Ludic numbers

From Rosetta Code
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 where all three numbers are also ludic numbers. Show all triplets of ludic numbers < 250 (Stretch goal)

ABAP

Works with NW 7.40 SP8 <lang ABAP>CLASS lcl_ludic DEFINITION CREATE PUBLIC.

 PUBLIC SECTION.
   TYPES: t_ludics TYPE SORTED TABLE OF i WITH UNIQUE KEY table_line.
   TYPES: BEGIN OF t_triplet,
            i1 TYPE i,
            i2 TYPE i,
            i3 TYPE i,
          END OF t_triplet.
   TYPES: t_triplets TYPE STANDARD TABLE OF t_triplet WITH EMPTY KEY.
   CLASS-METHODS:
     ludic_up_to
       IMPORTING i_int           TYPE i
       RETURNING VALUE(r_ludics) TYPE t_ludics,
     get_triplets
       IMPORTING i_ludics          TYPE t_ludics
       RETURNING VALUE(r_triplets) TYPE t_triplets.
   "RETURNING parameters (CallByValue) only used for readability of the demo
   "in "Real Life" you should use EXPORTING (CallByRef) for tables

ENDCLASS.

cl_demo_output=>begin_section( 'First 25 Ludics' ). cl_demo_output=>write( lcl_ludic=>ludic_up_to( 110 ) ).

cl_demo_output=>begin_section( 'Ludics up to 1000' ). cl_demo_output=>write( lines( lcl_ludic=>ludic_up_to( 1000 ) ) ).

cl_demo_output=>begin_section( '2000th - 2005th Ludics' ). DATA(ludics) = lcl_ludic=>ludic_up_to( 22000 ). cl_demo_output=>write( VALUE lcl_ludic=>t_ludics( FOR i = 2000 WHILE i <= 2005 ( ludics[ i ] ) ) ).

cl_demo_output=>begin_section( 'Triplets up to 250' ). cl_demo_output=>write( lcl_ludic=>get_triplets( lcl_ludic=>ludic_up_to( 250 ) ) ).

cl_demo_output=>display( ).

CLASS lcl_ludic IMPLEMENTATION.

 METHOD ludic_up_to.
   r_ludics = VALUE #( FOR i = 2 WHILE i <= i_int ( i ) ).
   DATA(cursor) = 0.
   WHILE cursor < lines( r_ludics ).
     cursor = cursor + 1.
     DATA(this_ludic) = r_ludics[ cursor ].
     DATA(remove_cursor) = cursor + this_ludic.
     WHILE remove_cursor <= lines( r_ludics ).
       DELETE r_ludics INDEX remove_cursor.
       remove_cursor = remove_cursor + this_ludic - 1.
     ENDWHILE.
   ENDWHILE.
   INSERT 1 INTO TABLE r_ludics.  "add one as the first Ludic number (per definition)
 ENDMETHOD.
 METHOD get_triplets.
   DATA(i) = 0.
   WHILE i < lines( i_ludics ) - 2.
     i = i + 1.
     DATA(this_ludic) = i_ludics[ i ].
     IF  line_exists( i_ludics[ table_line = this_ludic + 2 ] )
     AND line_exists( i_ludics[ table_line = this_ludic + 6 ] ).
       r_triplets = VALUE #(
          BASE r_triplets
          ( i1 = i_ludics[ table_line = this_ludic ]
            i2 = i_ludics[ table_line = this_ludic + 2 ]
            i3 = i_ludics[ table_line = this_ludic + 6 ]
          )
       ).
     ENDIF.
   ENDWHILE.
 ENDMETHOD.

ENDCLASS.</lang>

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 

AutoHotkey

Works with: AutoHotkey 1.1

<lang AutoHotkey>#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 }</lang>

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) 

C

<lang c>#include <stdio.h>

  1. 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; }</lang>

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)

C++

<lang cpp>

  1. include <vector>
  2. 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" );

} </lang>

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)

Clojure

<lang 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)))))</lang>
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))

D

opApply Version

Translation of: Python
Translation of: Perl 6

<lang d>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);

}</lang>

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.

Range Version

This is the same code modified to be a Range. <lang d>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);

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

Range Generator Version

<lang d>void main() {

   import std.stdio, std.range, std.algorithm, std.concurrency;
   Generator!T ludics(T)() {
       return new typeof(return)({
           T[] rotor, taken = [T(1)];
           yield(taken[0]);
           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 {
                   yield(i);
                   taken ~= i;
                   rotor ~= taken[j + 1];
               }
           }
       });
   }
   const L = ludics!uint.take(2005).array;
   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);

}</lang> The result is the same.

Eiffel

<lang Eiffel> class LUDIC_NUMBERS

create make

feature

make (n: INTEGER) -- Initialized arrays for find_ludic_numbers. require n_positive: n > 0 local i: INTEGER do create initial.make_filled (0, 1, n - 1) create ludic_numbers.make_filled (1, 1, 1) from i := 2 until i > n loop initial.put (i, i - 1) i := i + 1 end find_ludic_numbers end

ludic_numbers: ARRAY [INTEGER]

feature {NONE}

initial: ARRAY [INTEGER]

find_ludic_numbers -- Ludic numbers in array ludic_numbers. local count: INTEGER new_array: ARRAY [INTEGER] last: INTEGER do create new_array.make_from_array (initial) last := initial.count from count := 1 until count > last loop if ludic_numbers [ludic_numbers.count] /= new_array [1] then ludic_numbers.force (new_array [1], count + 1) end new_array := delete_i_elements (new_array) count := count + 1 end end

delete_i_elements (ar: ARRAY [INTEGER]): ARRAY [INTEGER] --- Array with all multiples of 'ar[1]' deleted. require ar_not_empty: ar.count > 0 local s_array: ARRAY [INTEGER] i, k: INTEGER length: INTEGER do create s_array.make_empty length := ar.count from i := 0 k := 1 until i = length loop if (i) \\ (ar [1]) /= 0 then s_array.force (ar [i + 1], k) k := k + 1 end i := i + 1 end if s_array.count = 0 then Result := ar else Result := s_array end ensure not_empty: not Result.is_empty end

end </lang> Test: <lang Eiffel> class APPLICATION

create make

feature

make local k, count: INTEGER do create ludic.make (22000) io.put_string ("%NLudic numbers up to 25. %N") across ludic.ludic_numbers.subarray (1, 25) as ld loop io.put_string (ld.item.out + "%N") end io.put_string ("%NLudic numbers from 2000 ... 2005. %N") across ludic.ludic_numbers.subarray (2000, 2005) as ld loop io.put_string (ld.item.out + "%N") end io.put_string ("%NNumber of Ludic numbers smaller than 1000. %N") from k := 1 until ludic.ludic_numbers [k] >= 1000 loop k := k + 1 count := count + 1 end io.put_integer (count) end

ludic: LUDIC_NUMBERS

end </lang>

Output:
Ludic numbers up to 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

Ludic numbers from 2000 ... 2005.
21475
21481
21487
21493
21503
21511

Number of Ludic numbers smaller than 1000.
142

Fortran

Works with: Fortran version 95 and later

<lang fortran>program ludic_numbers

 implicit none
 
 integer, parameter :: nmax = 25000
 logical :: ludic(nmax) = .true.
 integer :: i, j, n
 do i = 2, nmax / 2
   if (ludic(i)) then
     n = 0
     do j = i+1, nmax
       if(ludic(j)) n = n + 1
       if(n == i) then
         ludic(j) = .false.
         n = 0
       end if
     end do
   end if
 end do
 write(*, "(a)", advance = "no") "First 25 Ludic numbers: "
 n = 0
 do i = 1, nmax
   if(ludic(i)) then
     write(*, "(i0, 1x)", advance = "no") i
     n = n + 1
   end if
   if(n == 25) exit
 end do  
  
 write(*, "(/, a)", advance = "no") "Ludic numbers below 1000: "
 write(*, "(i0)") count(ludic(:999))

 write(*, "(a)", advance = "no") "Ludic numbers 2000 to 2005: " 
 n = 0
 do i = 1, nmax
   if(ludic(i)) then
      n = n + 1
      if(n >= 2000) then
        write(*, "(i0, 1x)", advance = "no") i
        if(n == 2005) exit
      end if
    end if
 end do  
 write(*, "(/, a)", advance = "no") "Ludic Triplets below 250: "
 do i = 1, 243
   if(ludic(i) .and. ludic(i+2) .and. ludic(i+6)) then
      write(*, "(a, 2(i0, 1x), i0, a, 1x)", advance = "no") "[", i, i+2, i+6, "]"
   end if  
 end do

end program</lang> 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 
Ludic numbers below 1000: 142
Ludic numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 
Ludic Triplets 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]

Go

<lang go>package main

import "fmt"

// Ludic returns a slice of Ludic numbers stopping after // either n entries or when max is exceeded. // Either argument may be <=0 to disable that limit. func Ludic(n int, max int) []uint32 { const maxInt32 = 1<<31 - 1 // i.e. math.MaxInt32 if max > 0 && n < 0 { n = maxInt32 } if n < 1 { return nil } if max < 0 { max = maxInt32 } sieve := make([]uint32, 10760) // XXX big enough for 2005 Ludics sieve[0] = 1 sieve[1] = 2 if n > 2 { // We start with even numbers already removed for i, j := 2, uint32(3); i < len(sieve); i, j = i+1, j+2 { sieve[i] = j } // We leave the Ludic numbers in place, // k is the index of the next Ludic for k := 2; k < n; k++ { l := int(sieve[k]) if l >= max { n = k break } i := l l-- // last is the last valid index last := k + i - 1 for j := k + i + 1; j < len(sieve); i, j = i+1, j+1 { last = k + i sieve[last] = sieve[j] if i%l == 0 { j++ } } // Truncate down to only the valid entries if last < len(sieve)-1 { sieve = sieve[:last+1] } } } if n > len(sieve) { panic("program error") // should never happen } return sieve[:n] }

func has(x []uint32, v uint32) bool { for i := 0; i < len(x) && x[i] <= v; i++ { if x[i] == v { return true } } return false }

func main() { // Ludic() is so quick we just call it repeatedly fmt.Println("First 25:", Ludic(25, -1)) fmt.Println("Numner of Ludics below 1000:", len(Ludic(-1, 1000))) fmt.Println("Ludic 2000 to 2005:", Ludic(2005, -1)[1999:])

fmt.Print("Tripples below 250:") x := Ludic(-1, 250) for i, v := range x[:len(x)-2] { if has(x[i+1:], v+2) && has(x[i+2:], v+6) { fmt.Printf(", (%d %d %d)", v, v+2, v+6) } } fmt.Println() }</lang> Run in Go Playground.

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]
Numner of Ludics below 1000: 142
Ludic 2000 to 2005: [21475 21481 21487 21493 21503 21511]
Tripples 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)

Haskell

<lang 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</lang>
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. <lang haskell>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</lang>

Icon and Unicon

This is inefficient, but was fun to code as a cascade of filters. Works in both languages. <lang unicon>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</lang>

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]
->

J

Solution (naive / brute force):<lang j> ludic =: _1 |.!.1 [: {."1 [: (#~ 0 ~: {. | i.@#)^:a: 2 + i.</lang> Examples:<lang j> # 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</lang>

Java

Works with: Java version 1.5+

This example uses pre-calculated ranges for the first and third task items (noted in comments). <lang java5>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))); } }</lang>

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]]

Julia

<lang Julia> function ludic_filter{T<:Integer}(n::T)

   0 < n || throw(DomainError())
   slud = trues(n)
   for i in 2:(n-1)
       slud[i] || continue
       x = 0
       for j in (i+1):n
           slud[j] || continue
           x += 1
           x %= i
           x == 0 || continue
           slud[j] = false
       end
   end
   return slud

end

ludlen = 10^5 slud = ludic_filter(ludlen) ludics = collect(1:ludlen)[slud]

n = 25 println("Generate and show here the first ", n, " ludic numbers.") print(" ") crwid = 76 wid = 0 for i in 1:(n-1)

   s = @sprintf "%d, " ludics[i]
   wid += length(s)
   if crwid < wid
       print("\n    ")
       wid = 0
   end
   print(s)

end println(ludics[n])

n = 10^3 println() println("How many ludic numbers are there less than or equal to ", n, "?") println(" ", sum(slud[1:n]))

lo = 2000 hi = lo+5 println() println("Show the ", lo, "..", hi, "'th ludic numbers.") for i in lo:hi

   println("    Ludic(", i, ") = ", ludics[i])

end

n = 250 println() println("Show all triplets of ludic numbers < ", n) for i = 1:n-7

   slud[i] || continue
   j = i+2
   slud[j] || continue
   k = i+6
   slud[k] || continue
   println("    ", i, ", ", j, ", ", k)

end </lang>

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.
    Ludic(2000) = 21475
    Ludic(2001) = 21481
    Ludic(2002) = 21487
    Ludic(2003) = 21493
    Ludic(2004) = 21503
    Ludic(2005) = 21511

Show 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

Mathematica

<lang 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];</lang>

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}}

Oforth

<lang Oforth>func: ludic(n) { | ludics l p |

  ListBuffer newSize(n) seqFrom(2, n) over addAll ->l
  ListBuffer newSize(n) dup add(1) ->ludics
  while(l notEmpty) [
     l removeFirst dup ludics add ->p  
     l size p / p * while(dup 1 > ) [ dup l removeAt drop p - ] drop
     ] 
  ludics

}

func: ludics { | l i |

  ludic(22000) ->l
  "First 25     : " print l left(25) println
  "Below 1000   : " print l filter(#[ 1000 < ]) size println
  "2000 to 2005 : " print l extract(2000, 2005) println
  250 loop: i [
     l include(i) ifFalse: [ continue ]
     l include(i 2 +) ifFalse: [ continue ]
     l include(i 6 +) ifFalse: [ continue ]
     i print ", " print i 2 + print ", " print i 6 + println
     ]

}</lang>

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]
Below 1000   : 142
2000 to 2005 : [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

Perl

The "ludic" subroutine caches the longest generated sequence so far. It also generates the candidates only if no candidates remain. <lang perl>#!/usr/bin/perl use warnings; use strict; use feature qw{ say };

{ my @ludic = (1);

   my $max = 3;
   my @candidates;
   sub sieve {
       my $l = shift;
       for (my $i = 0; $i <= $#candidates; $i += $l) {
           splice @candidates, $i, 1;
       }
   }
   sub ludic {
       my ($type, $n) = @_;
       die "Arg0 Type must be 'count' or 'max'\n"
            unless grep $_ eq $type, qw( count max );
       die "Arg1 Number must be > 0\n" if 0 >= $n;
       return (@ludic[ 0 .. $n - 1 ]) if 'count' eq $type and @ludic >= $n;
       return (grep $_ <= $n, @ludic) if 'max'   eq $type and $ludic[-1] >= $n;
       while (1) {
           if (@candidates) {
               last if ('max' eq $type and $candidates[0] > $n)
                    or ($n == @ludic);
               push @ludic, $candidates[0];
               sieve($ludic[-1] - 1);
           } else {
               $max *= 2;
               @candidates = 2 .. $max;
               for my $l (@ludic) {
                   sieve($l - 1) unless 1 == $l;
               }
           }
       }
       return (@ludic)
   }

}

my @triplet; my %ludic; undef @ludic{ ludic(max => 250) }; for my $i (keys %ludic) {

   push @triplet, $i if exists $ludic{ $i + 2 } and exists $ludic { $i + 6 };

}

say 'First 25: ', join ' ', ludic(count => 25); say 'Count < 1000: ', scalar ludic(max => 1000); say '2000..2005th: ', join ' ', (ludic(count => 2005))[1999 .. 2004]; say 'triplets < 250: ', join ' ',

                       map { '(' . join(' ',$_, $_ + 2, $_ + 6) . ')' }
                       sort { $a <=> $b } @triplet;</lang>
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 < 1000:   142
2000..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)

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... :-) <lang perl6>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;
   }</lang>
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>

PicoLisp

<lang PicoLisp>(de drop (Lst)

  (let N (car Lst)
     (make
        (for (I . X) (cdr Lst)
           (unless (=0 (% I N)) (link X)) ) ) ) )

(de comb (M Lst)

  (cond
     ((=0 M) '(NIL))
     ((not Lst))
     (T
        (conc
           (mapcar
              '((Y) (cons (car Lst) Y))
              (comb (dec M) (cdr Lst)) )
           (comb M (cdr Lst)) ) ) ) )
          

(de ludic (N)

  (let Ludic (range 1 100000)
     (make
        (link (pop 'Ludic))
        (do (dec N)
           (link (car Ludic))
           (setq Ludic (drop Ludic)) ) ) ) )

(let L (ludic 2005)

  (println (head 25 L))
  (println (cnt '((X) (< X 1000)) L))
  (println (tail 6 L))
  (println
     (filter
        '((Lst)
           (and
             (= (+ 2 (car Lst)) (cadr Lst))
             (= (+ 6 (car Lst)) (caddr Lst)) ) )
        (comb
           3
           (filter '((X) (< X 250)) L) ) ) ) )

(bye)</lang>

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)

((1 3 7) (5 7 11) (11 13 17) (23 25 29) (41 43 47) (173 175 179) (221 223 227) (233 235 239))

PL/I

This example is incorrect. Please fix the code and remove this message.

Details: Missing Triplet 1,3,7.

<lang PL/I>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;</lang> 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)

PL/SQL

<lang plsql>SET SERVEROUTPUT ON DECLARE

 c_limit CONSTANT PLS_INTEGER := 25000;
 TYPE t_nums IS TABLE OF PLS_INTEGER INDEX BY PLS_INTEGER;
 v_nums t_nums;
 v_ludic t_nums;
 v_count_ludic PLS_INTEGER;
 v_count_pos PLS_INTEGER;
 v_pos PLS_INTEGER;
 v_next_ludic PLS_INTEGER;
 FUNCTION is_ludic(p_num PLS_INTEGER) RETURN BOOLEAN IS
 BEGIN
   FOR i IN 1..v_ludic.COUNT LOOP
     EXIT WHEN v_ludic(i) > p_num;
     IF v_ludic(i) = p_num THEN
       RETURN TRUE;
     END IF;
   END LOOP;
   RETURN FALSE;
 END;

BEGIN

 FOR i IN 1..c_limit LOOP
   v_nums(i) := i;
 END LOOP;
 v_count_ludic := 1;
 v_next_ludic := 1;
 v_ludic(v_count_ludic) := v_next_ludic;
 v_nums.DELETE(1);
 WHILE v_nums.COUNT > 0 LOOP
   v_pos := v_nums.FIRST;
   v_next_ludic := v_nums(v_pos);
   v_count_ludic := v_count_ludic + 1;
   v_ludic(v_count_ludic) := v_next_ludic;
   v_count_pos := 0;
   WHILE v_pos IS NOT NULL LOOP
     IF MOD(v_count_pos, v_next_ludic) = 0 THEN
       v_nums.DELETE(v_pos);
     END IF;
     v_pos := v_nums.NEXT(v_pos);
     v_count_pos := v_count_pos + 1;
   END LOOP;
 END LOOP;
 dbms_output.put_line('Generate and show here the first 25 ludic numbers.');
 FOR i IN 1..25 LOOP
   dbms_output.put(v_ludic(i) || ' ');
 END LOOP;
 dbms_output.put_line();
 dbms_output.put_line('How many ludic numbers are there less than or equal to 1000?');
 v_count_ludic := 0;
 FOR i IN 1..v_ludic.COUNT LOOP
   EXIT WHEN v_ludic(i) > 1000;
   v_count_ludic := v_count_ludic + 1;
 END LOOP;
 dbms_output.put_line(v_count_ludic);
 dbms_output.put_line('Show the 2000..2005th ludic numbers.');
 FOR i IN 2000..2005 LOOP
   dbms_output.put(v_ludic(i) || ' ');
 END LOOP;
 dbms_output.put_line();
 dbms_output.put_line('A triplet is any three numbers x, x + 2, x + 6 where all three numbers are also ludic numbers.');
 dbms_output.put_line('Show all triplets of ludic numbers < 250 (Stretch goal)');
 FOR i IN 1..v_ludic.COUNT LOOP
   EXIT WHEN (v_ludic(i)+6) >= 250;
   IF is_ludic(v_ludic(i)+2) AND is_ludic(v_ludic(i)+6) THEN
     dbms_output.put_line(v_ludic(i) || ', ' || (v_ludic(i)+2) || ', ' || (v_ludic(i)+6));
   END IF;
 END LOOP;

END; / </lang>

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.
21475 21481 21487 21493 21503 21511 
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

Python

Python: Fast

<lang python>def ludic(nmax=100000):

   yield 1
   lst = list(range(2, nmax + 1))
   while lst:
       yield lst[0]
       del lst[::lst[0]]

ludics = [l for l in 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))</lang>
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)]

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. <lang python>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]</lang>

Output is the same as for the fast version.

Racket

<lang 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))))</lang>
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

REXX

<lang 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.*/</lang>

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►

Ruby

<lang 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</lang>
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]]

Seed7

<lang 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;</lang>
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)

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. <lang tcl>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]

}

  1. 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]}

}}]

  1. 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 ,]"</lang>

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)

zkl

This solution builds a stack of iterators, one for each Ludic number, each extending the previous iterator. A "master" iterator sits atop the stack and provides the interface to the stack. When the next Ludic number is requested, a "pulse train" ripples down and up and down ... the stack as numbers are crossed off the list(s) (figure of speech, no numbers are cached). <lang zkl>fcn dropNth(n,seq){

  if(n==2) return(seq.tweak(fcn(n){ if(n.isEven) Void.Skip else n })); // uggg, special case
  seq.tweak(fcn(n,skipper,w){ if(0==(w.idx+1)%skipper) Void.Skip else n }.fp1(n),

Void,Void,True); } fcn ludic{ //-->Walker

  Walker(fcn(rw){ w:=rw.value; n:=w.next(); rw.set(dropNth(n,w)); n }.fp(Ref([2..]))).push(1);

}</lang> <lang zkl>ludic().walk(25).toString(*).println(); ludic().reduce(fcn(sum,n){ if(n<1000) return(sum+1); return(Void.Stop,sum); },0).println(); ludic().drop(1999).walk(6).println(); // Ludic's between 2000 & 2005

ls:=ludic().filter(fcn(n){ (n<250) and True or Void.Stop }); // Ludic's < 250 ls.filter('wrap(n){ ls.holds(n+2) and ls.holds(n+6) }).apply(fcn(n){ T(n,n+2,n+6) }).println();</lang>

Output:
L(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
L(21475,21481,21487,21493,21503,21511)
L(L(1,3,7),L(5,7,11),L(11,13,17),L(23,25,29),L(41,43,47),L(173,175,179),L(221,223,227),L(233,235,239))