Ludic numbers

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Task
Ludic numbers
You are encouraged to solve this task according to the task description, using any language you may know.

Ludic numbers are related to prime numbers as they are generated by a sieve quite like the Sieve of Eratosthenes is used to generate prime numbers.

The first ludic number is 1.
To generate succeeding ludic numbers create an array of increasing integers starting from 2

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 ...

(Loop)

  • Take the first member of the resultant array as the next Ludic number 2.
  • Remove every 2'nd indexed item from the array (including the first).
2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 ...
  • (Unrolling a few loops...)
  • Take the first member of the resultant array as the next Ludic number 3.
  • Remove every 3'rd indexed item from the array (including the first).
3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 ...
  • Take the first member of the resultant array as the next Ludic number 5.
  • Remove every 5'th indexed item from the array (including the first).
5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59 61 65 67 71 73 77 ...
  • Take the first member of the resultant array as the next Ludic number 7.
  • Remove every 7'th indexed item from the array (including the first).
7 11 13 17 23 25 29 31 37 41 43 47 53 55 59 61 67 71 73 77 83 85 89 91 97 ...
  • ...
  • Take the first member of the current array as the next Ludic number L.
  • Remove every L'th indexed item from the array (including the first).
  • ...
Task
  • Generate and show here the first 25 ludic numbers.
  • How many ludic numbers are there less than or equal to 1000?
  • Show the 2000..2005'th ludic numbers.
  • A triplet is any three numbers x, x + 2, x + 6 where all three numbers are also ludic numbers. Show all triplets of ludic numbers < 250 (Stretch goal)

Contents

[edit] AutoHotkey

Works with: AutoHotkey 1.1
#NoEnv
SetBatchLines, -1
Ludic := LudicSieve(22000)
 
Loop, 25 ; the first 25 ludic numbers
Task1 .= Ludic[A_Index] " "
 
for i, Val in Ludic ; the number of ludic numbers less than or equal to 1000
if (Val <= 1000)
Task2++
else
break
 
Loop, 6 ; the 2000..2005'th ludic numbers
Task3 .= Ludic[1999 + A_Index] " "
 
for i, Val in Ludic { ; all triplets of ludic numbers < 250
if (Val + 6 > 249)
break
if (Ludic[i + 1] = Val + 2 && Ludic[i + 2] = Val + 6 || i = 1)
Task4 .= "(" Val " " Val + 2 " " Val + 6 ") "
}
 
MsgBox, % "First 25:`t`t" Task1
. "`nLudics below 1000:`t" Task2
. "`nLudic 2000 to 2005:`t" Task3
. "`nTriples below 250:`t" Task4
return
 
LudicSieve(Limit) {
Arr := [], Ludic := []
Loop, % Limit
Arr.Insert(A_Index)
Ludic.Insert(Arr.Remove(1))
while Arr.MaxIndex() != 1 {
Ludic.Insert(n := Arr.Remove(1))
, Removed := 0
Loop, % Arr.MaxIndex() // n {
Arr.Remove(A_Index * n - Removed)
, Removed++
}
}
Ludic.Insert(Arr[1])
return Ludic
}
Output:
First 25:		1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107 
Ludics below 1000:	142
Ludic 2000 to 2005:	21475 21481 21487 21493 21503 21511 
Triples below 250:	(1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239) 

[edit] C

#include <stdio.h>
#include <stdlib.h>
 
typedef unsigned uint;
typedef struct { uint i, v; } filt_t;
 
// ludics with at least so many elements and reach at least such value
uint* ludic(uint min_len, uint min_val, uint *len)
{
uint cap, i, v, active = 1, nf = 0;
filt_t *f = calloc(cap = 2, sizeof(*f));
f[1].i = 4;
 
for (v = 1; ; ++v) {
for (i = 1; i < active && --f[i].i; i++);
 
if (i < active)
f[i].i = f[i].v;
else if (nf == f[i].i)
f[i].i = f[i].v, ++active; // enable one more filter
else {
if (nf >= cap)
f = realloc(f, sizeof(*f) * (cap*=2));
f[nf] = (filt_t){ v + nf, v };
if (++nf >= min_len && v >= min_val) break;
}
}
 
// pack the sequence into a uint[]
// filt_t struct was used earlier for cache locality in loops
uint *x = (void*) f;
for (i = 0; i < nf; i++) x[i] = f[i].v;
x = realloc(x, sizeof(*x) * nf);
 
*len = nf;
return x;
}
 
int find(uint *a, uint v)
{
uint i;
for (i = 0; a[i] <= v; i++)
if (v == a[i]) return 1;
return 0;
}
 
int main(void)
{
uint len, i, *x = ludic(2005, 1000, &len);
 
printf("First 25:");
for (i = 0; i < 25; i++) printf(" %u", x[i]);
putchar('\n');
 
for (i = 0; x[i] <= 1000; i++);
printf("Ludics below 1000: %u\n", i);
 
printf("Ludic 2000 to 2005:");
for (i = 2000; i <= 2005; i++) printf(" %u", x[i - 1]);
putchar('\n');
 
printf("Triples below 250:");
for (i = 0; x[i] + 6 <= 250; i++)
if (find(x, x[i] + 2) && find(x, x[i] + 6))
printf(" (%u %u %u)", x[i], x[i] + 2, x[i] + 6);
 
putchar('\n');
 
free(x);
return 0;
}
Output:
First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107
Ludics below 1000: 142
Ludic 2000 to 2005: 21475 21481 21487 21493 21503 21511
Triples below 250: (1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239)

[edit] C++

 
#include <vector>
#include <iostream>
using namespace std;
 
class ludic
{
public:
void ludicList()
{
_list.push_back( 1 );
 
vector<int> v;
for( int x = 2; x < 22000; x++ )
v.push_back( x );
 
while( true )
{
vector<int>::iterator i = v.begin();
int z = *i;
_list.push_back( z );
 
while( true )
{
i = v.erase( i );
if( distance( i, v.end() ) <= z - 1 ) break;
advance( i, z - 1 );
}
if( v.size() < 1 ) return;
}
}
 
void show( int s, int e )
{
for( int x = s; x < e; x++ )
cout << _list[x] << " ";
}
 
void findTriplets( int e )
{
int lu, x = 0;
while( _list[x] < e )
{
lu = _list[x];
if( inList( lu + 2 ) && inList( lu + 6 ) )
cout << "(" << lu << " " << lu + 2 << " " << lu + 6 << ")\n";
x++;
}
}
 
int count( int e )
{
int x = 0, c = 0;
while( _list[x++] <= 1000 ) c++;
return c;
}
 
private:
bool inList( int lu )
{
for( int x = 0; x < 250; x++ )
if( _list[x] == lu ) return true;
return false;
}
 
vector<int> _list;
};
 
int main( int argc, char* argv[] )
{
ludic l;
l.ludicList();
cout << "first 25 ludic numbers:" << "\n";
l.show( 0, 25 );
cout << "\n\nThere are " << l.count( 1000 ) << " ludic numbers <= 1000" << "\n";
cout << "\n2000 to 2005'th ludic numbers:" << "\n";
l.show( 1999, 2005 );
cout << "\n\nall triplets of ludic numbers < 250:" << "\n";
l.findTriplets( 250 );
cout << "\n\n";
return system( "pause" );
}
 
Output:
first 25 ludic numbers:
1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107

There are 142 ludic numbers <= 1000

2000 to 2005'th ludic numbers:
21475 21481 21487 21493 21503 21511

all triplets of ludic numbers < 250:
(1 3 7)
(5 7 11)
(11 13 17)
(23 25 29)
(41 43 47)
(173 175 179)
(221 223 227)
(233 235 239)

[edit] Clojure

(defn ints-from [n]
(cons n (lazy-seq (ints-from (inc n)))))
 
(defn drop-nth [n seq]
(cond
(zero? n) seq
(empty? seq) []
 :else (concat (take (dec n) seq) (lazy-seq (drop-nth n (drop n seq))))))
 
(def ludic ((fn ludic
([] (ludic 1))
([n] (ludic n (ints-from (inc n))))
([n [f & r]] (cons n (lazy-seq (ludic f (drop-nth f r))))))))
 
(defn ludic? [n] (= (first (filter (partial <= n) ludic)) n))
 
(print "First 25: ")
(println (take 25 ludic))
(print "Count below 1000: ")
(println (count (take-while (partial > 1000) ludic)))
(print "2000th through 2005th: ")
(println (map (partial nth ludic) (range 1999 2005)))
(print "Triplets < 250: ")
(println (filter (partial every? ludic?)
(for [i (range 250)] (list i (+ i 2) (+ i 6)))))
Output:
First 25: (1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107)
Count below 1000: 142
2000th through 2005th: (21475 21481 21487 21493 21503 21511)
Triplets < 250: ((1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239))

[edit] D

[edit] opApply Version

Translation of: Python
Translation of: Perl 6
struct Ludics(T) {
int opApply(int delegate(in ref T) dg) {
int result;
T[] rotor, taken = [T(1)];
result = dg(taken[0]);
if (result) return result;
 
for (T i = 2; ; i++) { // Shoud be stopped if T has a max.
size_t j = 0;
for (; j < rotor.length; j++)
if (!--rotor[j])
break;
 
if (j < rotor.length) {
rotor[j] = taken[j + 1];
} else {
result = dg(i);
if (result) return result;
taken ~= i;
rotor ~= taken[j + 1];
}
}
}
}
 
void main() {
import std.stdio, std.range, std.algorithm;
 
// std.algorithm.take can't be used here.
uint[] L;
foreach (const x; Ludics!uint())
if (L.length < 2005)
L ~= x;
else
break;
 
writeln("First 25 ludic primes:\n", L.take(25));
writefln("\nThere are %d ludic numbers <= 1000.",
L.until!q{ a > 1000 }.walkLength);
 
writeln("\n2000'th .. 2005'th ludic primes:\n", L[1999 .. 2005]);
 
enum m = 250;
const triplets = L.filter!(x => x + 6 < m &&
L.canFind(x + 2) && L.canFind(x + 6))
// Ugly output:
//.map!(x => tuple(x, x + 2, x + 6)).array;
.map!(x => [x, x + 2, x + 6]).array;
writefln("\nThere are %d triplets less than %d:\n%s",
triplets.length, m, triplets);
}
Output:
First 25 ludic primes:
[1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107]

There are 142 ludic numbers <= 1000.

2000'th .. 2005'th ludic primes:
[21475, 21481, 21487, 21493, 21503, 21511]

There are 8 triplets less than 250:
[[1, 3, 7], [5, 7, 11], [11, 13, 17], [23, 25, 29], [41, 43, 47], [173, 175, 179], [221, 223, 227], [233, 235, 239]]

The run-time is about 0.03 seconds or less. It takes about 2.0 seconds to generate 50_000 Ludic numbers with ldc2 compiler.

[edit] Range Version

This is the same code modified to be a Range.

struct Ludics(T) {
T[] rotor, taken = [T(1)];
T i;
size_t j;
T front = 1; // = taken[0];
bool running = false;
static immutable bool empty = false;
 
void popFront() pure nothrow @safe {
if (running)
goto RESUME;
else
running = true;
 
i = 2;
while (true) {
j = 0;
 
while (j < rotor.length) {
rotor[j]--;
if (!rotor[j])
break;
j++;
}
if (j < rotor.length) {
rotor[j] = taken[j + 1];
} else {
front = i;
return;
RESUME:
taken ~= i;
rotor ~= taken[j + 1];
}
i++; // Could overflow if T has a max.
}
}
}
 
void main() {
import std.stdio, std.range, std.algorithm, std.array;
 
Ludics!uint L;
writeln("First 25 ludic primes:\n", L.take(25));
writefln("\nThere are %d ludic numbers <= 1000.",
L.until!q{ a > 1000 }.walkLength);
 
writeln("\n2000'th .. 2005'th ludic primes:\n", L.drop(1999).take(6));
 
enum uint m = 250;
const few = L.until!(x => x > m).array;
const triplets = few.filter!(x => x + 6 < m && few.canFind(x + 2)
&& few.canFind(x + 6))
// Ugly output:
//.map!(x => tuple(x, x + 2, x + 6)).array;
.map!(x => [x, x + 2, x + 6]).array;
writefln("\nThere are %d triplets less than %d:\n%s",
triplets.length, m, triplets);
}

The output is the same. This version is slower, it takes about 3.3 seconds to generate 50_000 Ludic numbers with ldc2 compiler.

[edit] Haskell

import Data.List (unfoldr, genericSplitAt)
 
ludic :: [Integer]
ludic = 1 : unfoldr (\xs@(x:_) -> Just (x, dropEvery x xs)) [2..] where
dropEvery n = concat . map tail . unfoldr (Just . genericSplitAt n)
 
main :: IO ()
main = do
print $ take 25 $ ludic
print $ length $ takeWhile (<= 1000) $ ludic
print $ take 6 $ drop 1999 $ ludic
-- haven't done triplets task yet
Output:
[1,2,3,5,7,11,13,17,23,25,29,37,41,43,47,53,61,67,71,77,83,89,91,97,107]
142
[21475,21481,21487,21493,21503,21511]

The filter for dropping every n-th number can be delayed until it's needed, which speeds up the generator, more so when a longer sequence is taken.

ludic = 1:2 : f 3 [3..] [(4,2)] where
f n (x:xs) yy@((i,y):ys)
| n == i = f n (dropEvery y xs) ys
| otherwise = x : f (1+n) xs (yy ++ [(n+x, x)])
 
dropEvery n s = a ++ dropEvery n (tail b) where
(a,b) = splitAt (n-1) s
 
main = print $ ludic !! 10000

[edit] Icon and Unicon

This is inefficient, but was fun to code as a cascade of filters. Works in both languages.

global num, cascade, sieve, nfilter
 
procedure main(A)
lds := ludic(2005) # All we need for the four tasks.
every writes("First 25:" | (" "||!lds)\25 | "\n")
every (n := 0) +:= (!lds < 1000, 1)
write("There are ",n," Ludic numbers < 1000.")
every writes("2000th through 2005th: " | (lds[2000 to 20005]||" ") | "\n")
writes("Triplets:")
every (250 > (x := !lds)) & (250 > (x+2 = !lds)) & (250 > (x+6 = !lds)) do
writes(" [",x,",",x+2,",",x+6,"]")
write()
end
 
procedure ludic(limit)
candidates := create seq(2)
put(cascade := [], create {
repeat {
report(l := num, limit)
put(cascade, create (cnt:=0, repeat ((cnt+:=1)%l=0, @sieve) | @@nfilter))
cascade[-2] :=: cascade[-1] # keep this sink as the last filter
@sieve
}
})
sieve := create while num := @candidates do @@(nfilter := create !cascade)
report(1, limit)
return @sieve
end
 
procedure report(ludic, limit)
static count, lds
initial {count := 0; lds := []}
if (count +:= 1) > limit then lds@&main
put(lds, ludic)
end

Output:

->ludic    
First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107
There are 142 Ludic numbers < 1000.
2000th through 20005th: 21475 21481 21487 21493 21503 21511 
Triplets: [1,3,7] [5,7,11] [11,13,17] [23,25,29] [41,43,47] [173,175,179] [221,223,227] [233,235,239]
->

[edit] J

Solution (naive / brute force):
   ludic =: _1 |.!.1 [: {."1 [: (#~ 0 ~: {. | i.@#)^:a: 2 + i.
Examples:
   # ludic 110  NB. 110 is sufficient to generate 25 Ludic numbers
25
ludic 110 NB. First 25 Ludic numbers
1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107
 
#ludic 1000 NB. 142 Ludic numbers <= 1000
142
 
# ludic 22000 NB. 22000 is sufficient to generate > 2005 Ludic numbers
2042
(2000+i.6) { ludic 22000 NB. Ludic numbers 2000-2005
21481 21487 21493 21503 21511 21523
 
0 2 6 (] (*./ .e.~ # |:@]) +/) ludic 250 NB. Ludic triplets <= 250
1 3 7
5 7 11
11 13 17
23 25 29
41 43 47
173 175 179
221 223 227
233 235 239

[edit] Java

Works with: Java version 1.5+

This example uses pre-calculated ranges for the first and third task items (noted in comments).

import java.util.ArrayList;
import java.util.List;
 
public class Ludic{
public static List<Integer> ludicUpTo(int n){
List<Integer> ludics = new ArrayList<Integer>(n);
for(int i = 1; i <= n; i++){ //fill the initial list
ludics.add(i);
}
 
//start at index 1 because the first ludic number is 1 and we don't remove anything for it
for(int cursor = 1; cursor < ludics.size(); cursor++){
int thisLudic = ludics.get(cursor); //the first item in the list is a ludic number
int removeCursor = cursor + thisLudic; //start removing that many items later
while(removeCursor < ludics.size()){
ludics.remove(removeCursor); //remove the next item
removeCursor = removeCursor + thisLudic - 1; //move the removal cursor up as many spaces as we need to
//then back one to make up for the item we just removed
}
}
return ludics;
}
 
public static List<List<Integer>> getTriplets(List<Integer> ludics){
List<List<Integer>> triplets = new ArrayList<List<Integer>>();
for(int i = 0; i < ludics.size() - 2; i++){ //only need to check up to the third to last item
int thisLudic = ludics.get(i);
if(ludics.contains(thisLudic + 2) && ludics.contains(thisLudic + 6)){
List<Integer> triplet = new ArrayList<Integer>(3);
triplet.add(thisLudic);
triplet.add(thisLudic + 2);
triplet.add(thisLudic + 6);
triplets.add(triplet);
}
}
return triplets;
}
 
public static void main(String[] srgs){
System.out.println("First 25 Ludics: " + ludicUpTo(110)); //110 will get us 25 numbers
System.out.println("Ludics up to 1000: " + ludicUpTo(1000).size());
System.out.println("2000th - 2005th Ludics: " + ludicUpTo(22000).subList(1999, 2005)); //22000 will get us 2005 numbers
System.out.println("Triplets up to 250: " + getTriplets(ludicUpTo(250)));
}
}
Output:
First 25 Ludics: [1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107]
Ludics up to 1000: 142
2000th - 2005th Ludics: [21475, 21481, 21487, 21493, 21503, 21511]
Triplets up to 250: [[1, 3, 7], [5, 7, 11], [11, 13, 17], [23, 25, 29], [41, 43, 47], [173, 175, 179], [221, 223, 227], [233, 235, 239]]

[edit] Mathematica

n=10^5;
Ludic={1};
seq=Range[2,n];
ClearAll[DoStep]
DoStep[seq:{f_,___}]:=Module[{out=seq},
AppendTo[Ludic,f];
out[[;;;;f]]=Sequence[];
out
]
Nest[DoStep,seq,2500];
Output:
Ludic[[;; 25]]
LengthWhile[Ludic, # < 1000 &]
Ludic[[2000 ;; 2005]]
Select[Subsets[Select[Ludic, # < 250 &], {3}], Differences[#] == {2, 4} &]

{1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107}
142
{21475, 21481, 21487, 21493, 21503, 21511}
{{1, 3, 7}, {5, 7, 11}, {11, 13, 17}, {23, 25, 29}, {41, 43, 47}, {173, 175, 179}, {221, 223, 227}, {233, 235, 239}}

[edit] Perl 6

This implementation has no arbitrary upper limit, since it can keep adding new rotors on the fly. It just gets slower and slower instead... :-)

constant ludic = gather {
my @taken = take 1;
my @rotor;
 
for 2..* -> $i {
loop (my $j = 0; $j < @rotor; $j++) {
--@rotor[$j] or last;
}
if $j < @rotor {
@rotor[$j] = @taken[$j+1];
}
else {
push @taken, take $i;
push @rotor, @taken[$j+1];
}
}
}
 
say ludic[^25];
say "Number of Ludic numbers <= 1000: ", +(ludic ...^ * > 1000);
say "Ludic numbers 2000..2005: ", ludic[1999..2004];
 
my \l250 = set ludic ...^ * > 250;
say "Ludic triples < 250: ", gather
for l250.keys -> $a {
my $b = $a + 2;
my $c = $a + 6;
take "<$a $b $c>" if $b ∈ l250 and $c ∈ l250;
}
Output:
1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107
Number of Ludic numbers <= 1000: 142
Ludic numbers 2000..2005: 21475 21481 21487 21493 21503 21511
Ludic triples < 250: <1 3 7> <5 7 11> <11 13 17> <23 25 29> <41 43 47> <173 175 179> <221 223 227> <233 235 239>

[edit] PL/I

This example is incorrect. Missing Triplet 1,3,7. Please fix the code and remove this message.
Ludic_numbers: procedure options (main);                /* 18 April 2014 */
declare V(2:22000) fixed, L(2200) fixed;
declare (step, i, j, k, n) fixed binary;
 
Ludic: procedure;
n = hbound(V,1); k = 1; L(1) = 1;
do i = 2 to n; V(i) = i; end;
 
do forever;
 
k = k + 1; L(k), step = V(2);
 
do i = 2 to n by step;
V(i) = 0;
end;
call compress;
if L(k) >= 21511 then leave;
end;
 
put skip list ('The first 25 Ludic numbers are:');
put skip edit ( (L(i) do i = 1 to 25) ) (F(4));
 
k = 0;
do i = 1 by 1;
if L(i) < 1000 then k = k + 1; else leave;
end;
 
put skip list ('There are ' || trim(k) || ' Ludic numbers < 1000');
put skip list ('Six Ludic numbers from the 2000-th:');
put skip edit ( (L(i) do i = 2000 to 2005) ) (f(7));
/* Triples are values of the form x, x+2, x+6 */
put skip list ('Triples are:');
put skip;
i = 1;
do i = 1 by 1 while (L(i+2) <= 250);
if (L(i) = L(i+1) - 2) & (L(i) = L(i+2) - 6) then
put edit ('(', L(i), L(i+1), L(i+2), ') ' ) (A, 3 F(4), A);
end;
 
compress: procedure;
j = 2;
do i = 2 to n;
if V(i) ^= 0 then do; V(j) = V(i); j = j + 1; end;
end;
n = j-1;
end compress;
 
end Ludic;
 
call Ludic;
 
end Ludic_numbers;

Output:

The first 25 Ludic numbers are: 
   1   2   3   5   7  11  13  17  23  25  29  37  41  43  47
  53  61  67  71  77  83  89  91  97 107
There are 142 Ludic numbers < 1000 
Six Ludic numbers from the 2000-th: 
  21475  21481  21487  21493  21503  21511
Triples are: 
(   5   7  11) (  11  13  17) (  23  25  29) (  41  43  47)
( 173 175 179) ( 221 223 227) ( 233 235 239)

[edit] Python

[edit] Python: Fast

def ludic(nmax=100000):
yield 1
lst = list(range(2, nmax + 1))
while lst:
yield lst[0]
del lst[::lst[0]]
 
ludics = [l for i,l in zip(range(2005), ludic())]
 
print('First 25 ludic primes:')
print(ludics[:25])
print("\nThere are %i ludic numbers <= 1000"
 % sum(1 for l in ludics if l <= 1000))
print("\n2000'th..2005'th ludic primes:")
print(ludics[2000-1: 2005])
 
n = 250
triplets = [(x, x+2, x+6)
for x in ludics
if x+6 < n and x+2 in ludics and x+6 in ludics]
print('\nThere are %i triplets less than %i:\n  %r'
 % (len(triplets), n, triplets))
Output:
First 25 ludic primes:
[1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107]

There are 142 ludic numbers <= 1000

2000'th..2005'th ludic primes:
[21475, 21481, 21487, 21493, 21503, 21511]

There are 8 triplets less than 250:
  [(1, 3, 7), (5, 7, 11), (11, 13, 17), (23, 25, 29), (41, 43, 47), (173, 175, 179), (221, 223, 227), (233, 235, 239)]

[edit] Python: No set maximum

The following version of function ludic will return ludic numbers until reaching system limits. It is less efficient than the fast version as all lucid numbers so far are cached; on exhausting the current lst a new list of twice the size is created and the previous deletions applied before continuing.

def ludic(nmax=64):
yield 1
taken = []
while True:
lst, nmax = list(range(2, nmax + 1)), nmax * 2
for t in taken:
del lst[::t]
while lst:
t = lst[0]
taken.append(t)
yield t
del lst[::t]

Output is the same as for the fast version.

[edit] Racket

#lang racket
(define lucid-sieve-size 25000) ; this should be enough to do me!
(define lucid?
(let ((lucid-bytes-sieve
(delay
(define sieve-bytes (make-bytes lucid-sieve-size 1))
(bytes-set! sieve-bytes 0 0) ; not a lucid number
(define (sieve-pass L)
(let loop ((idx (add1 L)) (skip (sub1 L)))
(cond
[(= idx lucid-sieve-size)
(for/first ((rv (in-range (add1 L) lucid-sieve-size))
#:unless (zero? (bytes-ref sieve-bytes rv))) rv)]
[(zero? (bytes-ref sieve-bytes idx))
(loop (add1 idx) skip)]
[(= skip 0)
(bytes-set! sieve-bytes idx 0)
(loop (add1 idx) (sub1 L))]
[else (loop (add1 idx) (sub1 skip))])))
(let loop ((l 2))
(when l (loop (sieve-pass l))))
sieve-bytes)))
 
(λ (n) (= 1 (bytes-ref (force lucid-bytes-sieve) n)))))
 
(define (dnl . things) (for-each displayln things))
 
(dnl
"Generate and show here the first 25 ludic numbers."
(for/list ((_ 25) (l (sequence-filter lucid? (in-naturals)))) l)
"How many ludic numbers are there less than or equal to 1000?"
(for/sum ((n 1001) #:when (lucid? n)) 1)
"Show the 2000..2005'th ludic numbers."
(for/list ((i 2006) (l (sequence-filter lucid? (in-naturals))) #:when (>= i 2000)) l)
#<<EOS
A triplet is any three numbers x, x + 2, x + 6 where all three numbers are
also ludic numbers. Show all triplets of ludic numbers < 250 (Stretch goal)
EOS
(for/list ((x (in-range 250)) #:when (and (lucid? x) (lucid? (+ x 2)) (lucid? (+ x 6))))
(list x (+ x 2) (+ x 6))))
Output:
Generate and show here the first 25 ludic numbers.
(1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107)
How many ludic numbers are there less than or equal to 1000?
142
Show the 2000..2005'th ludic numbers.
(21481 21487 21493 21503 21511 21523)
A triplet is any three numbers x, x + 2, x + 6 where all three numbers are
also ludic numbers. Show all triplets of ludic numbers < 250 (Stretch goal)
((1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239))
cpu time: 18753 real time: 18766 gc time: 80

[edit] REXX

/*REXX program to display (a range of) ludic numbers, or a count of same*/
parse arg N count bot top triples . /*obtain optional parameters/args*/
if N=='' then N=25 /*Not specified? Use the default.*/
if count=='' then count=1000 /* " " " " " */
if bot=='' then bot=2000 /* " " " " " */
if top=='' then top=2005 /* " " " " " */
if triples=='' then triples=250-1 /* " " " " " */
say 'The first ' N " ludic numbers: " ludic(n)
say
say "There are " words(ludic(-count)) ' ludic numbers from 1───►'count " (inclusive)."
say
say "The " bot ' to ' top " ludic numbers are: " ludic(bot,top)
$=ludic(-triples) 0 0; #=0; @=
say
do j=1 for words($); _=word($,j) /*it is known that ludic _ exists*/
if wordpos(_+2,$)==0 | wordpos(_+6,$)==0 then iterate /*¬triple.*/
#=#+1; @=@ '◄'_ _+2 _+6"► " /*bump triple counter, and ··· */
end /*j*/ /* [↑] append found triple ──► @*/
 
if @=='' then say 'From 1──►'triples", no triples found."
else say 'From 1──►'triples", " # ' triples found:' @
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────LUDIC subroutine────────────────────*/
ludic: procedure; parse arg m 1 mm,h; am=abs(m); if h\=='' then am=h
$=1 2; @= /*$=ludic #s superset, @=# series*/
/* [↓] construct a ludic series.*/
do j=3 by 2 to am * max(1,15*((m>0)|h\=='')); @=@ j; end; @=@' '
/* [↑] high limit: approx|exact */
do while words(@)\==0 /* [↓] examine the first word. */
f=word(@,1); $=$ f /*append this first word to list.*/
do d=1 by f while d<=words(@) /*use 1st #, elide all occurances*/
@=changestr(' 'word(@,d)" ",@, ' . ') /*delete the # in the seq#*/
end /*d*/ /* [↑] done eliding "1st" number*/
@=translate(@,,.) /*translate periods to blanks. */
end /*forever*/ /* [↑] done eliding ludic #s. */
@=space(@) /*remove extra blanks from list. */
 
if h=='' then return subword($,1,am) /*return a range of ludic numbers*/
return subword($,m,h-m+1) /*return a section of a range.*/

Some older REXXes don't have a changestr bif, so one is included here ──► CHANGESTR.REX.

output   using the defaults for input:

The first  25  ludic numbers:  1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107

There are  142  ludic numbers from 1───►1000  (inclusive).

The  2000  to  2005  ludic numbers are:  21475 21481 21487 21493 21503 21511

From 1──►249,  8  triples found:  ◄1 3 7►  ◄5 7 11►  ◄11 13 17►  ◄23 25 29►  ◄41 43 47►  ◄173 175 179►  ◄221 223 227►  ◄233 235 239►

[edit] Ruby

def ludic(nmax=100000)
Enumerator.new do |y|
y << 1
ary = *2..nmax
until ary.empty?
y << (n = ary.first)
(0...ary.size).step(n){|i| ary[i] = nil}
ary.compact!
end
end
end
 
puts "First 25 Ludic numbers:", ludic.first(25).to_s
 
puts "Ludics below 1000:", ludic(1000).count
 
puts "Ludic numbers 2000 to 2005:", ludic.first(2005).last(6).to_s
 
ludics = ludic(250).to_a
puts "Ludic triples below 250:",
ludics.select{|x| ludics.include?(x+2) and ludics.include?(x+6)}.map{|x| [x, x+2, x+6]}.to_s
Output:
First 25 Ludic numbers:
[1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107]
Ludics below 1000:
142
Ludic numbers 2000 to 2005:
[21475, 21481, 21487, 21493, 21503, 21511]
Ludic triples below 250:
[[1, 3, 7], [5, 7, 11], [11, 13, 17], [23, 25, 29], [41, 43, 47], [173, 175, 179], [221, 223, 227], [233, 235, 239]]

[edit] Seed7

$ include "seed7_05.s7i";
 
const func set of integer: ludicNumbers (in integer: n) is func
result
var set of integer: ludicNumbers is {1};
local
var set of integer: sieve is EMPTY_SET;
var integer: ludicNumber is 0;
var integer: number is 0;
var integer: count is 0;
begin
sieve := {2 .. n};
while sieve <> EMPTY_SET do
ludicNumber := min(sieve);
incl(ludicNumbers, ludicNumber);
count := 0;
for number range sieve do
if count rem ludicNumber = 0 then
excl(sieve, number);
end if;
incr(count);
end for;
end while;
end func;
 
const integer: limit is 22000;
const set of integer: ludicNumbers is ludicNumbers(limit);
 
const proc: main is func
local
var integer: number is 0;
var integer: count is 0;
begin
write("First 25:");
for number range ludicNumbers until count = 25 do
write(" " <& number);
incr(count);
end for;
writeln;
count := 0;
for number range ludicNumbers until number > 1000 do
incr(count);
end for;
writeln("Ludics below 1000: " <& count);
write("Ludic 2000 to 2005:");
count := 0;
for number range ludicNumbers until count >= 2005 do
incr(count);
if count >= 2000 then
write(" " <& number);
end if;
end for;
writeln;
write("Triples below 250:");
for number range ludicNumbers until number > 250 do
if number + 2 in ludicNumbers and number + 6 in ludicNumbers then
write(" (" <& number <& ", " <& number + 2 <& ", " <& number + 6 <& ")");
end if;
end for;
writeln;
end func;
Output:
First 25: 1 2 3 5 7 11 13 17 23 25 29 37 41 43 47 53 61 67 71 77 83 89 91 97 107
Ludics below 1000: 142
Ludic 2000 to 2005: 21475 21481 21487 21493 21503 21511
Triples below 250: (1, 3, 7) (5, 7, 11) (11, 13, 17) (23, 25, 29) (41, 43, 47) (173, 175, 179) (221, 223, 227) (233, 235, 239)

[edit] Tcl

Works with: Tcl version 8.6

The limit on the number of values generated is the depth of stack; this can be set to arbitrarily deep to go as far as you want. Provided you are prepared to wait for the values to be generated.

package require Tcl 8.6
 
proc ludic n {
global ludicList ludicGenerator
for {} {[llength $ludicList] <= $n} {lappend ludicList $i} {
set i [$ludicGenerator]
set ludicGenerator [coroutine L_$i apply {{gen k} {
yield [info coroutine]
while true {
set val [$gen]
if {[incr i] == $k} {set i 0} else {yield $val}
}
}} $ludicGenerator $i]
}
return [lindex $ludicList $n]
}
# Bootstrap the generator sequence
set ludicList [list 1]
set ludicGenerator [coroutine L_1 apply {{} {
set n 1
yield [info coroutine]
while true {yield [incr n]}
}}]
 
# Default of 1000 is not enough
interp recursionlimit {} 5000
 
for {set i 0;set l {}} {$i < 25} {incr i} {lappend l [ludic $i]}
puts "first25: [join $l ,]"
 
for {set i 0} {[ludic $i] <= 1000} {incr i} {}
puts "below=1000: $i"
 
for {set i 1999;set l {}} {$i < 2005} {incr i} {lappend l [ludic $i]}
puts "2000-2005: [join $l ,]"
 
for {set i 0} {[ludic $i] < 256} {incr i} {set isl([ludic $i]) $i}
for {set i 1;set l {}} {$i < 250} {incr i} {
if {[info exists isl($i)] && [info exists isl([expr {$i+2}])] && [info exists isl([expr {$i+6}])]} {
lappend l ($i,[expr {$i+2}],[expr {$i+6}])
}
}
puts "triplets: [join $l ,]"
Output:
first25: 1,2,3,5,7,11,13,17,23,25,29,37,41,43,47,53,61,67,71,77,83,89,91,97,107
below=1000: 142
2000-2005: 21475,21481,21487,21493,21503,21511
triplets: (1,3,7),(5,7,11),(11,13,17),(23,25,29),(41,43,47),(173,175,179),(221,223,227),(233,235,239)
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