == 0; n /= p) {
if (++count > 1)
return false;
}
}
return true;
}
void print_square_free_numbers(integer from, integer to) {
std::cout << "Square-free numbers between " << from
<< " and " << to << ":\n";
std::string line;
for (integer i = from; i <= to; ++i) {
if (square_free(i)) {
if (!line.empty())
line += ' ';
line += std::to_string(i);
if (line.size() >= 80) {
std::cout << line << '\n';
line.clear();
}
}
}
if (!line.empty())
std::cout << line << '\n';
}
void print_square_free_count(integer from, integer to) {
integer count = 0;
for (integer i = from; i <= to; ++i) {
if (square_free(i))
++count;
}
std::cout << "Number of square-free numbers between "
<< from << " and " << to << ": " << count << '\n';
}
int main() {
print_square_free_numbers(1, 145);
print_square_free_numbers(1000000000000LL, 1000000000145LL);
print_square_free_count(1, 100);
print_square_free_count(1, 1000);
print_square_free_count(1, 10000);
print_square_free_count(1, 100000);
print_square_free_count(1, 1000000);
return 0;
}
- 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 Number of square-free numbers between 1 and 100: 61 Number of square-free numbers between 1 and 1000: 608 Number of square-free numbers between 1 and 10000: 6083 Number of square-free numbers between 1 and 100000: 60794 Number of square-free numbers between 1 and 1000000: 607926
D
import std.array;
import std.math;
import std.stdio;
long[] sieve(long limit) {
long[] primes = [2];
bool[] c = uninitializedArray!(bool[])(cast(size_t)(limit + 1));
long p = 3;
while (true) {
long p2 = p * p;
if (p2 > limit) {
break;
}
for (long i = p2; i <= limit; i += 2 * p) {
c[cast(size_t)i] = true;
}
while (true) {
p += 2;
if (!c[cast(size_t)p]) {
break;
}
}
}
for (long i = 3; i <= limit; i += 2) {
if (!c[cast(size_t)i]) {
primes ~= i;
}
}
return primes;
}
long[] squareFree(long from, long to) {
long limit = cast(long)sqrt(cast(real)to);
auto primes = sieve(limit);
long[] results;
outer:
for (long i = from; i <= to; i++) {
foreach (p; primes) {
long p2 = p * p;
if (p2 > i) {
break;
}
if (i % p2 == 0) {
continue outer;
}
}
results ~= i;
}
return results;
}
enum TRILLION = 1_000_000_000_000L;
void main() {
writeln("Square-free integers from 1 to 145:");
auto sf = squareFree(1, 145);
foreach (i,v; sf) {
if (i > 0 && i % 20 == 0) {
writeln;
}
writef("%4d", v);
}
writefln("\n\nSquare-free integers from %d to %d:", TRILLION, TRILLION + 145);
sf = squareFree(TRILLION, TRILLION + 145);
foreach (i,v; sf) {
if (i > 0 && i % 5 == 0) {
writeln;
}
writef("%14d", v);
}
writeln("\n\nNumber of square-free integers:");
foreach (to; [100, 1_000, 10_000, 100_000, 1_000_000]) {
writefln(" from 1 to %d = %d", to, squareFree(1, to).length);
}
}
- 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.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
- 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
Forth
: square_free? ( n -- ? )
dup 4 mod 0= if drop false exit then
3
begin
2dup dup * >=
while
0 >r
begin
2dup mod 0=
while
r> 1+ dup 1 > if
2drop drop false exit
then
>r
tuck / swap
repeat
rdrop
2 +
repeat
2drop true ;
\ print square-free numbers from n3 to n2, n1 per line
: print_square_free_numbers ( n1 n2 n3 -- )
2dup
." Square-free integers between "
1 .r ." and " 1 .r ." :" cr
0 -rot
swap 1+ swap do
i square_free? if
i 3 .r space
1+
dup 2 pick mod 0= if cr then
then
loop 2drop cr ;
: count_square_free_numbers ( n1 n2 -- n )
0 -rot
swap 1+ swap do
i square_free? if 1+ then
loop ;
: main
20 145 1 print_square_free_numbers cr
5 1000000000145 1000000000000 print_square_free_numbers cr
." Number of square-free integers:" cr
100
begin
dup 1000000 <=
while
dup 1
2dup ." from " 1 .r ." to " 7 .r ." : "
count_square_free_numbers . cr
10 *
repeat drop ;
main
bye
- Output:
Square-free integers 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 integers 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 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
FreeBASIC
' version 06-07-2018
' compile with: fbc -s console
Const As ULongInt trillion = 1000000000000ull
Const 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) = 1
Next
For 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 If
Next
For i = 2 To max ' move primes to a list
If sieve(i) = 0 Then
c += 1
prime(c) = i
End If
Next
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
Next
Next
k = 0
For 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 If
Next
Print : Print
sieve(0) = 1 ' = trillion
For 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
Next
Next
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) = 1
Next
k = 0
For i = 0 To 145
If sieve(i) = 0 Then
Print Using "################"; (trillion + i);
k += 1
If k Mod 5 = 0 Then print
End If
Next
Print : Print
a = 1 : b = 100 : k = 0
Do 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 *= 10
Loop
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
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 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
Haskell
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 -> Bool
isSquareFree = 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 left
countSqrFree=: +/@:isSqrFree
thru=: <. + i.@(+ *)@-~ NB. helper verb
Required Examples:
isSqrFree filter 1 thru 145 NB. returns all results, but not all are displayed
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...
100 list isSqrFree filter 1000000000000 thru 1000000000145 NB. ensure that all results are displayed
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
countSqrFree 1 thru 100
61
countSqrFree 1 thru 1000
608
1 countSqrFree@thru&> 10 ^ 2 3 4 5 6 NB. count square free ints for 1 to each of 100, 1000, 10000, 10000, 100000 and 1000000
61 608 6083 60794 607926
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());
}
}
}
- 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 tasks
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 ))";
The specific tasks
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
// 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 true
end
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
Maple
with(NumberTheory):
with(ArrayTools):
squareFree := proc(n::integer)
if mul(PrimeFactors(n)) = n then return true;
else return false; end if;
return true;
end proc:
sfintegers := Array([]):
for count from 1 to 145 do
if squareFree(count) then Append(sfintegers, count); end if;
end do:
print(sfintegers):
sfintegers := Array([]):
for count from 10^12 to 10^12+145 do
if squareFree(count) then Append(sfintegers, count); end if;
end do:
print(sfintegers):
sfgroups := Array([]):
sfcount := 0:
for number from 1 to 100 do
if squareFree(number) then sfcount += 1: end if:
end do:
Append(sfgroups, sfcount):
for expon from 3 to 6 do
for number from 10^(expon - 1) to 10^expon do
if squareFree(number) then sfcount += 1: end if:
end do:
Append(sfgroups, sfcount):
end do:
seq(cat(sfgroups[i], " from 1 to ", 10^(i+1)), i = 1..5);- 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 from 1 to 100, 608 from 1 to 1000, 6083 from 1 to 10000,
60794 from 1 to 100000, 607926 from 1 to 1000000
Mathematica /Wolfram Language
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
Nanoquery
def is_square_free(n)
if n < 2^2
return true
end
for root in range(2, int(sqrt(n)))
if (n % (root ^ 2)) = 0
return false
end
end
return true
end
def square_free(low, high, show_values)
print format("%d-%d: ", low, high)
leng = len(str(low)) + len(str(high)) + 3
count = 0
for i in range(low, high)
if is_square_free(i)
count += 1
if show_values
if leng > 110
println
leng = 0
end
print format("%d ", i)
leng += len(str(i)) + 1
end
end
end
print format("count=%d\n\n", count)
end
square_free(1, 145, true)
square_free(10^12, 10^12 + 145, true)
square_free(1, 100, false)
square_free(1, 1000, false)
square_free(1, 10000, false)
square_free(1, 100000, false)
square_free(1, 1000000, false)- Output:
1-145: 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 count=90 1000000000000-1000000000145: 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 count=89 1-100: count=61 1-1000: count=608 1-10000: count=6083 1-100000: count=60794 1-1000000: count=607926
Nim
import math, strutils
proc sieve(limit: Natural): seq[int] =
result = @[2]
var c = newSeq[bool](limit + 1) # Composite = true.
# No need to process even numbers > 2.
var p = 3
while true:
let p2 = p * p
if p2 > limit: break
for i in countup(p2, limit, 2 * p):
c[i] = true
while true:
inc p, 2
if not c[p]: break
for i in countup(3, limit, 2):
if not c[i]: result.add i
proc squareFree(fromVal, toVal: Natural): seq[int] =
let limit = int(sqrt(toVal.toFloat))
let primes = sieve(limit)
for i in fromVal..toVal:
block check:
for p in primes:
let p2 = p * p
if p2 > i: break
if i mod p2 == 0:
break check # Not square free.
result.add i
when isMainModule:
const Trillion = 1_000_000_000_000
echo "Square-free integers from 1 to 145:"
var sf = squareFree(1, 145)
for i, val in sf:
if i > 0 and i mod 20 == 0:
echo()
stdout.write ($val).align(4)
echo()
echo "\nSquare-free integers from $1 to $2:\n".format(Trillion, Trillion + 145)
sf = squareFree(Trillion, Trillion + 145)
for i, val in sf:
if i > 0 and i mod 5 == 0:
echo()
stdout.write ($val).align(14)
echo()
echo "\nNumber of square-free integers:\n"
for n in [100, 1_000, 10_000, 100_000, 1_000_000]:
echo " from $1 to $2 = $3".format(1, n, squareFree(1, n).len)
- 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
OCaml
let squarefree (number: int) : bool =
let max = Float.of_int number |> sqrt |> Float.to_int |> (fun x -> x + 2) in
let rec inner i number2 =
if i == max
then true
else if number2 mod (i*i) == 0
then false
else inner (i+1) number2
in inner 2 number
;;
let list_squarefree_integers (x, y) =
let rec inner start finish output =
if start == finish
then output
else if squarefree start
then inner (start+1) finish (start :: output)
else inner (start+1) finish output
in inner x y []
;;
let print_squarefree_integers (x, y) =
let squarefrees_unrev = list_squarefree_integers (x, y) in
let squarefrees = List.rev squarefrees_unrev in
let rec inner sfs i count =
match sfs with
[] -> count
| h::t -> (if (i+1) mod 5 == 0 then (Printf.printf "%d\n" h) else (Printf.printf "%d " h); inner t (i+1) (count+1);)
in Printf.printf "\n\nTotal count of square-free numbers between %d and %d: %d\n" x y (inner squarefrees 0 0)
;;
let () =
print_squarefree_integers (1, 146);
print_squarefree_integers (1000000000000, 1000000000146)
;;
- 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 Total count of square-free numbers between 1 and 146: 90 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 1000000000146: 89
Pascal
As so often, marking/sieving multiples of square of primes to speed things up.
program SquareFree;
{$IFDEF FPC}
{$MODE DELPHI}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
const
//needs 1e10 Byte = 10 Gb maybe someone got 128 Gb :-) nearly linear time
BigLimit = 10*1000*1000*1000;
TRILLION = 1000*1000*1000*1000;
primeLmt = trunc(sqrt(TRILLION+150));
var
primes : array of byte;
sieve : array of byte;
procedure initPrimes;
var
i,lmt,dp :NativeInt;
Begin
setlength(primes,80000);
setlength(sieve,primeLmt);
sieve[0] := 1;
sieve[1] := 1;
i := 2;
repeat
IF sieve[i] = 0 then
Begin
lmt:= i*i;
while lmt<primeLmt do
Begin
sieve[lmt] := 1;
inc(lmt,i);
end;
end;
inc(i);
until i*i>=primeLmt;
//extract difference of primes
i := 0;
lmt := 0;
dp := 0;
repeat
IF sieve[i] = 0 then
Begin
primes[lmt] := dp;
dp := 0;
inc(lmt);
end;
inc(dp);
inc(i);
until i >primeLmt;
setlength(sieve,0);
setlength(Primes,lmt+1);
end;
procedure SieveSquares;
//mark all powers >=2 of prime => all powers = 2 is sufficient
var
pSieve : pByte;
i,sq,k,prime : NativeInt;
Begin
pSieve := @sieve[0];
prime := 0;
For i := 0 to High(primes) do
Begin
prime := prime+primes[i];
sq := prime*prime;
k := sq;
if sq > BigLimit then
break;
repeat
pSieve[k] := 1;
inc(k,sq);
until k> BigLimit;
end;
end;
procedure Count_x10;
var
pSieve : pByte;
i,lmt,cnt: NativeInt;
begin
writeln(' square free count');
writeln('[1 to limit]');
pSieve := @sieve[0];
lmt := 10;
i := 1;
cnt := 0;
repeat
while i <= lmt do
Begin
inc(cnt,ORD(pSieve[i] = 0));
inc(i);
end;
writeln(lmt:12,' ',cnt:12);
IF lmt >= BigLimit then
BREAK;
lmt := lmt*10;
IF lmt >BigLimit then
lmt := BigLimit;
until false;
end;
function TestSquarefree(N:Uint64):boolean;
var
i,prime,sq : NativeUint;
Begin
prime := 0;
result := false;
For i := 0 to High(primes) do
Begin
prime := prime+primes[i];
sq := sqr(prime);
IF sq> N then
BREAK;
IF N MOD sq = 0 then
EXIT;
end;
result := true;
end;
var
i,k : NativeInt;
Begin
InitPrimes;
setlength(sieve,BigLimit+1);
SieveSquares;
writeln('Square free numbers from 1 to 145');
k := 80 div 4;
For i := 1 to 145 do
If sieve[i] = 0 then
Begin
write(i:4);
dec(k);
IF k = 0 then
Begin
writeln;
k := 80 div 4;
end;
end;
writeln;writeln;
writeln('Square free numbers from ',TRILLION,' to ',TRILLION+145);
k := 4;
For i := TRILLION to TRILLION+145 do
Begin
if TestSquarefree(i) then
Begin
write(i:20);
dec(k);
IF k = 0 then
Begin
writeln;
k := 4;
end;
end;
end;
writeln;writeln;
Count_x10;
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 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
square free count
[1 to limit]
10 7
100 61
1000 608
10000 6083
100000 60794
1000000 607926
10000000 6079291
100000000 60792694
1000000000 607927124
10000000000 6079270942
real 0m13,908s
Perl
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
Phix
with javascript_semantics function square_frees(atom start, finish) sequence res = {} integer maxprime = get_maxprime(finish) while start<=finish do if square_free(start,maxprime) then res &= start end if start += 1 end while return res end function procedure show_range(atom start, finish, integer jb, string fmt) sequence res = square_frees(start, finish) printf(1,"There are %d square-free integers from %,d to %,d:\n%s\n", {length(res),start,finish,join_by(res,1,jb,"","\n",fmt)}) end procedure show_range(1,145,20,"%4d") show_range(1e12,1e12+145,5,"%14d") printf(1,"\nNumber of square-free integers:\n"); for i=2 to 6 do integer lim = power(10,i), len = length(square_frees(1,lim)) printf(1," from %,d to %,d = %,d\n", {1,lim,len}) end for
- 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
Raku
(formerly Perl 6)
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 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 ) {
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 routine
sub is-square-free (Int $n) { my @v = $n.&prime-factors.Bag.values; @v.sum/@v <= 1 }
# The Task
# Parts 1 & 2
for 1, 145, 1e12.Int, 145+1e12.Int -> $start, $end {
say "\nSquare─free numbers between $start and $end:\n",
($start .. $end).hyper(:4batch).grep( &is-square-free ).list.fmt("%3d").comb(84).join("\n");
}
# Part 3
for 1e2, 1e3, 1e4, 1e5, 1e6 {
say "\nThe number of square─free numbers between 1 and {$_} (inclusive) is: ",
+(1 .. .Int).race.grep: &is-square-free;
}
- 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
/*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. */
# = 0 /*count of square─free numbers found. */
$= /*variable that holds a line of numbers*/
do j=LO to abs(HI) /*process all integers between LO & HI.*/
if \isSquareFree(j) then iterate /*Not square─free? Then skip this #. */
#= # + 1 /*bump the count of square─free numbers*/
if HI<0 then iterate /*Only counting 'em? Then look for more*/
if length($ || 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: ' #
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isSquareFree: procedure; parse arg x; if x<1 then return 0 /*is the number too small?*/
odd= x//2 /*ODD=1 if X is odd, ODD=0 if even.*/
do k=2+odd to iSqrt(x) by 1+odd /*use all numbers, or just odds*/
if x // k**2 == 0 then return 0 /*Is X 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
RPL
≪ DUP √ CEIL → n max
≪ 1 SF
2 max FOR root
IF n root SQ MOD NOT THEN 1 CF max 'root' STO END
NEXT
1 FS?
≫ ≫ 'SQFREE?' STO
≪ {} ROT ROT FOR n IF n SQFREE? THEN n + END NEXT
≫ 'TASK1' STO
≪ 0 ROT ROT FOR n IF n SQFREE? THEN 1 + END NEXT
≫ 'TASK2' STO
1 145 TASK1 1 1000 TASK2
- Output:
2: { 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 }
1: 608
Ruby
require "prime"
class Integer
def square_free?
prime_division.none?{|pr, exp| exp > 1}
end
end
puts (1..145).select(&:square_free?).each_slice(20).map{|a| a.join(" ")}
puts
m = 10**12
puts (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
Rust
fn square_free(mut n: usize) -> bool {
if n & 3 == 0 {
return false;
}
let mut p: usize = 3;
while p * p <= n {
let mut count = 0;
while n % p == 0 {
count += 1;
if count > 1 {
return false;
}
n /= p;
}
p += 2;
}
true
}
fn print_square_free_numbers(from: usize, to: usize) {
println!("Square-free numbers between {} and {}:", from, to);
let mut line = String::new();
for i in from..=to {
if square_free(i) {
if !line.is_empty() {
line.push_str(" ");
}
line.push_str(&i.to_string());
if line.len() >= 80 {
println!("{}", line);
line.clear();
}
}
}
if !line.is_empty() {
println!("{}", line);
}
}
fn print_square_free_count(from: usize, to: usize) {
let mut count = 0;
for i in from..=to {
if square_free(i) {
count += 1;
}
}
println!(
"Number of square-free numbers between {} and {}: {}",
from, to, count
)
}
fn main() {
print_square_free_numbers(1, 145);
print_square_free_numbers(1000000000000, 1000000000145);
let mut n: usize = 100;
while n <= 1000000 {
print_square_free_count(1, n);
n *= 10;
}
}
- 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 Number of square-free numbers between 1 and 100: 61 Number of square-free numbers between 1 and 1000: 608 Number of square-free numbers between 1 and 10000: 6083 Number of square-free numbers between 1 and 100000: 60794 Number of square-free numbers between 1 and 1000000: 607926
Rust FP
fn square_free(x: usize) -> bool {
fn iter(x: usize, start: usize, prev: usize) -> bool {
let limit = (x as f64).sqrt().ceil() as usize;
match (start..=limit).skip_while(|i| x % i > 0).next() {
Some(v) => if v == prev {false}
else {iter(x / v, v, v)},
None => x != prev
}
}
iter(x, 2, 0)
}
fn main() {
for (up, to, nl_limit) in vec!((1, 145, 20), (1000000000000, 1000000000145, 6)) {
let free_nums = (up..=to).into_iter().filter(|&sf| square_free(sf));
println!("square_free numbers between {} and {}:", up, to);
for (index, free_num) in free_nums.enumerate() {
let spnl = if (index + 1) % nl_limit > 0 {' '} else {'\n'};
print!("{:3}{}", free_num, spnl)
}
println!("\n");
}
for limit in (2..7).map(|e| 10usize.pow(e)) {
let start = std::time::Instant::now();
let number = (1..=limit).into_iter().filter(|&sf| square_free(sf)).count();
let duration = start.elapsed().as_millis();
println!("Number of square-free numbers between 1 and {:7}: {:6} [time(ms) {:5}]",
limit, number, duration)
}
}
- 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 Number of square-free numbers between 1 and 100: 61 [time(ms) 0] Number of square-free numbers between 1 and 1000: 608 [time(ms) 0] Number of square-free numbers between 1 and 10000: 6083 [time(ms) 2] Number of square-free numbers between 1 and 100000: 60794 [time(ms) 62] Number of square-free numbers between 1 and 1000000: 607926 [time(ms) 1638]
Scala
This example uses a brute-force approach to check whether a number is square-free, and builds a lazily evaluated list of all square-free numbers with a simple filter. To get the large square-free numbers, it avoids computing the beginning of the list by starting the list at a given number.
import spire.math.SafeLong
import spire.implicits._
import scala.annotation.tailrec
object SquareFreeNums {
def main(args: Array[String]): Unit = {
println(
s"""|1 - 145:
|${formatTable(sqrFree.takeWhile(_ <= 145).toVector, 10)}
|
|1T - 1T+145:
|${formatTable(sqrFreeInit(1000000000000L).takeWhile(_ <= 1000000000145L).toVector, 6)}
|
|Square-Free Counts...
|100: ${sqrFree.takeWhile(_ <= 100).length}
|1000: ${sqrFree.takeWhile(_ <= 1000).length}
|10000: ${sqrFree.takeWhile(_ <= 10000).length}
|100000: ${sqrFree.takeWhile(_ <= 100000).length}
|1000000: ${sqrFree.takeWhile(_ <= 1000000).length}
|""".stripMargin)
}
def chkSqr(num: SafeLong): Boolean = !LazyList.iterate(SafeLong(2))(_ + 1).map(_.pow(2)).takeWhile(_ <= num).exists(num%_ == 0)
def sqrFreeInit(init: SafeLong): LazyList[SafeLong] = LazyList.iterate(init)(_ + 1).filter(chkSqr)
def sqrFree: LazyList[SafeLong] = sqrFreeInit(1)
def formatTable(lst: Vector[SafeLong], rlen: Int): String = {
@tailrec
def fHelper(ac: Vector[String], src: Vector[String]): String = {
if(src.nonEmpty) fHelper(ac :+ src.take(rlen).mkString, src.drop(rlen))
else ac.mkString("\n")
}
val maxLen = lst.map(n => f"${n.toBigInt}%,d".length).max
val formatted = lst.map(n => s"%,${maxLen + 2}d".format(n.toBigInt))
fHelper(Vector[String](), formatted)
}
}
- Output:
1 - 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
1T - 1T+145:
1,000,000,000,001 1,000,000,000,002 1,000,000,000,003 1,000,000,000,005 1,000,000,000,006 1,000,000,000,007
1,000,000,000,009 1,000,000,000,011 1,000,000,000,013 1,000,000,000,014 1,000,000,000,015 1,000,000,000,018
1,000,000,000,019 1,000,000,000,021 1,000,000,000,022 1,000,000,000,023 1,000,000,000,027 1,000,000,000,029
1,000,000,000,030 1,000,000,000,031 1,000,000,000,033 1,000,000,000,037 1,000,000,000,038 1,000,000,000,039
1,000,000,000,041 1,000,000,000,042 1,000,000,000,043 1,000,000,000,045 1,000,000,000,046 1,000,000,000,047
1,000,000,000,049 1,000,000,000,051 1,000,000,000,054 1,000,000,000,055 1,000,000,000,057 1,000,000,000,058
1,000,000,000,059 1,000,000,000,061 1,000,000,000,063 1,000,000,000,065 1,000,000,000,066 1,000,000,000,067
1,000,000,000,069 1,000,000,000,070 1,000,000,000,073 1,000,000,000,074 1,000,000,000,077 1,000,000,000,078
1,000,000,000,079 1,000,000,000,081 1,000,000,000,082 1,000,000,000,085 1,000,000,000,086 1,000,000,000,087
1,000,000,000,090 1,000,000,000,091 1,000,000,000,093 1,000,000,000,094 1,000,000,000,095 1,000,000,000,097
1,000,000,000,099 1,000,000,000,101 1,000,000,000,102 1,000,000,000,103 1,000,000,000,105 1,000,000,000,106
1,000,000,000,109 1,000,000,000,111 1,000,000,000,113 1,000,000,000,114 1,000,000,000,115 1,000,000,000,117
1,000,000,000,118 1,000,000,000,119 1,000,000,000,121 1,000,000,000,122 1,000,000,000,123 1,000,000,000,126
1,000,000,000,127 1,000,000,000,129 1,000,000,000,130 1,000,000,000,133 1,000,000,000,135 1,000,000,000,137
1,000,000,000,138 1,000,000,000,139 1,000,000,000,141 1,000,000,000,142 1,000,000,000,145
Square-Free Counts...
100: 61
1000: 608
10000: 6083
100000: 60794
1000000: 607926
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
Swift
import BigInt
import Foundation
extension BinaryInteger {
@inlinable
public var isSquare: Bool {
var x = self / 2
var seen = Set([x])
while x * x != self {
x = (x + (self / x)) / 2
if seen.contains(x) {
return false
}
seen.insert(x)
}
return true
}
@inlinable
public var isSquareFree: Bool {
return factors().dropFirst().reduce(true, { $0 && !$1.isSquare })
}
@inlinable
public func factors() -> [Self] {
let maxN = Self(Double(self).squareRoot())
var res = Set<Self>()
for factor in stride(from: 1, through: maxN, by: 1) where self % factor == 0 {
res.insert(factor)
res.insert(self / factor)
}
return res.sorted()
}
}
let sqFree1to145 = (1...145).filter({ $0.isSquareFree })
print("Square free numbers in range 1...145: \(sqFree1to145)")
let sqFreeBig = (BigInt(1_000_000_000_000)...BigInt(1_000_000_000_145)).filter({ $0.isSquareFree })
print("Square free numbers in range 1_000_000_000_000...1_000_000_000_045: \(sqFreeBig)")
var count = 0
for n in 1...1_000_000 {
if n.isSquareFree {
count += 1
}
switch n {
case 100, 1_000, 10_000, 100_000, 1_000_000:
print("Square free numbers between 1...\(n): \(count)")
case _:
break
}
}
- Output:
Square free numbers in range 1...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 in range 1_000_000_000_000...1_000_000_000_045: [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...100: 61 Square free numbers between 1...1000: 608 Square free numbers between 1...10000: 6083 Square free numbers between 1...100000: 60794 Square free numbers between 1...1000000: 607926
Tcl
proc isSquarefree {n} {
for {set d 2} {($d * $d) <= $n} {set d [expr {($d+1)|1}]} {
if {0 == ($n % $d)} {
set n [expr {$n / $d}]
if {0 == ($n % $d)} {
return 0 ;# no, just found dup divisor
}
}
}
return 1 ;# yes, no dup divisor found
}
proc unComma {str {comma ,}} {
return [string map [list $comma {}] $str]
}
proc showRange {lo hi} {
puts "Square-free integers in range $lo..$hi are:"
set lo [unComma $lo]
set hi [unComma $hi]
set L [string length $hi]
set perLine 5
while {($perLine * 2 * ($L+1)) <= 80} {
set perLine [expr {$perLine * 2}]
}
set k 0
for {set n $lo} {$n <= $hi} {incr n} {
if {[isSquarefree $n]} {
puts -nonewline " [format %${L}s $n]"
incr k
if {$k >= $perLine} {
puts "" ; set k 0
}
}
}
if {$k > 0} {
puts ""
}
}
proc showCount {lo hi} {
set rangtxt "$lo..$hi"
set lo [unComma $lo]
set hi [unComma $hi]
set k 0
for {set n $lo} {$n <= $hi} {incr n} {
incr k [isSquarefree $n]
}
puts "Counting [format %6s $k] square-free integers in range $rangtxt"
}
showRange 1 145
showRange 1,000,000,000,000 1,000,000,000,145
foreach H {100 1000 10000 100000 1000000} {
showCount 1 $H
}
- Output:
Square-free integers in range 1..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 Square-free integers in range 1,000,000,000,000..1,000,000,000,145 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 Counting 61 square-free integers in range 1..100 Counting 608 square-free integers in range 1..1000 Counting 6083 square-free integers in range 1..10000 Counting 60794 square-free integers in range 1..100000 Counting 607926 square-free integers in range 1..1000000
Uiua
Formatting the output is left as an exercise for the reader.
Sqf ← ≠0/↧◿⊸(×.↘2⇡+1⌊√)
⟜⧻▽⊸≡Sqf +1⇡145
⟜⧻▽⊸≡Sqf +1e12+1⇡145
# Find all squares and their multiples, and mask them out.
▽:⇡⟜(¬±°⊚/◇⊂≡(□×⇡+1)⌊÷:⟜(.×.↘2⇡√.))1000000
⊙◌≡(/+<)⇌[⍥(×10.)4 100]¤- Output:
90 [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] 89 [1000000000001 1000000000002 1000000000003 1000000000005 1000000000006 1000000000007 1000000000009 ...etc... 1000000000142 1000000000145] [61 608 6083 60794 607926]
Visual Basic .NET
Module Module1
Function Sieve(limit As Long) As List(Of Long)
Dim primes As New List(Of Long) From {2}
Dim c(limit + 1) As Boolean
Dim p = 3L
While True
Dim p2 = p * p
If p2 > limit Then
Exit While
End If
For i = p2 To limit Step 2 * p
c(i) = True
Next
While True
p += 2
If Not c(p) Then
Exit While
End If
End While
End While
For i = 3 To limit Step 2
If Not c(i) Then
primes.Add(i)
End If
Next
Return primes
End Function
Function SquareFree(from As Long, to_ As Long) As List(Of Long)
Dim limit = CType(Math.Sqrt(to_), Long)
Dim primes = Sieve(limit)
Dim results As New List(Of Long)
Dim i = from
While i <= to_
For Each p In primes
Dim p2 = p * p
If p2 > i Then
Exit For
End If
If (i Mod p2) = 0 Then
i += 1
Continue While
End If
Next
results.Add(i)
i += 1
End While
Return results
End Function
ReadOnly TRILLION As Long = 1_000_000_000_000
Sub Main()
Console.WriteLine("Square-free integers from 1 to 145:")
Dim sf = SquareFree(1, 145)
For index = 0 To sf.Count - 1
Dim v = sf(index)
If index > 1 AndAlso (index Mod 20) = 0 Then
Console.WriteLine()
End If
Console.Write("{0,4}", v)
Next
Console.WriteLine()
Console.WriteLine()
Console.WriteLine("Square-free integers from {0} to {1}:", TRILLION, TRILLION + 145)
sf = SquareFree(TRILLION, TRILLION + 145)
For index = 0 To sf.Count - 1
Dim v = sf(index)
If index > 1 AndAlso (index Mod 5) = 0 Then
Console.WriteLine()
End If
Console.Write("{0,14}", v)
Next
Console.WriteLine()
Console.WriteLine()
Console.WriteLine("Number of square-free integers:")
For Each to_ In {100, 1_000, 10_000, 100_000, 1_000_000}
Console.WriteLine(" from 1 to {0} = {1}", to_, SquareFree(1, to_).Count)
Next
End Sub
End Module
- 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
Wren
import "./fmt" for Fmt
var isSquareFree = Fn.new { |n|
var i = 2
while (i * i <= n) {
if (n%(i*i) == 0) return false
i = (i > 2) ? i + 2 : i + 1
}
return true
}
var ranges = [ [1..145, 3, 20], [1e12..1e12+145, 12, 5] ]
for (r in ranges) {
System.print("The square-free integers between %(r[0].min) and %(r[0].max) inclusive are:")
var count = 0
for (i in r[0]) {
if (isSquareFree.call(i)) {
count = count + 1
System.write("%(Fmt.d(r[1], i)) ")
if (count %r[2] == 0) System.print()
}
}
System.print("\n")
}
System.print("Counts of square-free integers:")
var count = 0
var lims = [0, 100, 1000, 1e4, 1e5, 1e6]
for (i in 1...lims.count) {
System.write(" from 1 to (inclusive) %(Fmt.d(-7, lims[i])) = ")
for (j in lims[i-1]+1..lims[i]) {
if (isSquareFree.call(j)) count = count + 1
}
System.print(count)
}
- Output:
The square-free integers between 1 and 145 inclusive 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 square-free integers between 1000000000000 and 1000000000145 inclusive 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 Counts of square-free integers: from 1 to (inclusive) 100 = 61 from 1 to (inclusive) 1000 = 608 from 1 to (inclusive) 10000 = 6083 from 1 to (inclusive) 100000 = 60794 from 1 to (inclusive) 1000000 = 607926
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());
}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