EKG sequence convergence: Difference between revisions

From Rosetta Code
Content added Content deleted
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=={{header|J}}==
=={{header|J}}==
<lang j>
<lang j>
Until =: 2 :'u^:(0-:v)^:_'
Until =: 2 :'u^:(0-:v)^:_' NB. unused but so fun
prime_factors_of_tail =: ~.@:q:@:{:
prime_factors_of_tail =: ~.@:q:@:{:
numbers_not_in_list =: -.~ >:@:i.@:(>./)
numbers_not_in_list =: -.~ >:@:i.@:(>./)




ekg =: 3 :0 NB. return sequence with next term appended
ekg =: 3 :0 NB. return next sequence
if. 1 = # y do. NB. initialize prepend
if. 1 = # y do. NB. initialize
1 , y
1 , y
return.
return.
Line 272: Line 272:
return.
return.
end.
end.
NB. otherwise extend the list's range
NB. otherwise extend the list
y , >:Until(1 e. a e. q:) >: >./ y
b =. >: >./ y
while. 1 -.@:e. a e. q: b do.
b =. >: b
end.
y , b
)
)



ekg^:9&>2 5 7 9 10
ekg^:9&>2 5 7 9 10
Line 283: Line 286:
1 9 3 6 2 4 8 10 5 15
1 9 3 6 2 4 8 10 5 15
1 10 2 4 6 3 9 12 8 14
1 10 2 4 6 3 9 12 8 14





Line 289: Line 291:
assert (,2) -: prime_factors_of_tail 6 8 NB. (nub of)
assert (,2) -: prime_factors_of_tail 6 8 NB. (nub of)
assert 3 4 5 -: numbers_not_in_list 1 2 6
assert 3 4 5 -: numbers_not_in_list 1 2 6
</lang>
Somewhat shorter is ekg2,
<lang j>
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&,)`g@.(1<#)

assert (3 -: index_of_lowest { numbers_not_in_list)1 2 4 6

assert (ekg^:9&> -: ekg2^:9&>) 2 5 7 9 10
</lang>
</lang>



Revision as of 23:18, 3 March 2019

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

Translation of: Go

<lang c>#include <stdio.h>

  1. include <stdlib.h>
  1. define TRUE 1
  2. define FALSE 0
  3. 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;

}</lang>

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#

The Function

This task uses Extensible Prime Generator (F#) <lang fsharp> // 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) </lang>

The Task

<lang fsharp> 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) </lang>

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

<lang fsharp> prıntfn "%d" (EKG 2|>Seq.takeWhile(fun n->n<>104729) ((Seq.length n)+1) </lang>

Output:
203786
Real: 00:10:21.967, CPU: 00:10:25.300, GC gen0: 65296, gen1: 1

Go

<lang go>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")

}</lang>

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


J

<lang j> Until =: 2 :'u^:(0-:v)^:_' NB. unused but so fun prime_factors_of_tail =: ~.@:q:@:{: numbers_not_in_list =: -.~ >:@:i.@:(>./)


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 -.@: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 </lang> Somewhat shorter is ekg2, <lang j> 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&,)`g@.(1<#)

assert (3 -: index_of_lowest { numbers_not_in_list)1 2 4 6

assert (ekg^:9&> -: ekg2^:9&>) 2 5 7 9 10 </lang>

Julia

Translation of: Perl

<lang julia>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)) </lang>

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

Translation of: Go

<lang scala>// 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")

}</lang>

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

Perl

Translation of: Perl 6

<lang perl>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"</lang>

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

Perl 6

Works with: Rakudo Star version 2018.04.1

<lang perl6>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);

}</lang>

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

Phix

Translation of: C

<lang Phix>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)</lang>

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

Python: Using math.gcd

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. <lang python>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))}!")</lang>
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. <lang python># 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)]))</lang>

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])]

REXX

<lang rexx>/*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*/</lang>
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

Translation of: Perl 6

<lang ruby>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}"

}</lang>

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

zkl

Using gcd hint from Go. <lang zkl>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)

}</lang> <lang zkl>foreach n in (T(2,5,7,9,10)){ println("EKG(%2d): %s".fmt(n,ekgW(n).walk(10).concat(","))) }</lang>

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

<lang zkl>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);</lang>

Output:
EKG(5,7) converge at term 21
EKG(2,5,7,9,10) converge at term 45