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EKG sequence convergence

From Rosetta Code
Task
EKG sequence convergence
You are encouraged to solve this task according to the task description, using any language you may know.

The sequence is from the natural numbers and is defined by:

  • a(1) = 1;
  • a(2) = Start = 2;
  • for n > 2, a(n) shares at least one prime factor with a(n-1) and is the smallest such natural number not already used.


The sequence is called the EKG sequence (after its visual similarity to an electrocardiogram when graphed).

Variants of the sequence can be generated starting 1, N where N is any natural number larger than one. For the purposes of this task let us call:

  • The sequence described above , starting 1, 2, ... the EKG(2) sequence;
  • the sequence starting 1, 3, ... the EKG(3) sequence;
  • ... the sequence starting 1, N, ... the EKG(N) sequence.


Convergence

If an algorithm that keeps track of the minimum amount of numbers and their corresponding prime factors used to generate the next term is used, then this may be known as the generators essential state. Two EKG generators with differing starts can converge to produce the same sequence after initial differences.
EKG(N1) and EKG(N2) are said to to have converged at and after generation a(c) if state_of(EKG(N1).a(c)) == state_of(EKG(N2).a(c)).


Task
  1. Calculate and show here the first 10 members of EKG(2).
  2. Calculate and show here the first 10 members of EKG(5).
  3. Calculate and show here the first 10 members of EKG(7).
  4. Calculate and show here the first 10 members of EKG(9).
  5. Calculate and show here the first 10 members of EKG(10).
  6. Calculate and show here at which term EKG(5) and EKG(7) converge   (stretch goal).
Related Tasks
  1. Greatest common divisor
  2. Sieve of Eratosthenes


Reference



C[edit]

Translation of: Go
#include <stdio.h>
#include <stdlib.h>
 
#define TRUE 1
#define FALSE 0
#define LIMIT 100
 
typedef int bool;
 
int compareInts(const void *a, const void *b) {
int aa = *(int *)a;
int bb = *(int *)b;
return aa - bb;
}
 
bool contains(int a[], int b, size_t len) {
int i;
for (i = 0; i < len; ++i) {
if (a[i] == b) return TRUE;
}
return FALSE;
}
 
int gcd(int a, int b) {
while (a != b) {
if (a > b)
a -= b;
else
b -= a;
}
return a;
}
 
bool areSame(int s[], int t[], size_t len) {
int i;
qsort(s, len, sizeof(int), compareInts);
qsort(t, len, sizeof(int), compareInts);
for (i = 0; i < len; ++i) {
if (s[i] != t[i]) return FALSE;
}
return TRUE;
}
 
int main() {
int s, n, i;
int starts[5] = {2, 5, 7, 9, 10};
int ekg[5][LIMIT];
for (s = 0; s < 5; ++s) {
ekg[s][0] = 1;
ekg[s][1] = starts[s];
for (n = 2; n < LIMIT; ++n) {
for (i = 2; ; ++i) {
// a potential sequence member cannot already have been used
// and must have a factor in common with previous member
if (!contains(ekg[s], i, n) && gcd(ekg[s][n - 1], i) > 1) {
ekg[s][n] = i;
break;
}
}
}
printf("EKG(%2d): [", starts[s]);
for (i = 0; i < 30; ++i) printf("%d ", ekg[s][i]);
printf("\b]\n");
}
 
// now compare EKG5 and EKG7 for convergence
for (i = 2; i < LIMIT; ++i) {
if (ekg[1][i] == ekg[2][i] && areSame(ekg[1], ekg[2], i)) {
printf("\nEKG(5) and EKG(7) converge at term %d\n", i + 1);
return 0;
}
}
printf("\nEKG5(5) and EKG(7) do not converge within %d terms\n", LIMIT);
return 0;
}
Output:
EKG( 2): [1 2 4 6 3 9 12 8 10 5 15 18 14 7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36]
EKG( 5): [1 5 10 2 4 6 3 9 12 8 14 7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG( 7): [1 7 14 2 4 6 3 9 12 8 10 5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32]
EKG( 9): [1 9 3 6 2 4 8 10 5 15 12 14 7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG(10): [1 10 2 4 6 3 9 12 8 14 7 21 15 5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32]

EKG(5) and EKG(7) converge at term 21

F#[edit]

The Function[edit]

This task uses Extensible Prime Generator (F#)

 
// Generate EKG Sequences. Nigel Galloway: December 6th., 2018
let EKG n=seq{
let fN,fG=let i=System.Collections.Generic.Dictionary<int,int>()
let fN g=(if not (i.ContainsKey g) then i.[g]<-g);(g,i.[g])
((fun e->i.[e]<-i.[e]+e), (fun l->l|>List.map fN))
let fU l= pCache|>Seq.takeWhile(fun n->n<=l)|>Seq.filter(fun n->l%n=0)|>List.ofSeq
let rec EKG l (α,β)=seq{let b=fU β in if (β=n||β<snd((fG b|>List.maxBy snd))) then fN α; yield! EKG l (fG l|>List.minBy snd)
else fN α;yield β;yield! EKG b (fG b|>List.minBy snd)}
yield! seq[1;n]; let g=fU n in yield! EKG g (fG g|>Seq.minBy snd)}
let EKGconv n g=Seq.zip(EKG n)(EKG g)|>Seq.skip 2|>Seq.scan(fun(n,i,g,e)(l,β)->(Set.add l n,Set.add β i,l,β))(set[1;n],set[1;g],0,0)|>Seq.takeWhile(fun(n,i,g,e)->g<>e||n<>i)
 

The Task[edit]

 
EKG 2 |> Seq.take 45 |> Seq.iter(printf "%2d, ")
EKG 3 |> Seq.take 45 |> Seq.iter(printf "%2d, ")
EKG 5 |> Seq.take 45 |> Seq.iter(printf "%2d, ")
EKG 7 |> Seq.take 45 |> Seq.iter(printf "%2d, ")
EKG 9 |> Seq.take 45 |> Seq.iter(printf "%2d, ")
EKG 10 |> Seq.take 45 |> Seq.iter(printf "%2d, ")
printfn "%d" (let n,_,_,_=EKGconv 2 5|>Seq.last in ((Set.count n)+1)
 
Output:
 1, 2, 4, 6, 3, 9,12, 8,10, 5,15,18,14, 7,21,24,16,20,22,11,33,27,30,25,35,28,26,13,39,36,32,34,17,51,42,38,19,57,45,40,44,46,23,69,48
 1, 3, 6, 2, 4, 8,10, 5,15, 9,12,14, 7,21,18,16,20,22,11,33,24,26,13,39,27,30,25,35,28,32,34,17,51,36,38,19,57,42,40,44,46,23,69,45,48
 1, 5,10, 2, 4, 6, 3, 9,12, 8,14, 7,21,15,18,16,20,22,11,33,24,26,13,39,27,30,25,35,28,32,34,17,51,36,38,19,57,42,40,44,46,23,69,45,48
45

Extra Credit[edit]

 
prıntfn "%d" (EKG 2|>Seq.takeWhile(fun n->n<>104729) ((Seq.length n)+1)
 
Output:
203786
Real: 00:10:21.967, CPU: 00:10:25.300, GC gen0: 65296, gen1: 1

Factor[edit]

Works with: Factor version 0.99 2019-10-06
USING: combinators.short-circuit formatting fry io kernel lists
lists.lazy math math.statistics prettyprint sequences
sequences.generalizations ;
 
: ekg? ( n seq -- ? )
{ [ member? not ] [ last gcd nip 1 > ] } 2&& ;
 
: (ekg) ( seq -- seq' )
2 lfrom over [ ekg? ] curry lfilter car suffix! ;
 
: ekg ( n limit -- seq )
[ 1 ] [ V{ } 2sequence ] [ 2 - [ (ekg) ] times ] tri* ;
 
: show-ekgs ( seq n -- )
'[ dup _ ekg "EKG(%d) = %[%d, %]\n" printf ] each ;
 
: converge-at ( n m max -- o )
tuck [ ekg [ cum-sum ] [ rest-slice ] bi ] [email protected]
[ swapd [ = ] [email protected] and ] 4 nfind 4drop dup [ 2 + ] when ;
 
{ 2 5 7 9 10 } 20 show-ekgs nl
"EKG(5) and EKG(7) converge at term " write
5 7 100 converge-at .
Output:
EKG(2) = { 1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 22, 11 }
EKG(5) = { 1, 5, 10, 2, 4, 6, 3, 9, 12, 8, 14, 7, 21, 15, 18, 16, 20, 22, 11, 33 }
EKG(7) = { 1, 7, 14, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 16, 20, 22, 11, 33, 21 }
EKG(9) = { 1, 9, 3, 6, 2, 4, 8, 10, 5, 15, 12, 14, 7, 21, 18, 16, 20, 22, 11, 33 }
EKG(10) = { 1, 10, 2, 4, 6, 3, 9, 12, 8, 14, 7, 21, 15, 5, 20, 16, 18, 22, 11, 33 }

EKG(5) and EKG(7) converge at term 21

Go[edit]

package main
 
import (
"fmt"
"sort"
)
 
func contains(a []int, b int) bool {
for _, j := range a {
if j == b {
return true
}
}
return false
}
 
func gcd(a, b int) int {
for a != b {
if a > b {
a -= b
} else {
b -= a
}
}
return a
}
 
func areSame(s, t []int) bool {
le := len(s)
if le != len(t) {
return false
}
sort.Ints(s)
sort.Ints(t)
for i := 0; i < le; i++ {
if s[i] != t[i] {
return false
}
}
return true
}
 
func main() {
const limit = 100
starts := [5]int{2, 5, 7, 9, 10}
var ekg [5][limit]int
 
for s, start := range starts {
ekg[s][0] = 1
ekg[s][1] = start
for n := 2; n < limit; n++ {
for i := 2; ; i++ {
// a potential sequence member cannot already have been used
// and must have a factor in common with previous member
if !contains(ekg[s][:n], i) && gcd(ekg[s][n-1], i) > 1 {
ekg[s][n] = i
break
}
}
}
fmt.Printf("EKG(%2d): %v\n", start, ekg[s][:30])
}
 
// now compare EKG5 and EKG7 for convergence
for i := 2; i < limit; i++ {
if ekg[1][i] == ekg[2][i] && areSame(ekg[1][:i], ekg[2][:i]) {
fmt.Println("\nEKG(5) and EKG(7) converge at term", i+1)
return
}
}
fmt.Println("\nEKG5(5) and EKG(7) do not converge within", limit, "terms")
}
Output:
EKG( 2): [1 2 4 6 3 9 12 8 10 5 15 18 14 7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36]
EKG( 5): [1 5 10 2 4 6 3 9 12 8 14 7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG( 7): [1 7 14 2 4 6 3 9 12 8 10 5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32]
EKG( 9): [1 9 3 6 2 4 8 10 5 15 12 14 7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG(10): [1 10 2 4 6 3 9 12 8 14 7 21 15 5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32]

EKG(5) and EKG(7) converge at term 21

Haskell[edit]

import Data.List (findIndex, isPrefixOf, tails)
import Data.Maybe (fromJust)
 
seqEKGRec :: Int -> Int -> [Int] -> [Int]
seqEKGRec _ 0 l = l
seqEKGRec k n [] = seqEKGRec k (n - 2) [k, 1]
seqEKGRec k n l@(h:t) =
seqEKGRec
k
(n - 1)
(head (filter (\i -> notElem i l && (gcd h i > 1)) [2 ..]) : l)
 
seqEKG :: Int -> Int -> [Int]
seqEKG k n = reverse (seqEKGRec k n [])
 
main :: IO ()
main =
mapM_
(\x ->
putStr "EKG (" >> (putStr . show $ x) >> putStr ") is " >>
print (seqEKG x 20))
[2, 5, 7, 9, 10] >>
putStr "EKG(5) and EKG(7) converge at " >>
print
((+ 1) $
fromJust $
findIndex
(isPrefixOf (replicate 20 True))
(tails (zipWith (==) (seqEKG 7 80) (seqEKG 5 80))))
Output:
EKG (2) is [1,2,4,6,3,9,12,8,10,5,15,18,14,7,21,24,16,20,22,11]
EKG (5) is [1,5,10,2,4,6,3,9,12,8,14,7,21,15,18,16,20,22,11,33]
EKG (7) is [1,7,14,2,4,6,3,9,12,8,10,5,15,18,16,20,22,11,33,21]
EKG (9) is [1,9,3,6,2,4,8,10,5,15,12,14,7,21,18,16,20,22,11,33]
EKG (10) is [1,10,2,4,6,3,9,12,8,14,7,21,15,5,20,16,18,22,11,33]
EKG(5) and EKG(7) converge at 21

J[edit]

 
Until =: 2 :'u^:(0-:v)^:_' NB. unused but so fun
prime_factors_of_tail =: [email protected]:q:@:{:
numbers_not_in_list =: -.~ >:@:[email protected]:(>./)
 
 
ekg =: 3 :0 NB. return next sequence
if. 1 = # y do. NB. initialize
1 , y
return.
end.
a =. prime_factors_of_tail y
b =. numbers_not_in_list y
index_of_lowest =. {. _ ,~ I. 1 e."1 a e."1 q:b
if. index_of_lowest < _ do. NB. if the list doesn't need extension
y , index_of_lowest { b
return.
end.
NB. otherwise extend the list
b =. >: >./ y
while. 1 [email protected]:e. a e. q: b do.
b =. >: b
end.
y , b
)
 
ekg^:9&>2 5 7 9 10
1 2 4 6 3 9 12 8 10 5
1 5 10 2 4 6 3 9 12 8
1 7 14 2 4 6 3 9 12 8
1 9 3 6 2 4 8 10 5 15
1 10 2 4 6 3 9 12 8 14
 
 
assert 9 -: >:Until(>&8) 2
assert (,2) -: prime_factors_of_tail 6 8 NB. (nub of)
assert 3 4 5 -: numbers_not_in_list 1 2 6
 

Somewhat shorter is ekg2,

 
index_of_lowest =: [: {. _ ,~ [: I. 1 e."1 prime_factors_of_tail e."1 q:@:numbers_not_in_list
 
g =: 3 :0 NB. return sequence with next term appended
a =. prime_factors_of_tail y
(, (index_of_lowest { numbers_not_in_list)`(([: >:Until(1 e. a e. q:) [: >: >./))@.(_ = index_of_lowest)) y
)
 
ekg2 =: (1&,)`[email protected](1<#)
 
assert (3 -: index_of_lowest { numbers_not_in_list)1 2 4 6
 
assert (ekg^:9&> -: ekg2^:9&>) 2 5 7 9 10
 

Java[edit]

 
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
 
public class EKGSequenceConvergence {
 
public static void main(String[] args) {
System.out.println("Calculate and show here the first 10 members of EKG[2], EKG[5], EKG[7], EKG[9] and EKG[10].");
for ( int i : new int[] {2, 5, 7, 9, 10} ) {
System.out.printf("EKG[%d] = %s%n", i, ekg(i, 10));
}
System.out.println("Calculate and show here at which term EKG[5] and EKG[7] converge.");
List<Integer> ekg5 = ekg(5, 100);
List<Integer> ekg7 = ekg(7, 100);
for ( int i = 1 ; i < ekg5.size() ; i++ ) {
if ( ekg5.get(i) == ekg7.get(i) && sameSeq(ekg5, ekg7, i)) {
System.out.printf("EKG[%d](%d) = EKG[%d](%d) = %d, and are identical from this term on%n", 5, i+1, 7, i+1, ekg5.get(i));
break;
}
}
}
 
// Same last element, and all elements in sequence are identical
private static boolean sameSeq(List<Integer> seq1, List<Integer> seq2, int n) {
List<Integer> list1 = new ArrayList<>(seq1.subList(0, n));
Collections.sort(list1);
List<Integer> list2 = new ArrayList<>(seq2.subList(0, n));
Collections.sort(list2);
for ( int i = 0 ; i < n ; i++ ) {
if ( list1.get(i) != list2.get(i) ) {
return false;
}
}
return true;
}
 
// Without HashMap to identify seen terms, need to examine list.
// Calculating 3000 terms in this manner takes 10 seconds
// With HashMap to identify the seen terms, calculating 3000 terms takes .1 sec.
private static List<Integer> ekg(int two, int maxN) {
List<Integer> result = new ArrayList<>();
result.add(1);
result.add(two);
Map<Integer,Integer> seen = new HashMap<>();
seen.put(1, 1);
seen.put(two, 1);
int minUnseen = two == 2 ? 3 : 2;
int prev = two;
for ( int n = 3 ; n <= maxN ; n++ ) {
int test = minUnseen - 1;
while ( true ) {
test++;
if ( ! seen.containsKey(test) && gcd(test, prev) > 1 ) {
 
result.add(test);
seen.put(test, n);
prev = test;
if ( minUnseen == test ) {
do {
minUnseen++;
} while ( seen.containsKey(minUnseen) );
}
break;
}
}
}
return result;
}
 
private static final int gcd(int a, int b) {
if ( b == 0 ) {
return a;
}
return gcd(b, a%b);
}
 
}
 
Output:
Calculate and show here the first 10 members of EKG[2], EKG[5], EKG[7], EKG[9] and EKG[10].
EKG[2] = [1, 2, 4, 6, 3, 9, 12, 8, 10, 5]
EKG[5] = [1, 5, 10, 2, 4, 6, 3, 9, 12, 8]
EKG[7] = [1, 7, 14, 2, 4, 6, 3, 9, 12, 8]
EKG[9] = [1, 9, 3, 6, 2, 4, 8, 10, 5, 15]
EKG[10] = [1, 10, 2, 4, 6, 3, 9, 12, 8, 14]
Calculate and show here at which term EKG[5] and EKG[7] converge.
EKG[5](21) = EKG[7](21) = 24, and are identical from this term on

jq[edit]

Adapted from Wren

Works with: jq

Works with gojq, the Go implementation of jq

A very small point of interest is that the appropriate width for printing the results neatly is determined dynamically based on the entire set of sequences.

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 ;
 
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
 
 

The Task

 
def areSame($s; $t):
($s|length) == ($t|length) and ($s|sort) == ($t|sort);
 
def task:
 
# compare EKG5 and EKG7 for convergence, assuming . has been constructed appropriately:
def compare:
first( range(2; .limit) as $i
| select(.ekg[1][$i] == .ekg[2][$i] and areSame(.ekg[1][0:$i]; .ekg[2][0:$i]))
| "\nEKG(5) and EKG(7) converge at term \($i+1)." )
// "\nEKG5(5) and EKG(7) do not converge within \(.limit) terms." ;
 
{ limit: 100,
starts: [2, 5, 7, 9, 10],
ekg: [],
width: 0 # keep track of the number of characters required to print the results neatly
}
| reduce range(0;4) as $i (.; .ekg[$i] = [range(0; .limit) | 0] )
| reduce range(0; .starts|length ) as $s (.;
.starts[$s] as $start
| .ekg[$s][0] = 1
| .ekg[$s][1] = $start
| reduce range( 2; .limit) as $n (.;
.i = 2
| .stop = false
| until( .stop;
# a potential sequence member cannot already have been used
# and must have a factor in common with previous member
.ekg[$s] as $ekg
| if (.i | IN( $ekg[0:$n][]) | not) and gcd($ekg[$n-1]; .i) > 1
then .ekg[$s][$n] = .i
| .width = ([.width, (.i|tostring|length)] | max)
| .stop = true
else .
end
| .i += 1) ) )
 
# Read out the results of interest:
| (range(0; .starts|length ) as $s
| .width as $width
| "EKG(\(.starts[$s]|lpad(2))): \(.ekg[$s][0:30]|map(lpad($width))|join(" "))" ),
compare
 ;
 
task
Output:
EKG( 2):   1   2   4   6   3   9  12   8  10   5  15  18  14   7  21  24  16  20  22  11  33  27  30  25  35  28  26  13  39  36
EKG( 5):   1   5  10   2   4   6   3   9  12   8  14   7  21  15  18  16  20  22  11  33  24  26  13  39  27  30  25  35  28  32
EKG( 7):   1   7  14   2   4   6   3   9  12   8  10   5  15  18  16  20  22  11  33  21  24  26  13  39  27  30  25  35  28  32
EKG( 9):   1   9   3   6   2   4   8  10   5  15  12  14   7  21  18  16  20  22  11  33  24  26  13  39  27  30  25  35  28  32
EKG(10):   1  10   2   4   6   3   9  12   8  14   7  21  15   5  20  16  18  22  11  33  24  26  13  39  27  30  25  35  28  32

EKG(5) and EKG(7) converge at term 21.

Julia[edit]

Translation of: Perl
using Primes
 
function ekgsequence(n, limit)
ekg::Array{Int,1} = [1, n]
while length(ekg) < limit
for i in 2:2<<18
if all(j -> j != i, ekg) && gcd(ekg[end], i) > 1
push!(ekg, i)
break
end
end
end
ekg
end
 
function convergeat(n, m, max = 100)
ekgn = ekgsequence(n, max)
ekgm = ekgsequence(m, max)
for i in 3:max
if ekgn[i] == ekgm[i] && sum(ekgn[1:i+1]) == sum(ekgm[1:i+1])
return i
end
end
warn("no converge in $max terms")
end
 
[println(rpad("EKG($i): ", 9), join(ekgsequence(i, 30), " ")) for i in [2, 5, 7, 9, 10]]
println("EKGs of 5 & 7 converge at term ", convergeat(5, 7))
 
Output:

EKG(2): 1 2 4 6 3 9 12 8 10 5 15 18 14 7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36 EKG(5): 1 5 10 2 4 6 3 9 12 8 14 7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32 EKG(7): 1 7 14 2 4 6 3 9 12 8 10 5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32 EKG(9): 1 9 3 6 2 4 8 10 5 15 12 14 7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32 EKG(10): 1 10 2 4 6 3 9 12 8 14 7 21 15 5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32 EKGs of 5 & 7 converge at term 21

Kotlin[edit]

Translation of: Go
// Version 1.2.60
 
fun gcd(a: Int, b: Int): Int {
var aa = a
var bb = b
while (aa != bb) {
if (aa > bb)
aa -= bb
else
bb -= aa
}
return aa
}
 
const val LIMIT = 100
 
fun main(args: Array<String>) {
val starts = listOf(2, 5, 7, 9, 10)
val ekg = Array(5) { IntArray(LIMIT) }
 
for ((s, start) in starts.withIndex()) {
ekg[s][0] = 1
ekg[s][1] = start
for (n in 2 until LIMIT) {
var i = 2
while (true) {
// a potential sequence member cannot already have been used
// and must have a factor in common with previous member
if (!ekg[s].slice(0 until n).contains(i) &&
gcd(ekg[s][n - 1], i) > 1) {
ekg[s][n] = i
break
}
i++
}
}
System.out.printf("EKG(%2d): %s\n", start, ekg[s].slice(0 until 30))
}
 
// now compare EKG5 and EKG7 for convergence
for (i in 2 until LIMIT) {
if (ekg[1][i] == ekg[2][i] &&
ekg[1].slice(0 until i).sorted() == ekg[2].slice(0 until i).sorted()) {
println("\nEKG(5) and EKG(7) converge at term ${i + 1}")
return
}
}
println("\nEKG5(5) and EKG(7) do not converge within $LIMIT terms")
}
Output:
EKG( 2): [1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 22, 11, 33, 27, 30, 25, 35, 28, 26, 13, 39, 36]
EKG( 5): [1, 5, 10, 2, 4, 6, 3, 9, 12, 8, 14, 7, 21, 15, 18, 16, 20, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32]
EKG( 7): [1, 7, 14, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 16, 20, 22, 11, 33, 21, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32]
EKG( 9): [1, 9, 3, 6, 2, 4, 8, 10, 5, 15, 12, 14, 7, 21, 18, 16, 20, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32]
EKG(10): [1, 10, 2, 4, 6, 3, 9, 12, 8, 14, 7, 21, 15, 5, 20, 16, 18, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32]

EKG(5) and EKG(7) converge at term 21

Mathematica/Wolfram Language[edit]

ClearAll[NextInSequence, EKGSequence]
NextInSequence[seq_List] := Module[{last, new = 1, holes, max, sel, found, i},
last = Last[seq];
max = Max[seq];
holes = Complement[Range[max], seq];
sel = SelectFirst[holes, Not[CoprimeQ[last, #]] &];
If[MissingQ[sel],
i = max;
found = False;
While[! found,
i++;
If[Not[CoprimeQ[last, i]],
found = True
]
];
Append[seq, i]
,
Append[seq, sel]
]
]
EKGSequence[start_Integer, n_] := Nest[NextInSequence, {1, start}, n - 2]
 
Table[EKGSequence[s, 10], {s, {2, 5, 7, 9, 10}}] // Grid
 
s = Reverse[Transpose[{EKGSequence[5, 1000], EKGSequence[7, 1000]}]];
len = LengthWhile[s, Apply[Equal]];
s //= Reverse[Drop[#, len]] &;
Length[s] + 1
Output:
1	2	4	6	3	9	12	8	10	5
1	5	10	2	4	6	3	9	12	8
1	7	14	2	4	6	3	9	12	8
1	9	3	6	2	4	8	10	5	15
1	10	2	4	6	3	9	12	8	14

21

Nim[edit]

import algorithm, math, sets, strformat, strutils
 
#---------------------------------------------------------------------------------------------------
 
iterator ekg(n, limit: Positive): (int, int) =
var values: HashSet[int]
doAssert n >= 2
yield (1, 1)
yield (2, n)
values.incl(n)
var i = 3
var prev = n
while i <= limit:
var val = 2
while true:
if val notin values and gcd(val, prev) != 1:
values.incl(val)
yield (i, val)
prev = val
break
inc val
inc i
 
#---------------------------------------------------------------------------------------------------
 
for n in [2, 5, 7, 9, 10]:
var result: array[1..10, int]
for i, val in ekg(n, 10): result[i] = val
let title = fmt"EKG({n}):"
echo fmt"{title:8} {result.join("", "")}"
 
var ekg5, ekg7: array[1..100, int]
for i, val in ekg(5, 100): ekg5[i] = val
for i, val in ekg(7, 100): ekg7[i] = val
var convIndex = 0
for i in 2..100:
if ekg5[i] == ekg7[i] and sorted(ekg5[1..<i]) == sorted(ekg7[1..<i]):
convIndex = i
break
if convIndex > 0:
echo fmt"EKG(5) and EKG(7) converge at index {convIndex}."
else:
echo "No convergence found in the first {convIndex} terms."
Output:
EKG(2):  1, 2, 4, 6, 3, 9, 12, 8, 10, 5
EKG(5):  1, 5, 10, 2, 4, 6, 3, 9, 12, 8
EKG(7):  1, 7, 14, 2, 4, 6, 3, 9, 12, 8
EKG(9):  1, 9, 3, 6, 2, 4, 8, 10, 5, 15
EKG(10): 1, 10, 2, 4, 6, 3, 9, 12, 8, 14
EKG(5) and EKG(7) converge at index 21.

Perl[edit]

Translation of: Raku
use List::Util qw(none sum);
 
sub gcd { my ($u,$v) = @_; $v ? gcd($v, $u%$v) : abs($u) }
sub shares_divisors_with { gcd( $_[0], $_[1]) > 1 }
 
sub EKG {
my($n,$limit) = @_;
my @ekg = (1, $n);
while (@ekg < $limit) {
for my $i (2..1e18) {
next unless none { $_ == $i } @ekg and shares_divisors_with($ekg[-1], $i);
push(@ekg, $i) and last;
}
}
@ekg;
}
 
sub converge_at {
my($n1,$n2) = @_;
my $max = 100;
my @ekg1 = EKG($n1,$max);
my @ekg2 = EKG($n2,$max);
do { return $_+1 if $ekg1[$_] == $ekg2[$_] && sum(@ekg1[0..$_]) == sum(@ekg2[0..$_])} for 2..$max;
return "(no convergence in $max terms)";
}
 
print "EKG($_): " . join(' ', EKG($_,10)) . "\n" for 2, 5, 7, 9, 10;
print "EKGs of 5 & 7 converge at term " . converge_at(5, 7) . "\n"
Output:
EKG(2): 1 2 4 6 3 9 12 8 10 5
EKG(5): 1 5 10 2 4 6 3 9 12 8
EKG(7): 1 7 14 2 4 6 3 9 12 8
EKG(9): 1 9 3 6 2 4 8 10 5 15
EKG(10): 1 10 2 4 6 3 9 12 8 14
EKGs of 5 & 7 converge at term 21

Phix[edit]

Translation of: C
constant LIMIT = 100
constant starts = {2, 5, 7, 9, 10}
sequence ekg = {}
string fmt = "EKG(%2d): ["&join(repeat("%d",min(LIMIT,30))," ")&"]\n"
for s=1 to length(starts) do
ekg = append(ekg,{1,starts[s]}&repeat(0,LIMIT-2))
for n=3 to LIMIT do
-- a potential sequence member cannot already have been used
-- and must have a factor in common with previous member
integer i = 2
while find(i,ekg[s])
or gcd(ekg[s][n-1],i)<=1 do
i += 1
end while
ekg[s][n] = i
end for
printf(1,fmt,starts[s]&ekg[s][1..min(LIMIT,30)])
end for
 
-- now compare EKG5 and EKG7 for convergence
constant EKG5 = find(5,starts),
EKG7 = find(7,starts)
string msg = sprintf("do not converge within %d terms", LIMIT)
for i=3 to LIMIT do
if ekg[EKG5][i]=ekg[EKG7][i]
and sort(ekg[EKG5][1..i-1])=sort(ekg[EKG7][1..i-1]) then
msg = sprintf("converge at term %d", i)
exit
end if
end for
printf(1,"\nEKG5(5) and EKG(7) %s\n", msg)
Output:
EKG( 2): [1 2 4 6 3 9 12 8 10 5 15 18 14 7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36]
EKG( 5): [1 5 10 2 4 6 3 9 12 8 14 7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG( 7): [1 7 14 2 4 6 3 9 12 8 10 5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32]
EKG( 9): [1 9 3 6 2 4 8 10 5 15 12 14 7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32]
EKG(10): [1 10 2 4 6 3 9 12 8 14 7 21 15 5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32]

EKG5(5) and EKG(7) converge at term 21

Python[edit]

Python: Using math.gcd[edit]

If this alternate definition of function EKG_gen is used then the output would be the same as above. Instead of keeping a cache of prime factors this calculates the gretest common divisor as needed.

from itertools import count, islice, takewhile
from math import gcd
 
def EKG_gen(start=2):
"""\
Generate the next term of the EKG together with the minimum cache of
numbers left in its production; (the "state" of the generator).
Using math.gcd
"""

c = count(start + 1)
last, so_far = start, list(range(2, start))
yield 1, []
yield last, []
while True:
for index, sf in enumerate(so_far):
if gcd(last, sf) > 1:
last = so_far.pop(index)
yield last, so_far[::]
break
else:
so_far.append(next(c))
 
def find_convergence(ekgs=(5,7)):
"Returns the convergence point or zero if not found within the limit"
ekg = [EKG_gen(n) for n in ekgs]
for e in ekg:
next(e) # skip initial 1 in each sequence
return 2 + len(list(takewhile(lambda state: not all(state[0] == s for s in state[1:]),
zip(*ekg))))
 
if __name__ == '__main__':
for start in 2, 5, 7, 9, 10:
print(f"EKG({start}):", str([n[0] for n in islice(EKG_gen(start), 10)])[1: -1])
print(f"\nEKG(5) and EKG(7) converge at term {find_convergence(ekgs=(5,7))}!")
Output:

(Same as above).

EKG(2): 1, 2, 4, 6, 3, 9, 12, 8, 10, 5
EKG(5): 1, 5, 10, 2, 4, 6, 3, 9, 12, 8
EKG(7): 1, 7, 14, 2, 4, 6, 3, 9, 12, 8
EKG(9): 1, 9, 3, 6, 2, 4, 8, 10, 5, 15
EKG(10): 1, 10, 2, 4, 6, 3, 9, 12, 8, 14

EKG(5) and EKG(7) converge at term 21!
Note

Despite EKG(5) and EKG(7) seeming to converge earlier, as seen above; their hidden states differ.
Here is those series out to 21 terms where you can see them diverge again before finally converging. The state is also shown.

# After running the above, in the terminal:
from pprint import pprint as pp
 
for start in 5, 7:
print(f"EKG({start}):\n[(<next>, [<state>]), ...]")
pp(([n for n in islice(EKG_gen(start), 21)]))

Generates:

EKG(5):
[(<next>, [<state>]), ...]
[(1, []),
 (5, []),
 (10, [2, 3, 4, 6, 7, 8, 9]),
 (2, [3, 4, 6, 7, 8, 9]),
 (4, [3, 6, 7, 8, 9]),
 (6, [3, 7, 8, 9]),
 (3, [7, 8, 9]),
 (9, [7, 8]),
 (12, [7, 8, 11]),
 (8, [7, 11]),
 (14, [7, 11, 13]),
 (7, [11, 13]),
 (21, [11, 13, 15, 16, 17, 18, 19, 20]),
 (15, [11, 13, 16, 17, 18, 19, 20]),
 (18, [11, 13, 16, 17, 19, 20]),
 (16, [11, 13, 17, 19, 20]),
 (20, [11, 13, 17, 19]),
 (22, [11, 13, 17, 19]),
 (11, [13, 17, 19]),
 (33, [13, 17, 19, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]),
 (24, [13, 17, 19, 23, 25, 26, 27, 28, 29, 30, 31, 32])]
EKG(7):
[(<next>, [<state>]), ...]
[(1, []),
 (7, []),
 (14, [2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13]),
 (2, [3, 4, 5, 6, 8, 9, 10, 11, 12, 13]),
 (4, [3, 5, 6, 8, 9, 10, 11, 12, 13]),
 (6, [3, 5, 8, 9, 10, 11, 12, 13]),
 (3, [5, 8, 9, 10, 11, 12, 13]),
 (9, [5, 8, 10, 11, 12, 13]),
 (12, [5, 8, 10, 11, 13]),
 (8, [5, 10, 11, 13]),
 (10, [5, 11, 13]),
 (5, [11, 13]),
 (15, [11, 13]),
 (18, [11, 13, 16, 17]),
 (16, [11, 13, 17]),
 (20, [11, 13, 17, 19]),
 (22, [11, 13, 17, 19, 21]),
 (11, [13, 17, 19, 21]),
 (33, [13, 17, 19, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]),
 (21, [13, 17, 19, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32]),
 (24, [13, 17, 19, 23, 25, 26, 27, 28, 29, 30, 31, 32])]

Raku[edit]

(formerly Perl 6)

Works with: Rakudo Star version 2018.04.1
sub infix:<shares-divisors-with> { ($^a gcd $^b) > 1 }
 
sub next-EKG ( *@s ) {
return first {
@s$_ and @s.tail shares-divisors-with $_
}, 2..*;
}
 
sub EKG ( Int $start ) { 1, $start, &next-EKG … * }
 
sub converge-at ( @ints ) {
my @ekgs = @ints.map: &EKG;
 
return (2 .. *).first: -> $i {
[==] @ekgs.map( *.[$i] ) and
[===] @ekgs.map( *.head($i).Set )
}
}
 
say "EKG($_): ", .&EKG.head(10) for 2, 5, 7, 9, 10;
 
for [5, 7], [2, 5, 7, 9, 10] -> @ints {
say "EKGs of (@ints[]) converge at term {$_+1}" with converge-at(@ints);
}
Output:
EKG(2): (1 2 4 6 3 9 12 8 10 5)
EKG(5): (1 5 10 2 4 6 3 9 12 8)
EKG(7): (1 7 14 2 4 6 3 9 12 8)
EKG(9): (1 9 3 6 2 4 8 10 5 15)
EKG(10): (1 10 2 4 6 3 9 12 8 14)
EKGs of (5 7) converge at term 21
EKGs of (2 5 7 9 10) converge at term 45

REXX[edit]

/*REXX program can  generate and display several  EKG  sequences  (with various starts).*/
parse arg nums start /*obtain optional arguments from the CL*/
if nums=='' | nums=="," then nums= 50 /*Not specified? Then use the default.*/
if start= '' | start= "," then start=2 5 7 9 10 /* " " " " " " */
 
do s=1 for words(start); $= /*step through the specified STARTs. */
second= word(start, s); say /*obtain the second integer in the seq.*/
 
do j=1 for nums
if j<3 then do; #=1; if j==2 then #=second; end /*handle 1st & 2nd number*/
else #= ekg(#)
$= $ right(#, max(2, length(#) ) ) /*append the EKG integer to the $ list.*/
end /*j*/ /* [↑] the RIGHT BIF aligns the numbers*/
say '(start' right(second, max(2, length(second) ) )"):"$ /*display EKG seq.*/
end /*s*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
add_: do while z//j == 0; z=z%j; _=_ j; w=w+1; end; return strip(_)
/*──────────────────────────────────────────────────────────────────────────────────────*/
ekg: procedure expose $; parse arg x 1 z,,_
w=0 /*W: number of factors.*/
do k=1 to 11 by 2; j=k; if j==1 then j=2 /*divide by low primes. */
if j==9 then iterate; call add_ /*skip ÷ 9; add to list.*/
end /*k*/
/*↓ skips multiples of 3*/
do y=0 by 2; j= j + 2 + y//4 /*increment J by 2 or 4.*/
parse var j '' -1 r; if r==5 then iterate /*divisible by five ? */
if j*j>x | j>z then leave /*passed the sqrt(x) ? */
_= add_() /*add a factor to list. */
end /*y*/
j=z; if z\==1 then _= add_() /*Z¬=1? Then add──►list.*/
if _='' then _=x /*Null? Then use prime. */
do j=3; done=1
do k=1 for w
if j // word(_, k)==0 then do; done=0; leave; end
end /*k*/
if done then iterate
if wordpos(j, $)==0 then return j /*return an EKG integer.*/
end /*j*/
output   when using the default inputs:
(start  2):  1  2  4  6  3  9 12  8 10  5 15 18 14  7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36 32 34 17 51 42 38 19 57 45 40 44 46 23 69 48 50 52 54 56 49

(start  5):  1  5 10  4  6  3  9 12  8 14  7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32 34 17 51 36 38 19 57 42 40 44 46 23 69 45 48 50 52 54 56 49 63

(start  7):  1  7 14  4  6  3  9 12  8 10  5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32 34 17 51 36 38 19 57 42 40 44 46 23 69 45 48 50 52 54 56 49 63

(start  9):  1  9  3  6  4  8 10  5 15 12 14  7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32 34 17 51 36 38 19 57 42 40 44 46 23 69 45 48 50 52 54 56 49 63

(start 10):  1 10  4  6  3  9 12  8 14  7 21 15  5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32 34 17 51 36 38 19 57 42 40 44 46 23 69 45 48 50 52 54 56 49 63

Sidef[edit]

Translation of: Raku
class Seq(terms, callback) {
method next {
terms += callback(terms)
}
 
method nth(n) {
while (terms.len < n) {
self.next
}
terms[n-1]
}
 
method first(n) {
while (terms.len < n) {
self.next
}
terms.first(n)
}
}
 
func next_EKG (s) {
2..Inf -> first {|k|
 !(s.contains(k) || s[-1].is_coprime(k))
}
}
 
func EKG (start) {
Seq([1, start], next_EKG)
}
 
func converge_at(ints) {
var ekgs = ints.map(EKG)
 
2..Inf -> first {|k|
(ekgs.map { .nth(k) }.uniq.len == 1) &&
(ekgs.map { .first(k).sort }.uniq.len == 1)
}
}
 
for k in [2, 5, 7, 9, 10] {
say "EKG(#{k}) = #{EKG(k).first(10)}"
}
 
for arr in [[5,7], [2, 5, 7, 9, 10]] {
var c = converge_at(arr)
say "EKGs of #{arr} converge at term #{c}"
}
Output:
EKG(2) = [1, 2, 4, 6, 3, 9, 12, 8, 10, 5]
EKG(5) = [1, 5, 10, 2, 4, 6, 3, 9, 12, 8]
EKG(7) = [1, 7, 14, 2, 4, 6, 3, 9, 12, 8]
EKG(9) = [1, 9, 3, 6, 2, 4, 8, 10, 5, 15]
EKG(10) = [1, 10, 2, 4, 6, 3, 9, 12, 8, 14]
EKGs of [5, 7] converge at term 21
EKGs of [2, 5, 7, 9, 10] converge at term 45

Wren[edit]

Translation of: Go
Library: Wren-sort
Library: Wren-math
Library: Wren-fmt
import "/sort" for Sort
import "/math" for Int
import "/fmt" for Fmt
 
var areSame = Fn.new { |s, t|
var le = s.count
if (le != t.count) return false
Sort.quick(s)
Sort.quick(t)
for (i in 0...le) if (s[i] != t[i]) return false
return true
}
 
var limit = 100
var starts = [2, 5, 7, 9, 10]
var ekg = List.filled(5, null)
for (i in 0..4) ekg[i] = List.filled(limit, 0)
var s = 0
for (start in starts) {
ekg[s][0] = 1
ekg[s][1] = start
for (n in 2...limit) {
var i = 2
while (true) {
// a potential sequence member cannot already have been used
// and must have a factor in common with previous member
if (!ekg[s].take(n).contains(i) && Int.gcd(ekg[s][n-1], i) > 1) {
ekg[s][n] = i
break
}
i = i + 1
}
}
Fmt.print("EKG($2d): $2d", start, ekg[s].take(30).toList)
s = s + 1
}
 
// now compare EKG5 and EKG7 for convergence
for (i in 2...limit) {
if (ekg[1][i] == ekg[2][i] && areSame.call(ekg[1][0...i], ekg[2][0...i])) {
System.print("\nEKG(5) and EKG(7) converge at term %(i+1).")
return
}
}
System.print("\nEKG5(5) and EKG(7) do not converge within %(limit) terms.")
Output:
EKG( 2):  1  2  4  6  3  9 12  8 10  5 15 18 14  7 21 24 16 20 22 11 33 27 30 25 35 28 26 13 39 36
EKG( 5):  1  5 10  2  4  6  3  9 12  8 14  7 21 15 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32
EKG( 7):  1  7 14  2  4  6  3  9 12  8 10  5 15 18 16 20 22 11 33 21 24 26 13 39 27 30 25 35 28 32
EKG( 9):  1  9  3  6  2  4  8 10  5 15 12 14  7 21 18 16 20 22 11 33 24 26 13 39 27 30 25 35 28 32
EKG(10):  1 10  2  4  6  3  9 12  8 14  7 21 15  5 20 16 18 22 11 33 24 26 13 39 27 30 25 35 28 32

EKG(5) and EKG(7) converge at term 21.

zkl[edit]

Using gcd hint from Go.

fcn ekgW(N){	// --> iterator
Walker.tweak(fcn(rp,buf,w){
foreach n in (w){
if(rp.value.gcd(n)>1)
{ rp.set(n); w.push(buf.xplode()); buf.clear(); return(n); }
buf.append(n); // save small numbers not used yet
}
}.fp(Ref(N),List(),Walker.chain([2..N-1],[N+1..]))).push(1,N)
}
foreach n in (T(2,5,7,9,10)){ println("EKG(%2d): %s".fmt(n,ekgW(n).walk(10).concat(","))) }
Output:
EKG( 2): 1,2,4,6,3,9,12,8,10,5
EKG( 5): 1,5,10,2,4,6,3,9,12,8
EKG( 7): 1,7,14,2,4,6,3,9,12,8
EKG( 9): 1,9,3,6,2,4,8,10,5,15
EKG(10): 1,10,2,4,6,3,9,12,8,14
fcn convergeAt(n1,n2,etc){ ns:=vm.arglist;
ekgWs:=ns.apply(ekgW); ekgWs.apply2("next"); // pop initial 1
ekgNs:=List()*vm.numArgs; // ( (ekg(n1)), (ekg(n2)) ...)
do(1_000){ // find convergence in this many terms or bail
ekgN:=ekgWs.apply("next"); // (ekg(n1)[n],ekg(n2)[n] ...)
ekgNs.zipWith(fcn(ns,n){ ns.merge(n) },ekgN); // keep terms sorted
// are all ekg[n]s == and both sequences have same terms?
if(not ekgN.filter1('!=(ekgN[0])) and not ekgNs.filter1('!=(ekgNs[0])) ){
println("EKG(", ns.concat(","), ") converge at term ",ekgNs[0].len() + 1);
return();
}
}
println(ns.concat(",")," don't converge");
}
convergeAt(5,7);
convergeAt(2,5,7,9,10);
Output:
EKG(5,7) converge at term 21
EKG(2,5,7,9,10) converge at term 45