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Coprime triplets

From Rosetta Code
Coprime triplets is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Starting from the sequence a(1)=1 and a(2)=2 find the next smallest number which is coprime to the last two predecessors and has not yet appeared yet in the sequence.
p and q are coprimes if they have no common factors other than 1.
Let p, q < 50

Action![edit]

INT FUNC Gcd(INT a,b)
INT tmp
 
IF a<b THEN
tmp=a a=b b=tmp
FI
 
WHILE b#0
DO
tmp=a MOD b
a=b b=tmp
OD
RETURN (a)
 
BYTE FUNC Contains(INT v INT ARRAY a INT count)
INT i
 
FOR i=0 TO count-1
DO
IF a(i)=v THEN
RETURN (1)
FI
OD
RETURN (0)
 
BYTE FUNC Skip(INT v INT ARRAY a INT count)
IF Contains(v,a,count) THEN
RETURN (1)
ELSEIF Gcd(v,a(count-1))>1 THEN
RETURN (1)
ELSEIF Gcd(v,a(count-2))>1 THEN
RETURN (1)
FI
RETURN (0)
 
BYTE FUNC CoprimeTriplets(INT limit INT ARRAY a)
INT i,count
 
a(0)=1 a(1)=2
count=2
 
DO
i=3
WHILE Skip(i,a,count)
DO
i==+1
OD
IF i>=limit THEN
RETURN (count)
FI
a(count)=i
count==+1
OD
RETURN (count)
 
PROC Main()
DEFINE LIMIT="50"
INT ARRAY a(LIMIT)
INT i,count
 
count=CoprimeTriplets(LIMIT,a)
FOR i=0 TO count-1
DO
PrintI(a(i)) Put(32)
OD
PrintF("%E%EThere are %I coprimes less than %I",count,LIMIT)
RETURN
Output:

Screenshot from Atari 8-bit computer

1 2 3 5 4 7 9 8 11 13 6 17 19 10 21 23 16 15 29 14 25 27 22 31 35 12 37 41 18 43 47 20 33 49 26 45

There are 36 coprimes less than 50

ALGOL 68[edit]

BEGIN # find members of the coprime triplets sequence: starting from 1, 2 the #
# subsequent members are the lowest number coprime to the previous two #
# that haven't appeared in the sequence yet #
# iterative Greatest Common Divisor routine, returns the gcd of m and n #
PROC gcd = ( INT m, n )INT:
BEGIN
INT a := ABS m, b := ABS n;
WHILE b /= 0 DO
INT new a = b;
b := a MOD b;
a := new a
OD;
a
END # gcd # ;
# returns an array of the coprime triplets up to n #
OP COPRIMETRIPLETS = ( INT n )[]INT:
BEGIN
[ 1 : n ]INT result;
IF n > 0 THEN
result[ 1 ] := 1;
IF n > 1 THEN
[ 1 : n ]BOOL used;
used[ 1 ] := used[ 2 ] := TRUE;
FOR i FROM 3 TO n DO used[ i ] := FALSE; result[ i ] := 0 OD;
result[ 2 ] := 2;
FOR i FROM 3 TO n DO
INT p1 = result[ i - 1 ];
INT p2 = result[ i - 2 ];
BOOL found := FALSE;
FOR j TO n WHILE NOT found DO
IF NOT used[ j ] THEN
found := gcd( p1, j ) = 1 AND gcd( p2, j ) = 1;
IF found THEN
used[ j ] := TRUE;
result[ i ] := j
FI
FI
OD
OD
FI
FI;
result
END # COPRIMETRIPLETS # ;
[]INT cps = COPRIMETRIPLETS 49;
INT printed := 0;
FOR i TO UPB cps DO
IF cps[ i ] /= 0 THEN
print( ( whole( cps[ i ], -3 ) ) );
printed +:= 1;
IF printed MOD 10 = 0 THEN print( ( newline ) ) FI
FI
OD;
print( ( newline, "Found ", whole( printed, 0 ), " coprime triplets up to ", whole( UPB cps, 0 ), newline ) )
END
Output:
  1  2  3  5  4  7  9  8 11 13
  6 17 19 10 21 23 16 15 29 14
 25 27 22 31 35 12 37 41 18 43
 47 20 33 49 26 45
Found 36 coprime triplets up to 49

ALGOL W[edit]

begin % find a sequence of coprime triplets, each element is coprime to the  %
 % two predeccessors and hasn't appeared in the list yet, the first two %
 % elements are 1 and 2  %
integer procedure gcd ( integer value m, n ) ;
begin
integer a, b;
a := abs m;
b := abs n;
while b not = 0 do begin
integer newA;
newA := b;
b  := a rem b;
a  := newA
end while_b_ne_0 ;
a
end gcd ;
 % construct the sequence %
integer array seq ( 1 :: 49 );
integer sCount;
seq( 1 ) := 1; seq( 2 ) := 2; for i := 3 until 49 do seq( i ) := 0;
for i := 3 until 49 do begin
integer s1, s2, lowest;
s1  := seq( i - 1 );
s2  := seq( i - 2 );
lowest := 2;
while begin logical candidate;
lowest  := lowest + 1;
candidate := gcd( s1, lowest ) = 1 and gcd( s2, lowest ) = 1;
if candidate then begin
 % lowest is coprime to the previous two elements %
 % check it hasn't appeared already  %
for pos := 1 until i - 1 do begin
candidate := candidate and lowest not = seq( pos );
end for_pos ;
if candidate then seq( i ) := lowest;
end if_lowest_coprime_to_s1_and_s2 ;
not candidate and lowest < 50
end do begin end while_not_found
end for_i ;
 % show the sequence %
sCount := 0;
for i := 1 until 49 do begin
if seq( i ) not = 0 then begin
writeon( i_w := 2, s_w := 0, " ", seq( i ) );
sCount := sCount + 1;
if sCount rem 10 = 0 then write()
end if_seq_i_ne_0
end for_i ;
write( i_w := 1, s_w := 0, sCount, " coprime triplets below 50" )
end.
Output:
  1  2  3  5  4  7  9  8 11 13
  6 17 19 10 21 23 16 15 29 14
 25 27 22 31 35 12 37 41 18 43
 47 20 33 49 26 45
36 coprime triplets below 50

AppleScript[edit]

on hcf(a, b)
repeat until (b = 0)
set x to a
set a to b
set b to x mod b
end repeat
 
if (a < 0) then return -a
return a
end hcf
 
on coprimeTriplets(max)
if (max < 3) then return {}
script o
property candidates : {}
property output : {1, 2}
end script
 
-- When repeatedly searching for lowest unused numbers, it's faster in
-- AppleScript to take numbers from a preset list of candidates which
-- grows shorter from at or near the low end as used numbers are removed
-- than it is to test increasing numbers of previous numbers each time
-- against a list that's growing longer with them.
-- Generate the list of candidates here.
repeat with i from 3 to max
set end of o's candidates to i
end repeat
set candidateCount to max - 2
set {p1, p2} to o's output
set ok to true
repeat while (ok) -- While suitable coprimes found and candidates left.
repeat with i from 1 to candidateCount
set q to item i of o's candidates
set ok to ((hcf(p1, q) is 1) and (hcf(p2, q) is 1))
if (ok) then -- q is coprime with both p1 and p2.
set end of o's output to q
set p1 to p2
set p2 to q
-- Remove q from the candidate list.
set item i of o's candidates to missing value
set o's candidates to o's candidates's numbers
set candidateCount to candidateCount - 1
set ok to (candidateCount > 0)
exit repeat
end if
end repeat
end repeat
 
return o's output
end coprimeTriplets
 
-- Task code:
return coprimeTriplets(49)
Output:
{1, 2, 3, 5, 4, 7, 9, 8, 11, 13, 6, 17, 19, 10, 21, 23, 16, 15, 29, 14, 25, 27, 22, 31, 35, 12, 37, 41, 18, 43, 47, 20, 33, 49, 26, 45}

Arturo[edit]

lst: [1 2]
 
while [true][
n: 3
prev2: lst\[dec dec size lst]
prev1: last lst
 
while -> any? @[
contains? lst n
1 <> gcd @[n prev2]
1 <> gcd @[n prev1]
] -> n: n + 1
 
if n >= 50 -> break
'lst ++ n
]
 
loop split.every:10 lst 'a ->
print map a => [pad to :string & 3]
Output:
  1   2   3   5   4   7   9   8  11  13 
  6  17  19  10  21  23  16  15  29  14 
 25  27  22  31  35  12  37  41  18  43 
 47  20  33  49  26  45

C[edit]

/*
************************
* *
* COPRIME TRIPLETS *
* *
************************
*/

/* Starting from the sequence a(1)=1 and a(2)=2 find the next smallest number
which is coprime to the last two predecessors and has not yet appeared in the
sequence.
p and q are coprimes if they have no common factors other than 1.
Let p, q < 50 */

 
#include <stdio.h>
 
int Gcd(int v1, int v2)
{
/* It evaluates the Greatest Common Divisor between v1 and v2 */
int a, b, r;
if (v1 < v2)
{
a = v2;
b = v1;
}
else
{
a = v1;
b = v2;
}
do
{
r = a % b;
if (r == 0)
{
break;
}
else
{
a = b;
b = r;
}
} while (1 == 1);
return b;
}
 
int NotInList(int num, int numtrip, int *tripletslist)
{
/* It indicates if the value num is already present in the list tripletslist of length numtrip */
for (int i = 0; i < numtrip; i++)
{
if (num == tripletslist[i])
{
return 0;
}
}
return 1;
}
 
int main()
{
int coprime[50];
int gcd1, gcd2;
int ntrip = 2;
int n = 3;
 
/* The first two values */
coprime[0] = 1;
coprime[1] = 2;
 
while ( n < 50)
{
gcd1 = Gcd(n, coprime[ntrip-1]);
gcd2 = Gcd(n, coprime[ntrip-2]);
/* if n is coprime of the previous two value
and it isn't already present in the list */

if (gcd1 == 1 && gcd2 == 1 && NotInList(n, ntrip, coprime))
{
coprime[ntrip++] = n;
/* It starts searching a new triplets */
n = 3;
}
else
{
/* Trying to find a triplet with the next value */
n++;
}
}
 
/* printing the list of coprime triplets */
printf("\n");
for (int i = 0; i < ntrip; i++)
{
printf("%2d ", coprime[i]);
if ((i+1) % 10 == 0)
{
printf("\n");
}
}
 
printf("\n\nNumber of elements in coprime triplets: %d\n\n", ntrip);
 
return 0;
}
Output:
 1  2  3  5  4  7  9  8 11 13
 6 17 19 10 21 23 16 15 29 14
25 27 22 31 35 12 37 41 18 43
47 20 33 49 26 45

Number of elements in coprime triplets: 36

Delphi[edit]

Translation of: Julia
 
program Coprime_triplets;
 
{$APPTYPE CONSOLE}
 
uses
System.SysUtils;
 
//https://rosettacode.org/wiki/Greatest_common_divisor#Pascal_.2F_Delphi_.2F_Free_Pascal
function Gcd(u, v: longint): longint;
begin
if v = 0 then
EXIT(u);
result := Gcd(v, u mod v);
end;
 
function IsIn(value: Integer; a: TArray<Integer>): boolean;
begin
for var e in a do
if e = value then
exit(true);
Result := false;
end;
 
function CoprimeTriplets(less_than: Integer = 50): TArray<Integer>;
var
cpt: TArray<Integer>;
_end: Integer;
begin
cpt := [1, 2];
_end := high(cpt);
 
while True do
begin
var m := 1;
while IsIn(m, cpt) or (gcd(m, cpt[_end]) <> 1) or (gcd(m, cpt[_end - 1]) <> 1) do
inc(m);
if m >= less_than then
exit(cpt);
SetLength(cpt, Length(cpt) + 1);
_end := high(cpt);
cpt[_end] := m;
end;
end;
 
begin
var trps := CoprimeTriplets();
writeln('Found ', length(trps), ' coprime triplets less than 50:');
for var i := 0 to High(trps) do
begin
write(trps[i]: 2, ' ');
if (i + 1) mod 10 = 0 then
writeln;
end;
{$IFNDEF UNIX} Readln; {$ENDIF}
end.

F#[edit]

 
// Coprime triplets: Nigel Galloway. May 12th., 2021
let rec fN g=function 0->g=1 |n->fN n (g%n)
let rec fG t n1 n2=seq{let n=seq{1..0x0FFFFFFF}|>Seq.find(fun n->not(List.contains n t) && fN n1 n && fN n2 n) in yield n; yield! cT(n::t) n2 n}
let cT=seq{yield 1; yield 2; yield! fG [1;2] 1 2}
cT|>Seq.takeWhile((>)50)|>Seq.iter(printf "%d "); printfn ""
 
Output:
1 2 3 5 4 7 9 8 11 13 6 17 19 10 21 23 16 15 29 14 25 27 22 31 35 12 37 41 18 43 47 20 33 49 26 45

Factor[edit]

Works with: Factor version 0.99 2021-02-05
USING: combinators.short-circuit.smart formatting grouping io
kernel make math prettyprint sequences sets ;
 
: coprime? ( m n -- ? ) simple-gcd 1 = ;
 
: coprime-both? ( m n o -- ? ) '[ _ coprime? ] both? ;
 
: triplet? ( hs m n o -- ? )
{ [ coprime-both? nip ] [ 2nip swap in? not ] } && ;
 
: next ( hs m n -- hs' m' n' )
0 [ 4dup triplet? ] [ 1 + ] until
nipd pick [ adjoin ] keepd ;
 
: (triplets-upto) ( n -- )
[ HS{ 1 2 } clone 1 , 1 2 ] dip
'[ 2dup [ _ < ] both? ] [ dup , next ] while 3drop ;
 
: triplets-upto ( n -- seq ) [ (triplets-upto) ] { } make ;
 
"Coprime triplets under 50:" print
50 triplets-upto
[ 9 group simple-table. nl ]
[ length "Found %d terms.\n" printf ] bi
Output:
Coprime triplets under 50:
1  2  3  5  4  7  9  8  11
13 6  17 19 10 21 23 16 15
29 14 25 27 22 31 35 12 37
41 18 43 47 20 33 49 26 45

Found 36 terms.

FreeBASIC[edit]

function gcd( a as uinteger, b as uinteger ) as uinteger
if b = 0 then return a
return gcd( b, a mod b )
end function
 
function num_in_array( array() as integer, num as integer ) as boolean
for i as uinteger = 1 to ubound(array)
if array(i) = num then return true
next i
return false
end function
 
redim as integer trips(1 to 2)
trips(1) = 1 : trips(2) = 2
dim as integer last
 
do
last = ubound(trips)
for q as integer = 1 to 49
if not num_in_array( trips(), q ) _
andalso gcd(q, trips(last)) = 1 _
andalso gcd(q, trips(last-1)) = 1 then
redim preserve as integer trips( 1 to last+1 )
trips(last+1) = q
continue do
end if
next q
exit do
loop
 
print using "Found ## terms:"; ubound(trips)
 
for i as integer = 1 to last
print trips(i);" ";
next i : print
Output:
Found 36 terms:
1  2  3  5  4  7  9  8  11  13  6  17  19  10  21  23  16  15  29  14  25  27  22  31  35  12  37  41  18  43  47  20  33  49  26  45

Go[edit]

Translation of: Wren
Library: Go-rcu
package main
 
import (
"fmt"
"rcu"
)
 
func contains(a []int, v int) bool {
for _, e := range a {
if e == v {
return true
}
}
return false
}
 
func main() {
const limit = 50
cpt := []int{1, 2}
for {
m := 1
l := len(cpt)
for contains(cpt, m) || rcu.Gcd(m, cpt[l-1]) != 1 || rcu.Gcd(m, cpt[l-2]) != 1 {
m++
}
if m >= limit {
break
}
cpt = append(cpt, m)
}
fmt.Printf("Coprime triplets under %d:\n", limit)
for i, t := range cpt {
fmt.Printf("%2d ", t)
if (i+1)%10 == 0 {
fmt.Println()
}
}
fmt.Printf("\n\nFound %d such numbers\n", len(cpt))
}
Output:
Coprime triplets under 50:
 1  2  3  5  4  7  9  8 11 13 
 6 17 19 10 21 23 16 15 29 14 
25 27 22 31 35 12 37 41 18 43 
47 20 33 49 26 45 

Found 36 such numbers

Haskell[edit]

import Data.List (find, transpose, unfoldr)
import Data.List.Split (chunksOf)
import qualified Data.Set as S
 
--------------------- COPRIME TRIPLES --------------------
 
coprimeTriples :: Integral a => [a]
coprimeTriples =
[1, 2] <> unfoldr go (S.fromList [1, 2], (1, 2))
where
go (seen, (a, b)) =
Just
(c, (S.insert c seen, (b, c)))
where
Just c =
find
( ((&&) . flip S.notMember seen)
<*> ((&&) . coprime a <*> coprime b)
)
[3 ..]
 
coprime :: Integral a => a -> a -> Bool
coprime a b = 1 == gcd a b
 
 
--------------------------- TEST -------------------------
main :: IO ()
main =
let xs = takeWhile (< 50) coprimeTriples
in putStrLn (show (length xs) <> " terms below 50:\n")
>> putStrLn
( spacedTable
justifyRight
(chunksOf 10 (show <$> xs))
)
 
 
-------------------------- FORMAT ------------------------
spacedTable ::
(Int -> Char -> String -> String) -> [[String]] -> String
spacedTable aligned rows =
unlines $
unwords
. zipWith
(`aligned` ' ')
(maximum . fmap length <$> transpose rows)
<$> rows
 
justifyRight :: Int -> Char -> String -> String
justifyRight n c = (drop . length) <*> (replicate n c <>)
Output:
36 terms below 50:

 1  2  3  5  4  7  9  8 11 13
 6 17 19 10 21 23 16 15 29 14
25 27 22 31 35 12 37 41 18 43
47 20 33 49 26 45

jq[edit]

Works with: jq

Works with gojq, the Go implementation of jq

Preliminaries

# jq optimizes the recursive call of _gcd in the following:
def gcd(a;b):
def _gcd:
if .[1] != 0 then [.[1], .[0] % .[1]] | _gcd else .[0] end;
[a,b] | _gcd ;
 
# Pretty-printing
def nwise($n):
def n: if length <= $n then . else .[0:$n] , (.[$n:] | n) end;
n;
 
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
 

The task

 
# Input: an upper bound greater than 2
# Output: the array of coprime triplets [1,2 ... n] where n is less than the upper bound
def coprime_triplets:
. as $less_than
| {cpt: [1, 2], m:0}
| until( .m >= $less_than;
.m = 1
| .cpt as $cpt
| until( (.m | IN($cpt[]) | not) and (gcd(.m; $cpt[-1]) == 1) and (gcd(.m; $cpt[-2]) == 1);
.m += 1 )
| .cpt = $cpt + [.m] )
| .cpt[:-1];
 
50 | coprime_triplets
| (nwise(10) | map(lpad(2)) | join(" "))
Output:
 1  2  3  5  4  7  9  8 11 13
 6 17 19 10 21 23 16 15 29 14
25 27 22 31 35 12 37 41 18 43
47 20 33 49 26 45

Julia[edit]

Translation of: Phix
function coprime_triplets(less_than = 50)
cpt = [1, 2]
while true
m = 1
while m in cpt || gcd(m, cpt[end]) != 1 || gcd(m, cpt[end - 1]) != 1
m += 1
end
m >= less_than && return cpt
push!(cpt, m)
end
end
 
trps = coprime_triplets()
println("Found $(length(trps)) coprime triplets less than 50:")
foreach(p -> print(rpad(p[2], 3), p[1] %10 == 0 ? "\n" : ""), enumerate(trps))
 
Output:

Found 36 coprime triplets less than 50: 1 2 3 5 4 7 9 8 11 13 6 17 19 10 21 23 16 15 29 14 25 27 22 31 35 12 37 41 18 43 47 20 33 49 26 45

Mathematica/Wolfram Language[edit]

ClearAll[NextTerm]
NextTerm[a_List] := Module[{pred1, pred2, cands},
{pred1, pred2} = Take[a, -2];
cands =
Select[Range[50], CoprimeQ[#, pred1] && CoprimeQ[#, pred2] &];
cands = Complement[cands, a];
If[Length[cands] > 0,
Append[a, First[cands]]
,
a
]
]
Nest[NextTerm, {1, 2}, 120]
Output:
{1, 2, 3, 5, 4, 7, 9, 8, 11, 13, 6, 17, 19, 10, 21, 23, 16, 15, 29, 14, 25, 27, 22, 31, 35, 12, 37, 41, 18, 43, 47, 20, 33, 49, 26, 45}

Nim[edit]

import math, strutils
 
var list = @[1, 2]
 
while true:
var n = 3
let prev2 = list[^2]
let prev1 = list[^1]
while n in list or gcd(n, prev2) != 1 or gcd(n, prev1) != 1:
inc n
if n >= 50: break
list.add n
 
echo list.join(" ")
Output:
1 2 3 5 4 7 9 8 11 13 6 17 19 10 21 23 16 15 29 14 25 27 22 31 35 12 37 41 18 43 47 20 33 49 26 45

Perl[edit]

Library: ntheory
use strict;
use warnings;
use feature <state say>;
use ntheory 'gcd';
use List::Util 'first';
use List::Lazy 'lazy_list';
use enum qw(False True);
use constant Inf => 1e5;
 
my $ct = lazy_list {
state @c = (1, 2);
state %seen = (1 => True, 2 => True);
state $min = 3;
my $g = $c[-2] * $c[-1];
my $n = first { !$seen{$_} and gcd($_,$g) == 1 } $min .. Inf;
$seen{$n} = True;
$min = first { !$seen{$_} } $min .. Inf;
push @c, $n;
shift @c
};
 
my @ct;
do { push @ct, $ct->next() } until $ct[-1] > 50; pop @ct;
say join ' ', @ct
Output:
1 2 3 5 4 7 9 8 11 13 6 17 19 10 21 23 16 15 29 14 25 27 22 31 35 12 37 41 18 43 47 20 33 49 26 45

Phix[edit]

function coprime_triplets(integer less_than=50)
    sequence cpt = {1,2}
    while true do
        integer m = 1
        while find(m,cpt) 
           or gcd(m,cpt[$])!=1
           or gcd(m,cpt[$-1])!=1 do
            m += 1
        end while
        if m>=less_than then exit end if
        cpt &= m
    end while
    return cpt
end function
sequence res = apply(true,sprintf,{{"%2d"},coprime_triplets()})
printf(1,"Found %d coprime triplets:\n%s\n",{length(res),join_by(res,1,10," ")})
Output:
Found 36 coprime triplets:
 1  2  3  5  4  7  9  8 11 13
 6 17 19 10 21 23 16 15 29 14
25 27 22 31 35 12 37 41 18 43
47 20 33 49 26 45

Python[edit]

 
########################
# #
# COPRIME TRIPLETS #
# #
########################
 
#Starting from the sequence a(1)=1 and a(2)=2 find the next smallest number
#which is coprime to the last two predecessors and has not yet appeared in
#the sequence.
#p and q are coprimes if they have no common factors other than 1.
#Let p, q < 50
 
#Function to find the Greatest Common Divisor between v1 and v2
def Gcd(v1, v2):
a, b = v1, v2
if (a < b):
a, b = v2, v1
r = 1
while (r != 0):
r = a % b
if (r != 0):
a = b
b = r
return b
 
#The first two values
a = [1, 2]
#The next value candidate to belong to a triplet
n = 3
 
while (n < 50):
gcd1 = Gcd(n, a[-1])
gcd2 = Gcd(n, a[-2])
 
#if n is coprime of the previous two value and isn't present in the list
if (gcd1 == 1 and gcd2 == 1 and not(n in a)):
#n is the next element of a triplet
a.append(n)
n = 3
else:
#searching a new triplet with the next value
n += 1
 
#printing the result
for i in range(0, len(a)):
if (i % 10 == 0):
print('')
print("%4d" % a[i], end = '');
 
 
print("\n\nNumber of elements in coprime triplets = " + str(len(a)), end = "\n")
 
Output:
   1   2   3   5   4   7   9   8  11  13
   6  17  19  10  21  23  16  15  29  14
  25  27  22  31  35  12  37  41  18  43
  47  20  33  49  26  45

Number of elements in coprime triplets = 36

Raku[edit]

my @coprime-triplets = 1, 2, {
state %seen = 1, True, 2, True;
state $min = 3;
my $g = $^a * $^b;
my $n = ($min .. *).first: { !%seen{$_} && ($_ gcd $g == 1) }
%seen{$n} = True;
if %seen.elems %% 100 { $min = ($min .. *).first: { !%seen{$_} } }
$n
}*;
 
put "Coprime triplets before first > 50:\n",
@coprime-triplets[^(@coprime-triplets.first: * > 50, :k)].batch(10)».fmt("%4d").join: "\n";
 
put "\nOr maybe, minimum Coprime triplets that encompass 1 through 50:\n",
@coprime-triplets[0..(@coprime-triplets.first: * == 42, :k)].batch(10)».fmt("%4d").join: "\n";
 
put "\nAnd for the heck of it: 1001st through 1050th Coprime triplet:\n",
@coprime-triplets[1000..1049].batch(10)».fmt("%4d").join: "\n";
Output:
Coprime triplets before first > 50:
   1    2    3    5    4    7    9    8   11   13
   6   17   19   10   21   23   16   15   29   14
  25   27   22   31   35   12   37   41   18   43
  47   20   33   49   26   45

Or maybe, minimum Coprime triplets that encompass 1 through 50:
   1    2    3    5    4    7    9    8   11   13
   6   17   19   10   21   23   16   15   29   14
  25   27   22   31   35   12   37   41   18   43
  47   20   33   49   26   45   53   28   39   55
  32   51   59   38   61   63   34   65   57   44
  67   69   40   71   73   24   77   79   30   83
  89   36   85   91   46   75   97   52   81   95
  56   87  101   50   93  103   58   99  107   62
 105  109   64  111  113   68  115  117   74  119
 121   48  125  127   42

And for the heck of it: 1001st through 1050th Coprime triplet:
 682 1293 1361  680 1287 1363  686 1299 1367  688
1305 1369  692 1311 1373  694 1317 1375  698 1323
1381  704 1329 1379  706 1335 1387  716 1341 1385
 712 1347 1391  700 1353 1399  710 1359 1393  718
1371 1397  722 1365 1403  724 1377 1405  728 1383

REXX[edit]

/*REXX program finds and display  coprime triplets  below a specified limit  (limit=50).*/
parse arg n cols . /*obtain optional arguments from the CL*/
if n=='' | n=="," then n= 50 /*Not specified? Then use the default.*/
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
w= max(3, length( commas(n) ) ) /*width of a number in any column. */
@copt= ' coprime triplets where N < ' commas(n)
if cols>0 then say ' index │'center(@copt, 1 + cols*(w+1) )
if cols>0 then say '───────┼'center("" , 1 + cols*(W+1), '─')
!.= 0; @.= !.; idx= 1; $= /*initialize some variables. */
do #=1
do j=1; if @.j then iterate /*J in list of coprime triplets? Skip.*/
if #<3 then leave /*First two entries not defined? Use it*/
a= # - 1; b= # - 2 /*get the last two indices of sequence.*/
if gcd(j, !.a)\==1 then iterate /*J not coprime with last number?*/
if gcd(j, !.b)\==1 then iterate /*" " " " penultimate " */
leave /*OK, we've found a new coprime triplet*/
end /*j*/
if j>=n then leave /*Have we exceeded the limit? Then quit*/
@.j= 1;  !.#= j /*flag a coprime triplet (two methods).*/
if cols==0 then iterate /*Not showing the numbers? Keep looking*/
$= $ right( commas(j), w) /*append coprime triplet to output list*/
if #//cols\==0 then iterate /*Is output line full? No, keep looking*/
say center(idx, 7)'│' substr($, 2); $= /*show output line of coprime triplets.*/
idx= idx + cols /*bump the index for the output line. */
end /*forever*/
 
if $\=='' then say center(idx, 7)'│' substr($, 2) /*show any residual output numbers*/
if cols>0 then say '───────┴'center("" , 1 + cols*(w+1), '─')
say
say 'Found ' commas(#-1) @copt
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
gcd: procedure; parse arg x,y; do until _==0; _= x//y; x= y; y= _; end; return x
output   when using the default inputs:
 index │    coprime triplets  where  N  <  50
───────┼─────────────────────────────────────────
   1   │   1   2   3   5   4   7   9   8  11  13
  11   │   6  17  19  10  21  23  16  15  29  14
  21   │  25  27  22  31  35  12  37  41  18  43
  31   │  47  20  33  49  26  45
───────┴─────────────────────────────────────────

Found  36  coprime triplets  where  N  <  50

Ring[edit]

 
see "working..." + nl
row = 2
numbers = 1:50
first = 1
second = 2
see "Coprime triplets are:" + nl
see "" + first + " " + second + " "
 
for n = 3 to len(numbers)
flag1 = 1
flag2 = 1
if first < numbers[n]
min = first
else
min = numbers[n]
ok
for m = 2 to min
if first%m = 0 and numbers[n]%m = 0
flag1 = 0
exit
ok
next
if second < numbers[n]
min = second
else
min = numbers[n]
ok
for m = 2 to min
if second%m = 0 and numbers[n]%m = 0
flag2 = 0
exit
ok
next
if flag1 = 1 and flag2 = 1
see "" + numbers[n] + " "
first = second
second = numbers[n]
del(numbers,n)
row = row+1
if row%10 = 0
see nl
ok
n = 2
ok
next
 
see nl + "Found " + row + " coprime triplets" + nl
see "done..." + nl
 
Output:
working...
Coprime triplets are:
1 2 3 5 4 7 9 8 11 13 
6 17 19 10 21 23 16 15 29 14 
25 27 22 31 35 12 37 41 18 43 
47 20 33 49 26 45 
Found 36 coprime triplets
done...

Ruby[edit]

list = [1, 2]
available = (1..50).to_a - list
 
loop do
i = available.index{|a| list.last(2).all?{|b| a.gcd(b) == 1}}
break if i.nil?
list << available.delete_at(i)
end
 
puts list.join(" ")
 
Output:
1 2 3 5 4 7 9 8 11 13 6 17 19 10 21 23 16 15 29 14 25 27 22 31 35 12 37 41 18 43 47 20 33 49 26 45

Sidef[edit]

func coprime_triplets(callback) {
 
var (
list = [1,2],
a = 1,
b = 2,
k = 3,
seen = Set()
)
 
loop {
for (var n = k; true; ++n) {
if (!seen.has(n) && is_coprime(n, a) && is_coprime(n, b)) {
 
list << n
seen << n
 
callback(list) && return list
 
(a, b) = (b, n)
 
while (seen.has(k)) {
seen.remove(k++)
}
 
break
}
}
}
}
 
say "Coprime triplets before first term is > 50:"
coprime_triplets({|list|
list.tail >= 50
}).first(-1).slices(10).each {%« '%4d' -> join(' ').say }
 
say "\nLeast Coprime triplets that encompass 1 through 50:"
coprime_triplets({|list|
list.sort.first(50) == @(1..50)
}).slices(10).each {%« '%4d' -> join(' ').say }
 
say "\n1001st through 1050th Coprime triplet:"
coprime_triplets({|list|
list.len == 1050
}).last(50).slices(10).each {%« '%4d' -> join(' ').say }
Output:
Coprime triplets before first term is > 50:
   1    2    3    5    4    7    9    8   11   13
   6   17   19   10   21   23   16   15   29   14
  25   27   22   31   35   12   37   41   18   43
  47   20   33   49   26   45

Least Coprime triplets that encompass 1 through 50:
   1    2    3    5    4    7    9    8   11   13
   6   17   19   10   21   23   16   15   29   14
  25   27   22   31   35   12   37   41   18   43
  47   20   33   49   26   45   53   28   39   55
  32   51   59   38   61   63   34   65   57   44
  67   69   40   71   73   24   77   79   30   83
  89   36   85   91   46   75   97   52   81   95
  56   87  101   50   93  103   58   99  107   62
 105  109   64  111  113   68  115  117   74  119
 121   48  125  127   42

1001st through 1050th Coprime triplet:
 682 1293 1361  680 1287 1363  686 1299 1367  688
1305 1369  692 1311 1373  694 1317 1375  698 1323
1381  704 1329 1379  706 1335 1387  716 1341 1385
 712 1347 1391  700 1353 1399  710 1359 1393  718
1371 1397  722 1365 1403  724 1377 1405  728 1383

Wren[edit]

Translation of: Phix
Library: Wren-math
Library: Wren-seq
Library: Wren-fmt
import "/math" for Int
import "/seq" for Lst
import "/fmt" for Fmt
 
var limit = 50
var cpt = [1, 2]
 
while (true) {
var m = 1
while (cpt.contains(m) || Int.gcd(m, cpt[-1]) != 1 || Int.gcd(m, cpt[-2]) != 1) {
m = m + 1
}
if (m >= limit) break
cpt.add(m)
}
System.print("Coprime triplets under %(limit):")
for (chunk in Lst.chunks(cpt, 10)) Fmt.print("$2d", chunk)
System.print("\nFound %(cpt.count) such numbers.")
Output:
Coprime triplets under 50:
 1  2  3  5  4  7  9  8 11 13
 6 17 19 10 21 23 16 15 29 14
25 27 22 31 35 12 37 41 18 43
47 20 33 49 26 45

Found 36 such numbers.