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 witha(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, ...
theEKG(2)
sequence; - the sequence starting
1, 3, ...
theEKG(3)
sequence; - ... the sequence starting
1, N, ...
theEKG(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
- Calculate and show here the first 10 members of
EKG(2)
. - Calculate and show here the first 10 members of
EKG(5)
. - Calculate and show here the first 10 members of
EKG(7)
. - Calculate and show here the first 10 members of
EKG(9)
. - Calculate and show here the first 10 members of
EKG(10)
. - Calculate and show here at which term
EKG(5)
andEKG(7)
converge (stretch goal).
- Related Tasks
- Reference
- The EKG Sequence and the Tree of Numbers. (Video).
C
<lang c>#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;
}</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
Factor
<lang factor>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 ] 2bi@ [ swapd [ = ] 2bi@ 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 .</lang>
- 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
<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
Haskell
<lang Haskell>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))))</lang>
- 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
<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>
Java
<lang java> 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); }
} </lang>
- 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
Julia
<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
<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 $LIMIT terms")
}</lang>
- Output:
EKG( 2): [1, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 14, 7, 21, 24, 16, 20, 22, 11, 33, 27, 30, 25, 35, 28, 26, 13, 39, 36] EKG( 5): [1, 5, 10, 2, 4, 6, 3, 9, 12, 8, 14, 7, 21, 15, 18, 16, 20, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32] EKG( 7): [1, 7, 14, 2, 4, 6, 3, 9, 12, 8, 10, 5, 15, 18, 16, 20, 22, 11, 33, 21, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32] EKG( 9): [1, 9, 3, 6, 2, 4, 8, 10, 5, 15, 12, 14, 7, 21, 18, 16, 20, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32] EKG(10): [1, 10, 2, 4, 6, 3, 9, 12, 8, 14, 7, 21, 15, 5, 20, 16, 18, 22, 11, 33, 24, 26, 13, 39, 27, 30, 25, 35, 28, 32] EKG(5) and EKG(7) converge at term 21
Perl
<lang perl>use List::Util qw(none sum);
sub gcd { my ($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
Phix
<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])]
Raku
(formerly Perl 6)
<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
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
<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