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 2nd indexed item from the array (including the first).
234567891011121314151617181920212223242526...
- (Unrolling a few loops...)
- Take the first member of the resultant array as the next ludic number 3.
- Remove every 3rd indexed item from the array (including the first).
35 7911 131517 192123 252729 313335 373941 434547 4951...
- Take the first member of the resultant array as the next ludic number 5.
- Remove every 5th indexed item from the array (including the first).
57 11 13 171923 25 29 313537 41 43 474953 55 59 616567 71 73 77 ...
- Take the first member of the resultant array as the next ludic number 7.
- Remove every 7th indexed item from the array (including the first).
711 13 17 23 25 293137 41 43 47 53 555961 67 71 73 77 838589 91 97 ...
- ...
- Take the first member of the current array as the next ludic number L.
- Remove every Lth 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..2005th ludic numbers.
- Stretch goal
Show all triplets of ludic numbers < 250.
- A triplet is any three numbers where all three numbers are also ludic numbers.
11l
F ludic(nmax = 100000)
V r = [1]
V lst = Array(2..nmax)
L !lst.empty
r.append(lst[0])
[Int] newlst
V step = lst[0]
L(i) 0 .< lst.len
I i % step != 0
newlst.append(lst[i])
lst = newlst
R r
V ludics = ludic()
print(‘First 25 ludic primes:’)
print(ludics[0.<25])
print("\nThere are #. ludic numbers <= 1000".format(sum(ludics.filter(l -> l <= 1000).map(l -> 1))))
print("\n2000'th..2005'th ludic primes:")
print(ludics[2000 - 1 .. 2004])
V n = 250
V triplets = ludics.filter(x -> x + 6 < :n &
x + 2 C :ludics &
x + 6 C :ludics).map(x -> (x, x + 2, x + 6))
print("\nThere are #. triplets less than #.:\n #.".format(triplets.len, 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)]
360 Assembly
* Ludic numbers 23/04/2016
LUDICN CSECT
USING LUDICN,R15 set base register
LH R9,NMAX r9=nmax
SRA R9,1 r9=nmax/2
LA R6,2 i=2
LOOPI1 CR R6,R9 do i=2 to nmax/2
BH ELOOPI1
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE ELOOPJ1
SR R8,R8 n=0
LA R7,1(R6) j=i+1
LOOPJ1 CH R7,NMAX do j=i+1 to nmax
BH ELOOPJ1
LA R1,LUDIC-1(R7) @ludic(j)
CLI 0(R1),X'01' if ludic(j)
BNE NOTJ1
LA R8,1(R8) n=n+1
NOTJ1 CR R8,R6 if n=i
BNE NDIFI
LA R1,LUDIC-1(R7) @ludic(j)
MVI 0(R1),X'00' ludic(j)=false
SR R8,R8 n=0
NDIFI LA R7,1(R7) j=j+1
B LOOPJ1
ELOOPJ1 LA R6,1(R6) i=i+1
B LOOPI1
ELOOPI1 XPRNT =C'First 25 ludic numbers:',23
LA R10,BUF @buf=0
SR R8,R8 n=0
LA R6,1 i=1
LOOPI2 CH R6,NMAX do i=1 to nmax
BH ELOOPI2
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE NOTI2
XDECO R6,XDEC i
MVC 0(4,R10),XDEC+8 output i
LA R10,4(R10) @buf=@buf+4
LA R8,1(R8) n=n+1
LR R2,R8 n
SRDA R2,32
D R2,=F'5' r2=mod(n,5)
LTR R2,R2 if mod(n,5)=0
BNZ NOTI2
XPRNT BUF,20
LA R10,BUF @buf=0
NOTI2 EQU *
CH R8,=H'25' if n=25
BE ELOOPI2
LA R6,1(R6) i=i+1
B LOOPI2
ELOOPI2 MVC BUF(25),=C'Ludic numbers below 1000:'
SR R8,R8 n=0
LA R6,1 i=1
LOOPI3 CH R6,=H'999' do i=1 to 999
BH ELOOPI3
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE NOTI3
LA R8,1(R8) n=n+1
NOTI3 LA R6,1(R6) i=i+1
B LOOPI3
ELOOPI3 XDECO R8,XDEC edit n
MVC BUF+25(6),XDEC+6 output n
XPRNT BUF,31 print buffer
MVC BUF(80),=CL80'Ludic numbers 2000 to 2005:'
LA R10,BUF+28 @buf=28
SR R8,R8 n=0
LA R6,1 i=1
LOOPI4 CH R6,NMAX do i=1 to nmax
BH ELOOPI4
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE NOTI4
LA R8,1(R8) n=n+1
CH R8,=H'2000' if n>=2000
BL NOTI4
XDECO R6,XDEC edit i
MVC 0(6,R10),XDEC+6 output i
LA R10,6(R10) @buf=@buf+6
CH R8,=H'2005' if n=2005
BE ELOOPI4
NOTI4 LA R6,1(R6) i=i+1
B LOOPI4
ELOOPI4 XPRNT BUF,80 print buffer
XPRNT =C'Ludic triplets below 250:',25
LA R6,1 i=1
LOOPI5 CH R6,=H'243' do i=1 to 243
BH ELOOPI5
LA R1,LUDIC-1(R6) @ludic(i)
CLI 0(R1),X'01' if ludic(i)
BNE ITERI5
LA R1,LUDIC+1(R6) @ludic(i+2)
CLI 0(R1),X'01' if ludic(i+2)
BNE ITERI5
LA R1,LUDIC+5(R6) @ludic(i+6)
CLI 0(R1),X'01' if ludic(i+6)
BNE ITERI5
MVC BUF+0(1),=C'[' [
XDECO R6,XDEC edit i
MVC BUF+1(4),XDEC+8 output i
LA R2,2(R6) i+2
XDECO R2,XDEC edit i+2
MVC BUF+5(4),XDEC+8 output i+2
LA R2,6(R6) i+6
XDECO R2,XDEC edit i+6
MVC BUF+9(4),XDEC+8 output i+6
MVC BUF+13(1),=C']' ]
XPRNT BUF,14 print buffer
ITERI5 LA R6,1(R6) i=i+1
B LOOPI5
ELOOPI5 XR R15,R15 set return code
BR R14 return to caller
LTORG
BUF DS CL80 buffer
XDEC DS CL12 decimal editor
NMAX DC H'25000' nmax
LUDIC DC 25000X'01' ludic(nmax)=true
YREGS
END LUDICN- 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]
ABAP
Works with NW 7.40 SP8
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.
- 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
Action!
Calculations on a real Atari 8-bit computer take quite long time. It is recommended to use an emulator capable with increasing speed of Atari CPU.
DEFINE NOTLUDIC="0"
DEFINE LUDIC="1"
DEFINE UNKNOWN="2"
PROC LudicSieve(BYTE ARRAY a INT count)
INT i,j,k
SetBlock(a,count,UNKNOWN)
a(0)=NOTLUDIC
a(1)=LUDIC
i=2
WHILE i<count
DO
IF a(i)=UNKNOWN THEN
a(i)=LUDIC
j=i k=0
WHILE j<count
DO
IF a(j)=UNKNOWN THEN
k==+1
IF k=i THEN
a(j)=NOTLUDIC
k=0
FI
FI
j==+1
OD
FI
i==+1
Poke(77,0) ;turn off the attract mode
OD
RETURN
PROC PrintLudicNumbers(BYTE ARRAY a INT count,first,last)
INT i,j
i=1 j=0
WHILE i<count AND j<=last
DO
IF a(i)=LUDIC THEN
IF j>=first THEN
PrintI(i) Put(32)
FI
j==+1
FI
i==+1
OD
PutE() PutE()
RETURN
INT FUNC CountLudicNumbers(BYTE ARRAY a INT max)
INT i,res
res=0
FOR i=1 TO max
DO
IF a(i)=LUDIC THEN
res==+1
FI
OD
RETURN (res)
PROC PrintLudicTriplets(BYTE ARRAY a INT max)
INT i,j
j=0
FOR i=0 TO max-6
DO
IF a(i)=LUDIC AND a(i+2)=LUDIC AND a(i+6)=LUDIC THEN
j==+1
PrintF("%I. %I-%I-%I%E",j,i,i+2,i+6)
FI
OD
RETURN
PROC Main()
DEFINE COUNT="22000"
BYTE ARRAY lud(COUNT+1)
INT i,n
PrintE("Please wait...")
LudicSieve(lud,COUNT+1)
Put(125) PutE() ;clear the screen
PrintE("First 25 ludic numbers:")
PrintLudicNumbers(lud,COUNT+1,0,24)
n=CountLudicNumbers(lud,1000)
PrintF("There are %I ludic numbers <= 1000%E%E",n)
PrintE("2000'th..2005'th ludic numbers:")
PrintLudicNumbers(lud,COUNT+1,1999,2004)
PrintE("Ludic triplets below 250")
PrintLudicTriplets(lud,249)
RETURN- Output:
Screenshot from Atari 8-bit computer
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'th..2005'th ludic numbers: 21475 21481 21487 21493 21503 21511 Ludic triplets below 250 1. 1-3-7 2. 5-7-11 3. 11-13-17 4. 23-25-29 5. 41-43-47 6. 173-175-179 7. 221-223-227 8. 233-235-239
Ada
with Ada.Text_IO;
with Ada.Containers.Vectors;
procedure Ludic_Numbers is
package Lucid_Lists is
new Ada.Containers.Vectors (Positive, Natural);
use Lucid_Lists;
List : Vector;
procedure Fill is
use type Ada.Containers.Count_Type;
Vec : Vector;
Lucid : Natural;
Index : Positive;
begin
Append (List, 1);
for I in 2 .. 22_000 loop
Append (Vec, I);
end loop;
loop
Lucid := First_Element (Vec);
Append (List, Lucid);
Index := First_Index (Vec);
loop
Delete (Vec, Index);
Index := Index + Lucid - 1;
exit when Index > Last_Index (Vec);
end loop;
exit when Length (Vec) <= 1;
end loop;
end Fill;
procedure Put_Lucid (First, Last : in Natural) is
use Ada.Text_IO;
begin
Put_Line ("Lucid numbers " & First'Image & " to " & Last'Image & ":");
for I in First .. Last loop
Put (Natural'(List (I))'Image);
end loop;
New_Line;
end Put_Lucid;
procedure Count_Lucid (Below : in Natural) is
Count : Natural := 0;
begin
for Lucid of List loop
if Lucid <= Below then
Count := Count + 1;
end if;
end loop;
Ada.Text_IO.Put_Line ("There are " & Count'Image & " lucid numbers <=" & Below'Image);
end Count_Lucid;
procedure Find_Triplets (Limit : in Natural) is
function Is_Lucid (Value : in Natural) return Boolean is
begin
for X in 1 .. Limit loop
if List (X) = Value then
return True;
end if;
end loop;
return False;
end Is_Lucid;
use Ada.Text_IO;
Index : Natural;
Lucid : Natural;
begin
Put_Line ("All triplets of lucid numbers <" & Limit'Image);
Index := First_Index (List);
while List (Index) < Limit loop
Lucid := List (Index);
if Is_Lucid (Lucid + 2) and Is_Lucid (Lucid + 6) then
Put ("(");
Put (Lucid'Image);
Put (Natural'(Lucid + 2)'Image);
Put (Natural'(Lucid + 6)'Image);
Put_Line (")");
end if;
Index := Index + 1;
end loop;
end Find_Triplets;
begin
Fill;
Put_Lucid (First => 1,
Last => 25);
Count_Lucid (Below => 1000);
Put_Lucid (First => 2000,
Last => 2005);
Find_Triplets (Limit => 250);
end Ludic_Numbers;
- Output:
Lucid numbers 1 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 There are 142 lucid numbers <= 1000 Lucid numbers 2000 to 2005: 21475 21481 21487 21493 21503 21511 All triplets of lucid 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)
ALGOL 68
# find some Ludic numbers #
# sieve the Ludic numbers up to 30 000 #
INT max number = 30 000;
[ 1 : max number ]INT candidates;
FOR n TO UPB candidates DO candidates[ n ] := n OD;
FOR n FROM 2 TO UPB candidates OVER 2 DO
IF candidates[ n ] /= 0 THEN
# have a ludic number #
INT number count := -1;
FOR remove pos FROM n TO UPB candidates DO
IF candidates[ remove pos ] /= 0 THEN
# have a number we haven't elminated yet #
number count +:= 1;
IF number count = n THEN
# this number should be removed #
candidates[ remove pos ] := 0;
number count := 0
FI
FI
OD
FI
OD;
# show some Ludic numbers and counts #
print( ( "Ludic numbers: " ) );
INT ludic count := 0;
FOR n TO UPB candidates DO
IF candidates[ n ] /= 0 THEN
# have a ludic number #
ludic count +:= 1;
IF ludic count < 26 THEN
# this is one of the first few Ludic numbers #
print( ( " ", whole( n, 0 ) ) );
IF ludic count = 25 THEN
print( ( " ...", newline ) )
FI
FI;
IF ludic count = 2000 THEN
print( ( "Ludic numbers 2000-2005: ", whole( n, 0 ) ) )
ELIF ludic count > 2000 AND ludic count < 2006 THEN
print( ( " ", whole( n, 0 ) ) );
IF ludic count = 2005 THEN
print( ( newline ) )
FI
FI
FI;
IF n = 1000 THEN
# count ludic numbers up to 1000 #
print( ( "There are ", whole( ludic count, 0 ), " Ludic numbers up to 1000", newline ) )
FI
OD;
# find the Ludic triplets below 250 #
print( ( "Ludic triplets below 250:", newline ) );
FOR n TO 250 - 6 DO
IF candidates[ n ] /= 0 AND candidates[ n + 2 ] /= 0 AND candidates[ n + 6 ] /= 0 THEN
# have a triplet #
print( ( " ", whole( n, -3 ), ", ", whole( n + 2, -3 ), ", ", whole( n + 6, -3 ), newline ) )
FI
OD- Output:
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 up to 1000
Ludic numbers 2000-2005: 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
AppleScript
-- Generate a list of the ludic numbers up to and including n.
on ludicsTo(n)
if (n < 1) then return {}
-- Start with an array of numbers from 2 to n and a ludic collection already containing 1.
script o
property array : {}
property ludics : {1}
end script
repeat with i from 2 to n
set end of o's array to i
end repeat
-- Collect ludics and sieve the array until a ludic matches or exceeds the remaining
-- array length, at which point the array contains just the remaining ludics.
set thisLudic to 2
set arrayLength to n - 1
repeat while (thisLudic < arrayLength)
set end of o's ludics to thisLudic
repeat with i from 1 to arrayLength by thisLudic
set item i of o's array to missing value
end repeat
set o's array to o's array's numbers
set thisLudic to beginning of o's array
set arrayLength to (count o's array)
end repeat
return (o's ludics) & (o's array)
end ludicsTo
on doTask()
script o
property ludics : ludicsTo(22000)
end script
set output to {}
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ", "
set end of output to "First 25 ludic numbers:"
set end of output to (items 1 thru 25 of o's ludics) as text
repeat with i from 1 to (count o's ludics)
if (item i of o's ludics > 1000) then exit repeat
end repeat
set end of output to "There are " & (i - 1) & " ludic numbers ≤ 1000."
set end of output to "2000th-2005th ludic numbers:"
set end of output to (items 2000 thru 2005 of o's ludics) as text
set end of output to "Triplets < 250:"
set triplets to {}
repeat with x in o's ludics
set x to x's contents
if (x > 243) then exit repeat
if ((x + 2) is in o's ludics) and ((x + 6) is in o's ludics) then
set end of triplets to "{" & {x, x + 2, x + 6} & "}"
end if
end repeat
set end of output to triplets as text
set AppleScript's text item delimiters to linefeed
set output to output as text
set AppleScript's text item delimiters to astid
return output
end doTask
return doTask()
- 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.
2000th-2005th ludic numbers:
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}"
Arturo
ludicGen: function [nmax][
result: [1]
lst: new 2..nmax+1
i: 0
worked: false
while [and? [not? empty? lst] [i < size lst]][
item: lst\[i]
result: result ++ item
del: 0
worked: false
while [del < size lst][
worked: true
remove 'lst .index del
del: dec del + item
]
if not? worked -> i: i + 1
]
return result
]
ludics: ludicGen 25000
print "The first 25 ludic numbers:"
print first.n: 25 ludics
leThan1000: select ludics => [& =< 1000]
print ["\nThere are" size leThan1000 "ludic numbers less than/or equal to 1000\n"]
print ["The ludic numbers from 2000th to 2005th are:" slice ludics 1999 2004 "\n"]
print "The triplets of ludic numbers less than 250 are:"
print map select ludics 'x [
all? @[ x < 250
contains? ludics x+2
contains? ludics x+6
]
] 't -> @[t, t+2, t+6]
- Output:
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 less than/or equal to 1000 The ludic numbers from 2000th to 2005th are: [21475 21481 21487 21493 21503 21511] The triplets of ludic numbers less than 250 are: [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
#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)
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)
C#
using System;
using System.Linq;
using System.Collections.Generic;
public class Program
{
public static void Main()
{
Console.WriteLine("First 25 ludic numbers:");
Console.WriteLine(string.Join(", ", LudicNumbers(150).Take(25)));
Console.WriteLine();
Console.WriteLine($"There are {LudicNumbers(1001).Count()} ludic numbers below 1000");
Console.WriteLine();
foreach (var ludic in LudicNumbers(22000).Skip(1999).Take(6)
.Select((n, i) => $"#{i+2000} = {n}")) {
Console.WriteLine(ludic);
}
Console.WriteLine();
Console.WriteLine("Triplets below 250:");
var queue = new Queue<int>(5);
foreach (int x in LudicNumbers(255)) {
if (queue.Count == 5) queue.Dequeue();
queue.Enqueue(x);
if (x - 6 < 250 && queue.Contains(x - 6) && queue.Contains(x - 4)) {
Console.WriteLine($"{x-6}, {x-4}, {x}");
}
}
}
public static IEnumerable<int> LudicNumbers(int limit) {
yield return 1;
//Like a linked list, but with value types.
//Create 2 extra entries at the start to avoid ugly index calculations
//and another at the end to avoid checking for index-out-of-bounds.
Entry[] values = Enumerable.Range(0, limit + 1).Select(n => new Entry(n)).ToArray();
for (int i = 2; i < limit; i = values[i].Next) {
yield return values[i].N;
int start = i;
while (start < limit) {
Unlink(values, start);
for (int step = 0; step < i && start < limit; step++)
start = values[start].Next;
}
}
}
static void Unlink(Entry[] values, int index) {
values[values[index].Prev].Next = values[index].Next;
values[values[index].Next].Prev = values[index].Prev;
}
}
struct Entry
{
public Entry(int n) : this() {
N = n;
Prev = n - 1;
Next = n + 1;
}
public int N { get; }
public int Prev { get; set; }
public int Next { get; set; }
}
- 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 below 1000 #2000 = 21475 #2001 = 21481 #2002 = 21487 #2003 = 21493 #2004 = 21503 #2005 = 21511 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
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)
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))
Common Lisp
(defun ludic-numbers (max &optional n)
(loop with numbers = (make-array (1+ max) :element-type 'boolean :initial-element t)
for i from 2 to max
until (and n (= num-results (1- n))) ; 1 will be added at the end
when (aref numbers i)
collect i into results
and count t into num-results
and do (loop for j from i to max
count (aref numbers j) into counter
when (= (mod counter i) 1)
do (setf (aref numbers j) nil))
finally (return (cons 1 results))))
(defun main ()
(format t "First 25 ludic numbers:~%")
(format t "~{~D~^ ~}~%" (ludic-numbers 100 25))
(terpri)
(format t "How many ludic numbers <= 1000?~%")
(format t "~D~%" (length (ludic-numbers 1000)))
(terpri)
(let ((numbers (ludic-numbers 30000 2005)))
(format t "~{#~D: ~D~%~}"
(mapcan #'list '(2000 2001 2002 2003 2004 2005) (nthcdr 1999 numbers))))
(terpri)
(loop with numbers = (ludic-numbers 250)
initially (format t "Triplets:~%")
for x in numbers
when (and (find (+ x 2) numbers)
(find (+ x 6) numbers))
do (format t "~3D ~3D ~3D~%" x (+ x 2) (+ x 6))))
- 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 How many ludic numbers <= 1000? 142 #2000: 21475 #2001: 21481 #2002: 21487 #2003: 21493 #2004: 21503 #2005: 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
D
opApply Version
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.
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.
Range Generator Version
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);
}
The result is the same.
Delphi
See Pascal.
EasyLang
proc initLudicArray n . res[] .
len res[] n
res[1] = 1
for i = 2 to n
k = 0
for j = i - 1 downto 2
k = k * res[j] div (res[j] - 1) + 1
.
res[i] = k + 2
.
.
initLudicArray 2005 arr[]
for i = 1 to 25
write arr[i] & " "
.
print ""
print ""
i = 1
while arr[i] <= 1000
cnt += 1
i += 1
.
print cnt
print ""
for i = 2000 to 2005
write arr[i] & " "
.
- 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
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
Test:
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
- 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
Elixir
defmodule Ludic do
def numbers(n \\ 100000) do
[h|t] = Enum.to_list(1..n)
numbers(t, [h])
end
defp numbers(list, nums) when length(list) < hd(list), do: Enum.reverse(nums, list)
defp numbers([h|_]=list, nums) do
Enum.drop_every(list, h) |> numbers([h | nums])
end
def task do
IO.puts "First 25 : #{inspect numbers(200) |> Enum.take(25)}"
IO.puts "Below 1000: #{length(numbers(1000))}"
tuple = numbers(25000) |> List.to_tuple
IO.puts "2000..2005th: #{ inspect for i <- 1999..2004, do: elem(tuple, i) }"
ludic = numbers(250)
triple = for x <- ludic, x+2 in ludic, x+6 in ludic, do: [x, x+2, x+6]
IO.puts "Triples below 250: #{inspect triple, char_lists: :as_lists}"
end
end
Ludic.task
- 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..2005th: [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]]
Factor
USING: formatting fry kernel make math math.ranges namespaces
prettyprint.config sequences sequences.extras ;
IN: rosetta-code.ludic-numbers
: next-ludic ( seq -- seq' )
dup first '[ nip _ mod zero? not ] filter-index ;
: ludics-upto-2005 ( -- a )
22,000 2 swap [a,b] [ ! 22k suffices to produce 2005 ludics
1 , [ building get length 2005 = ]
[ dup first , next-ludic ] until drop
] { } make ;
: ludic-demo ( -- )
100 margin set ludics-upto-2005
[ 6 tail* ] [ [ 1000 < ] count ] [ 25 head ] tri
"First 25 ludic numbers:\n%u\n\n"
"Count of ludic numbers less than 1000:\n%d\n\n"
"Ludic numbers 2000 to 2005:\n%u\n" [ printf ] tri@ ;
MAIN: ludic-demo
- 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 }
Count of ludic numbers less than 1000:
142
Ludic numbers 2000 to 2005:
{ 21475 21481 21487 21493 21503 21511 }
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
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]
FreeBASIC
' FB 1.05.0 Win64
' As it would be too expensive to actually remove elements from the array
' we instead set an element to 0 if it has been removed.
Sub ludic(n As Integer, lu() As Integer)
If n < 1 Then Return
Redim lu(1 To n)
lu(1) = 1
If n = 1 Then Return
Dim As Integer count = 1, count2
Dim As Integer i, j, k = 1
Dim As Integer ub = 22000 '' big enough to deal with up to 2005 ludic numbers
Dim a(2 To ub) As Integer
For i = 2 To ub : a(i) = i : Next
Do
k += 1
For i = k to ub
If a(i) > 0 Then
count += 1
lu(count) = a(i)
If n = count Then Return
a(i) = 0
k = i
Exit For
End If
Next
count2 = 0
j = k + 1
While j <= ub
If a(j) > 0 Then
count2 +=1
If count2 = k Then
a(j) = 0
count2 = 0
End If
End If
j += 1
Wend
Loop
End Sub
Dim i As Integer
Dim lu() As Integer
ludic(2005, lu())
Print "The first 25 Ludic numbers are :"
For i = 1 To 25
Print Using "###"; lu(i);
Print " ";
Next
Print
Dim As Integer Count = 0
For i = 1 To 1000
If lu(i) <= 1000 Then
count += 1
Else
Exit For
End If
Next
Print
Print "There are"; count; " Ludic numbers <= 1000"
Print
Print "The 2000th to 2005th Ludics are :"
For i = 2000 To 2005
Print lu(i); " ";
Next
Print : Print
Print "The Ludic triplets below 250 are : "
Dim As Integer j, k, ldc
Dim b As Boolean
For i = 1 To 248
ldc = lu(i)
If ldc >= 244 Then Exit For
b = False
For j = i + 1 To 249
If lu(j) = ldc + 2 Then
b = True
k = j
Exit For
ElseIf lu(j) > ldc + 2 Then
Exit For
End If
Next j
If b = False Then Continue For
For j = k + 1 To 250
If lu(j) = ldc + 6 Then
Print "("; Str(ldc); ","; ldc + 2; ","; ldc + 6; ")"
Exit For
ElseIf lu(j) > ldc + 6 Then
Exit For
End If
Next j
Next i
Erase lu
Print
Print "Press any key to quit"
Sleep
- 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 The 2000th to 2005th Ludics are : 21475 21481 21487 21493 21503 21511 The Ludic triplets below 250 are : (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
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()
}
- 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
import Data.List (unfoldr, genericSplitAt)
ludic :: [Integer]
ludic = 1 : unfoldr (\xs@(x:_) -> Just (x, dropEvery x xs)) [2 ..]
where
dropEvery n = concatMap 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
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] ->
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
Java
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]]
JavaScript
ES6
/**
* Boilerplate to simply get an array filled between 2 numbers
* @param {!number} s Start here (inclusive)
* @param {!number} e End here (inclusive)
*/
const makeArr = (s, e) => new Array(e + 1 - s).fill(s).map((e, i) => e + i);
/**
* Remove every n-th element from the given array
* @param {!Array} arr
* @param {!number} n
* @return {!Array}
*/
const filterAtInc = (arr, n) => arr.filter((e, i) => (i + 1) % n);
/**
* Generate ludic numbers
* @param {!Array} arr
* @param {!Array} result
* @return {!Array}
*/
const makeLudic = (arr, result) => {
const iter = arr.shift();
result.push(iter);
return arr.length ? makeLudic(filterAtInc(arr, iter), result) : result;
};
/**
* Our Ludic numbers. This is a bit of a cheat, as we already know beforehand
* up to where our seed array needs to go in order to exactly get to the
* 2005th Ludic number.
* @type {!Array<!number>}
*/
const ludicResult = makeLudic(makeArr(2, 21512), [1]);
// Below is just logging out the results.
/**
* Given a number, return a function that takes an array, and return the
* count of all elements smaller than the given
* @param {!number} n
* @return {!Function}
*/
const smallerThanN = n => arr => {
return arr.reduce((p,c) => {
return c <= n ? p + 1 : p
}, 0)
};
const smallerThan1K = smallerThanN(1000);
console.log('\nFirst 25 Ludic Numbers:');
console.log(ludicResult.filter((e, i) => i < 25).join(', '));
console.log('\nTotal Ludic numbers smaller than 1000:');
console.log(smallerThan1K(ludicResult));
console.log('\nThe 2000th to 2005th ludic numbers:');
console.log(ludicResult.filter((e, i) => i > 1998).join(', '));
console.log('\nTriplets smaller than 250:');
ludicResult.forEach(e => {
if (e + 6 < 250 && ludicResult.indexOf(e + 2) > 0 && ludicResult.indexOf(e + 6) > 0) {
console.log([e, e + 2, e + 6].join(', '));
}
});
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 Total Ludic numbers smaller than 1000: 142 The 2000th to 2005th ludic numbers: 21475, 21481, 21487, 21493, 21503, 21511 Triplets smaller 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
jq
In this entry, for each task, we do not assume any prior calculation of how big the initial sieve must be. That is, an adaptive approach is taken.
# This method for sieving turns out to be the fastest in jq.
# Input: an array to be sieved.
# Output: if the array length is less then $n then empty, else the sieved array.
def sieve($n):
if length<$n then empty
else . as $in
| reduce range(0;length) as $i ([]; if $i % $n == 0 then . else . + [$in[$i]] end)
end;
# Generate a stream of ludic numbers based on sieving range(2; $nmax+1)
def ludic($nmax):
def l:
.[0] as $next
| $next, (sieve($next)|l);
1, ([range(2; $nmax+1)] | l);
# Output: an array of the first . ludic primes (including 1)
def first_ludic_primes:
. as $n
| def l:
. as $k
| [ludic(10*$k)] as $a
| if ($a|length) >= $n then $a[:$n]
else (10*$k) | l
end;
l;
# Output: an array of the ludic numbers less than .
def ludic_primes:
. as $n
| def l:
. as $k
| [ludic(10*$k)] as $a
| if $a[-1] >= $n then $a | map(select(. < $n))
else (10*$k) | l
end;
l;
# Output; a stream of triplets of ludic numbers, where each member of the triplet is less than .
def triplets:
ludic_primes as $primes
| $primes[] as $p
| $primes
| bsearch($p) as $i
| if $i >= 0
then $primes[$i+1:]
| select( bsearch($p+2) >= 0 and
bsearch($p+6) >= 0)
| [$p, $p+2, $p+6]
else empty
end;
"First 25 ludic primes:", (25|first_ludic_primes),
"\nThere are \(1000|ludic_primes|length) ludic numbers <= 1000",
( "The \n2000th to 2005th ludic primes are:",
(2005|first_ludic_primes)[2000:]),
( [250 | triplets]
| "\nThere are \(length) triplets less than 250:",
.[] )- 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 2000th to 2005th ludic primes: [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]
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
- 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
Kotlin
// version 1.0.6
/* Rather than remove elements from a MutableList which would be a relatively expensive operation
we instead use two arrays:
1. An array of the Ludic numbers to be returned.
2. A 'working' array of a suitable size whose elements are set to 0 to denote removal. */
fun ludic(n: Int): IntArray {
if (n < 1) return IntArray(0)
val lu = IntArray(n) // array of Ludic numbers required
lu[0] = 1
if (n == 1) return lu
var count = 1
var count2: Int
var j: Int
var k = 1
var ub = n * 11 // big enough to deal with up to 2005 ludic numbers
val a = IntArray(ub) { it } // working array
while (true) {
k += 1
for (i in k until ub) {
if (a[i] > 0) {
count +=1
lu[count - 1] = a[i]
if (n == count) return lu
a[i] = 0
k = i
break
}
}
count2 = 0
j = k + 1
while (j < ub) {
if (a[j] > 0) {
count2 +=1
if (count2 == k) {
a[j] = 0
count2 = 0
}
}
j += 1
}
}
}
fun main(args: Array<String>) {
val lu: IntArray = ludic(2005)
println("The first 25 Ludic numbers are :")
for (i in 0 .. 24) print("%4d".format(lu[i]))
val count = lu.count { it <= 1000 }
println("\n\nThere are $count Ludic numbers <= 1000" )
println("\nThe 2000th to 2005th Ludics are :")
for (i in 1999 .. 2004) print("${lu[i]} ")
println("\n\nThe Ludic triplets below 250 are : ")
var k: Int = 0
var ldc: Int
var b: Boolean
for (i in 0 .. 247) {
ldc = lu[i]
if (ldc >= 244) break
b = false
for (j in i + 1 .. 248) {
if (lu[j] == ldc + 2) {
b = true
k = j
break
}
else if (lu[j] > ldc + 2) break
}
if (!b) continue
for (j in k + 1 .. 249) {
if (lu[j] == ldc + 6) {
println("($ldc, ${ldc + 2}, ${ldc + 6})")
break
}
else if (lu[j] > ldc + 6) break
}
}
}
- 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 The 2000th to 2005th Ludics are : 21475 21481 21487 21493 21503 21511 The Ludic triplets below 250 are : (1, 3, 7) (5, 7, 11) (11, 13, 17) (23, 25, 29) (41, 43, 47) (173, 175, 179) (221, 223, 227) (233, 235, 239)
Lua
-- Return table of ludic numbers below limit
function ludics (limit)
local ludList, numList, index = {1}, {}
for n = 2, limit do table.insert(numList, n) end
while #numList > 0 do
index = numList[1]
table.insert(ludList, index)
for key = #numList, 1, -1 do
if key % index == 1 then table.remove(numList, key) end
end
end
return ludList
end
-- Return true if n is found in t or false otherwise
function foundIn (t, n)
for k, v in pairs(t) do
if v == n then return true end
end
return false
end
-- Display msg followed by all values in t
function show (msg, t)
io.write(msg)
for _, v in pairs(t) do io.write(" " .. v) end
print("\n")
end
-- Main procedure
local first25, under1k, inRange, tripList, triplets = {}, 0, {}, {}, {}
for k, v in pairs(ludics(30000)) do
if k <= 25 then table.insert(first25, v) end
if v <= 1000 then under1k = under1k + 1 end
if k >= 2000 and k <= 2005 then table.insert(inRange, v) end
if v < 250 then table.insert(tripList, v) end
end
for _, x in pairs(tripList) do
if foundIn(tripList, x + 2) and foundIn(tripList, x + 6) then
table.insert(triplets, "\n{" .. x .. "," .. x+2 .. "," .. x+6 .. "}")
end
end
show("First 25:", first25)
print(under1k .. " are less than or equal to 1000\n")
show("2000th to 2005th:", inRange)
show("Triplets:", triplets)
- 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
142 are less than or equal to 1000
2000th to 2005th: 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}
Mathematica /Wolfram Language
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];
Ludic[[;; 25]]
LengthWhile[Ludic, # < 1000 &]
Ludic[[2000 ;; 2005]]
Select[Subsets[Select[Ludic, # < 250 &], {3}], Differences[#] == {2, 4} &]
- 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}}
Nim
Ludic number generation is inspired by Python lazy streaming generator. Note that to store the ludic numbers we have chosen to use an array rather than a sequence, which allows to use 1-based indexes.
import strutils
type LudicArray[N: static int] = array[1..N, int]
func initLudicArray[N: static int](): LudicArray[N] =
## Initialize an array of ludic numbers.
result[1] = 1
for i in 2..N:
var k = 0
for j in countdown(i - 1, 2):
k = k * result[j] div (result[j] - 1) + 1
result[i] = k + 2
proc print(text: string; list: openArray[int]) =
## Print a text followed by a list of ludic numbers.
var line = text
let start = line.len
for val in list:
line.addSep(", ", start)
line.add $val
echo line
func isLudic(ludicArray: LudicArray; n, start: Positive): bool =
## Check if a number "n" is ludic, starting search from index "start".
for idx in start..ludicArray.N:
let val = ludicArray[idx]
if n == val: return true
if n < val: break
when isMainModule:
let ludicArray = initLudicArray[2005]()
print "The 25 first ludic numbers are: ", ludicArray[1..25]
var count = 0
for n in ludicArray:
if n > 1000: break
inc count
echo "\nThere are ", count, " ludic numbers less or equal to 1000."
print "\nThe 2000th to 2005th ludic numbers are: ", ludicArray[2000..2005]
echo "\nThe triplets of ludic numbers less than 250 are:"
var line = ""
for i, n in ludicArray:
if n >= 244:
echo line
break
if ludicArray.isLudic(n + 2, i + 1) and ludicArray.isLudic(n + 6, i + 2):
line.addSep(", ")
line.add "($1, $2, $3)".format(n, n + 2, n + 6)
- Output:
The 25 first 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 less or equal to 1000. The 2000th to 2005th ludic numbers are: 21475, 21481, 21487, 21493, 21503, 21511 The triplets of ludic numbers less than 250 are: (1, 3, 7), (5, 7, 11), (11, 13, 17), (23, 25, 29), (41, 43, 47), (173, 175, 179), (221, 223, 227), (233, 235, 239)
Objeck
use Collection.Generic;
class Ludic {
function : Main(args : String[]) ~ Nil {
ludics := LudicUpTo(110);
Show("First 25 Ludics: ", ludics, 0, ludics->Size());
System.IO.Console->Print("Ludics up to 1000: ")->PrintLine(LudicUpTo(1000)->Size());
ludics := LudicUpTo(22000);
Show("2000th - 2005th Ludics: ", ludics, 1999, 2005);
Show("Triplets up to 250: ", GetTriplets(LudicUpTo(250)));
}
function : LudicUpTo(n : Int) ~ CompareVector<IntHolder> {
ludics := CompareVector->New()<IntHolder>;
for(i := 1; i <= n; i++;){
ludics->AddBack(i);
};
for(cursor := 1; cursor < ludics->Size(); cursor++;) {
thisLudic := ludics->Get(cursor);
removeCursor := cursor + thisLudic;
while(removeCursor < ludics->Size()){
ludics->Remove(removeCursor);
removeCursor := removeCursor + thisLudic - 1;
};
};
return ludics;
}
function : GetTriplets(ludics : CompareVector<IntHolder>) ~ Vector<CompareVector<IntHolder> > {
triplets := Vector->New()<CompareVector<IntHolder> >;
for(i := 0; i < ludics->Size() - 2; i++;){
thisLudic := ludics->Get(i);
if(ludics->Has(thisLudic + 2) & ludics->Has(thisLudic + 6)){
triplet := CompareVector->New()<IntHolder>;
triplet->AddBack(thisLudic);
triplet->AddBack(thisLudic + 2);
triplet->AddBack(thisLudic + 6);
triplets->AddBack(triplet);
};
};
return triplets;
}
function : Show(title : String, ludics : CompareVector<IntHolder>, start : Int, end : Int) ~ Nil {
title->Print();
'['->Print();
for(i := start; i < end; i +=1;) {
ludics->Get(i)->Get()->Print();
if(i + 1 < ludics->Size()) {
','->Print();
};
};
']'->PrintLine();
}
function : Show(title : String, triplets : Vector<CompareVector<IntHolder> >) ~ Nil {
title->PrintLine();
each(i : triplets) {
triplet := triplets->Get(i);
Show("\t", triplet, 0, triplet->Size());
};
}
}- 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]
Oforth
: ludic(n)
| ludics l p |
ListBuffer newSize(n) seqFrom(2, n) over addAll ->l
ListBuffer newSize(n) dup add(1) dup ->ludics
while(l notEmpty) [
l removeFirst dup ludics add ->p
l size p / p * while(dup 1 > ) [ dup l removeAt drop p - ] drop
] ;
: 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
] ;- 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
PARI/GP
Version #1. Creating vector of ludic numbers' flags, where the index of each flag=1 is the ludic number.
\\ Creating Vlf - Vector of ludic numbers' flags,
\\ where the index of each flag=1 is the ludic number.
\\ 2/28/16 aev
ludic(maxn)={my(Vlf=vector(maxn,z,1),n2=maxn\2,k,j1);
for(i=2,n2,
if(Vlf[i], k=0; j1=i+1;
for(j=j1,maxn, if(Vlf[j], k++); if(k==i, Vlf[j]=0; k=0))
);
);
return(Vlf);
}
{
\\ Required tests:
my(Vr,L=List(),k=0,maxn=25000);
Vr=ludic(maxn);
print("The first 25 Ludic numbers: ");
for(i=1,maxn, if(Vr[i]==1, k++; print1(i," "); if(k==25, break)));
print("");print("");
k=0;
for(i=1,999, if(Vr[i]==1, k++));
print("Ludic numbers below 1000: ",k);
print("");
k=0;
print("Ludic numbers 2000 to 2005: ");
for(i=1,maxn, if(Vr[i]==1, k++; if(k>=2000&&k<=2005, listput(L,i)); if(k>2005, break)));
for(i=1,6, print1(L[i]," "));
print(""); print("");
print("Ludic Triplets below 250: ");
for(i=1,250, if(Vr[i]&&Vr[i+2]&&Vr[i+6], print1("(",i," ",i+2," ",i+6,") ")));
}- Output:
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 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)
Version #2. Creating vector of ludic numbers.
Upgraded script from A003309 to meet task requirements.
\\ Creating Vl - Vector of ludic numbers.
\\ 2/28/16 aev
ludic2(maxn)={my(Vw=vector(maxn, x, x+1),Vl=Vec([1]),vwn=#Vw,i);
while(vwn>0, i=Vw[1]; Vl=concat(Vl,[i]);
Vw=vector((vwn*(i-1))\i,x,Vw[(x*i+i-2)\(i-1)]); vwn=#Vw
); return(Vl);
}
{
\\ Required tests:
my(Vr,L=List(),k=0,maxn=22000,vrs,vi);
Vr=ludic2(maxn); vrs=#Vr;
print("The first 25 Ludic numbers: ");
for(i=1,25, print1(Vr[i]," "));
print("");print("");
k=0;
for(i=1,vrs, if(Vr[i]<1000, k++, break));
print("Ludic numbers below 1000: ",k);
print("");
k=0;
print("Ludic numbers 2000 to 2005: ");
for(i=2000,2005, print1(Vr[i]," "));
print("");print("");
print("Ludic Triplets below 250: ");
for(i=1,vrs, vi=Vr[i]; if(i==1,print1("(",vi," ",vi+2," ",vi+6,") "); next); if(vi+6<250,if(Vr[i+1]==vi+2&&Vr[i+2]==vi+6, print1("(",vi," ",vi+2," ",vi+6,") "))));
}- Output:
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 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)
Pascal
Inspired by "rotors" of Raku. Runtime nearly quadratic: maxLudicCnt = 10000 -> 0.03 s =>maxLudicCnt= 100000 -> 3 s
program lucid;
{$IFDEF FPC}
{$MODE objFPC} // useful for x64
{$ENDIF}
const
//66164 -> last < 1000*1000;
maxLudicCnt = 2005;//must be > 1
type
tDelta = record
dNum,
dCnt : LongInt;
end;
tpDelta = ^tDelta;
tLudicList = array of tDelta;
tArrdelta =array[0..0] of tDelta;
tpLl = ^tArrdelta;
function isLudic(plL:tpLl;maxIdx:nativeInt):boolean;
var
i,
cn : NativeInt;
Begin
//check if n is 'hit' by a prior ludic number
For i := 1 to maxIdx do
with plL^[i] do
Begin
//Mask read modify write reread
//dec(dCnt);IF dCnt= 0
cn := dCnt;
IF cn = 1 then
Begin
dcnt := dNum;
isLudic := false;
EXIT;
end;
dcnt := cn-1;
end;
isLudic := true;
end;
procedure CreateLudicList(var Ll:tLudicList);
var
plL : tpLl;
n,LudicCnt : NativeUint;
begin
// special case 1
n := 1;
Ll[0].dNum := 1;
plL := @Ll[0];
LudicCnt := 0;
repeat
inc(n);
If isLudic(plL,LudicCnt ) then
Begin
inc(LudicCnt);
with plL^[LudicCnt] do
Begin
dNum := n;
dCnt := n;
end;
IF (LudicCnt >= High(LL)) then
BREAK;
end;
until false;
end;
procedure firstN(var Ll:tLudicList;cnt: NativeUint);
var
i : NativeInt;
Begin
writeln('First ',cnt,' ludic numbers:');
For i := 0 to cnt-2 do
write(Ll[i].dNum,',');
writeln(Ll[cnt-1].dNum);
end;
procedure triples(var Ll:tLudicList;max: NativeUint);
var
i,
chk : NativeUint;
Begin
// special case 1,3,7
writeln('Ludic triples below ',max);
write('(',ll[0].dNum,',',ll[2].dNum,',',ll[4].dNum,') ');
For i := 1 to High(Ll) do
Begin
chk := ll[i].dNum;
If chk> max then
break;
If (ll[i+2].dNum = chk+6) AND (ll[i+1].dNum = chk+2) then
write('(',ll[i].dNum,',',ll[i+1].dNum,',',ll[i+2].dNum,') ');
end;
writeln;
writeln;
end;
procedure LastLucid(var Ll:tLudicList;start,cnt: NativeUint);
var
limit,i : NativeUint;
Begin
dec(start);
limit := high(Ll);
IF cnt >= limit then
cnt := limit;
if start+cnt >limit then
start := limit-cnt;
writeln(Start+1,'.th to ',Start+cnt+1,'.th ludic number');
For i := 0 to cnt-1 do
write(Ll[i+start].dNum,',');
writeln(Ll[start+cnt].dNum);
writeln;
end;
function CountLudic(var Ll:tLudicList;Limit: NativeUint):NativeUint;
var
i,res : NativeUint;
Begin
res := 0;
For i := 0 to High(Ll) do begin
IF Ll[i].dnum <= Limit then
inc(res)
else
BREAK;
CountLudic:= res;
end;
end;
var
LudicList : tLudicList;
BEGIN
setlength(LudicList,maxLudicCnt);
CreateLudicList(LudicList);
firstN(LudicList,25);
writeln('There are ',CountLudic(LudicList,1000),' ludic numbers below 1000');
LastLucid(LudicList,2000,5);
LastLucid(LudicList,maxLudicCnt,5);
triples(LudicList,250);//all-> (LudicList,LudicList[High(LudicList)].dNum);
END.
- 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 below 1000 2000.th to 2005.th ludic number 21475,21481,21487,21493,21503,21511 99995.th to 100000.th ludic number 1561243,1561291,1561301,1561307,1561313,1561333 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) real 0m2.921s
Using an array of byte, each containing the distance to the next ludic number. 64-Bit needs only ~ 60% runtime of 32-Bit. Three times slower than the Version 1. Much space left for improvements, like memorizing the count of ludics of intervals of size 1024 or so, to do bigger steps.Something like skiplist.
program ludic;
{$IFDEF FPC}{$MODE DELPHI}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils;
const
MAXNUM =21511;// > 1
//1561333;-> 100000 ludic numbers
//1561243,1561291,1561301,1561307,1561313,1561333
type
tarrLudic = array of byte;
tLudics = array of LongWord;
var
Ludiclst : tarrLudic;
procedure Firsttwentyfive;
var
i,actLudic : NativeInt;
Begin
writeln('First 25 ludic numbers');
actLudic:= 1;
For i := 1 to 25 do
Begin
write(actLudic:3,',');
inc(actLudic,Ludiclst[actLudic]);
IF i MOD 5 = 0 then
writeln(#8#32);
end;
writeln;
end;
procedure CountBelowOneThousand;
var
cnt,actLudic : NativeInt;
Begin
write('Count of ludic numbers below 1000 = ');
actLudic:= 1;
cnt := 1;
while actLudic <= 1000 do
Begin
inc(actLudic,Ludiclst[actLudic]);
inc(cnt);
end;
dec(cnt);
writeln(cnt);writeln;
end;
procedure Show2000til2005;
var
cnt,actLudic : NativeInt;
Begin
writeln('ludic number #2000 to #2005');
actLudic:= 1;
cnt := 1;
while cnt < 2000 do
Begin
inc(actLudic,Ludiclst[actLudic]);
inc(cnt);
end;
while cnt < 2005 do
Begin
write(actLudic,',');
inc(actLudic,Ludiclst[actLudic]);
inc(cnt);
end;
writeln(actLudic);writeln;
end;
procedure ShowTriplets;
var
actLudic,lastDelta : NativeInt;
Begin
writeln('ludic numbers triplets below 250');
actLudic:= 1;
while actLudic < 250-5 do
Begin
IF (Ludiclst[actLudic] <> 0) AND
(Ludiclst[actLudic+2] <> 0) AND
(Ludiclst[actLudic+6] <> 0) then
writeln('{',actLudic,'|',actLudic+2,'|',actLudic+6,'} ');
inc(actLudic);
end;
writeln;
end;
procedure CheckMaxdist;
var
actLudic,Delta,MaxDelta : NativeInt;
Begin
MaxDelta := 0;
actLudic:= 1;
repeat
delta := Ludiclst[actLudic];
inc(actLudic,delta);
IF MAxDelta<delta then
MAxDelta:= delta;
until actLudic>= MAXNUM;
writeln('MaxDist ',MAxDelta);writeln;
end;
function GetLudics:tLudics;
//Array of byte containing the distance to next ludic number
//eliminated numbers are set to 0
var
i,actLudic,actcnt,delta,actPos,lastPos,ludicCnt: NativeInt;
Begin
setlength(Ludiclst,MAXNUM+1);
For i := MAXNUM downto 0 do
Ludiclst[i]:= 1;
actLudic := 1;
ludicCnt := 1;
repeat
inc(actLudic,Ludiclst[actLudic]);
IF actLudic> MAXNUM then
BREAK;
inc(ludicCnt);
actPos := actLudic;
actcnt := 0;
// Only if there are enough ludics left
IF MaxNum-ludicCnt-actPos > actPos then
Begin
//eliminate every element in actLudic-distance
//delta so i can set Ludiclst[actpos] to zero
delta := Ludiclst[actpos];
repeat
lastPos := actPos;
inc(actpos,delta);
if actPos>=MAXNUM then
BREAK;
delta := Ludiclst[actpos];
inc(actcnt);
IF actcnt= actLudic then
Begin
inc(Ludiclst[LastPos],delta);
//mark as not ludic
Ludiclst[actpos] := 0;
actcnt := 0;
end;
until false;
end;
until false;
writeln(ludicCnt,' ludic numbers upto ',MAXNUM,#13#10);
end;
BEGIN
GetLudics;
CheckMaxdist;
Firsttwentyfive;CountBelowOneThousand;Show2000til2005;ShowTriplets ;
setlength(Ludiclst,0)
END.
- Output:
2005 ludic numbers upto 21511
MaxDist 56
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
Count of ludic numbers below 1000 = 142
ludic number #2000 to #2005
21475,21481,21487,21493,21503,21511
ludic numbers 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}
real 0m0.003s
100000 ludic numbers upto 1561334
...
real 0m8.438s
Perl
The "ludic" subroutine caches the longest generated sequence so far. It also generates the candidates only if no candidates remain.
#!/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;
- 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)
Phix
constant LUMAX = 25000 sequence ludic = repeat(1,LUMAX) integer n for i=2 to LUMAX/2 do if ludic[i] then n = 0 for j=i+1 to LUMAX do n += ludic[j] if n=i then ludic[j] = 0 n = 0 end if end for end if end for sequence s = {} for i=1 to LUMAX do if ludic[i] then s &= i if length(s)=25 then exit end if end if end for printf(1,"First 25 Ludic numbers: %s\n",{sprint(s)}) printf(1,"Ludic numbers below 1000: %d\n",{sum(ludic[1..1000])}) s = {} n = 0 for i=1 to LUMAX do if ludic[i] then n += 1 if n>=2000 then s &= i if n=2005 then exit end if end if end if end for printf(1,"Ludic numbers 2000 to 2005: %s\n",{sprint(s)}) s = {} for i=1 to 243 do if ludic[i] and ludic[i+2] and ludic[i+6] then s = append(s,{i,i+2,i+6}) end if end for printf(1,"There are %d Ludic triplets below 250: %s\n",{length(s),sprint(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}
Ludic numbers below 1000: 142
Ludic numbers 2000 to 2005: {21475,21481,21487,21493,21503,21511}
There are 8 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}}
Picat
Recursion
ludic(N) = Ludic =>
ludic(2..N, [1], Ludic).
ludic([], Ludic0, Ludic) =>
Ludic = Ludic0.reverse().
ludic(T, Ludic0, Ludic) =>
T2 = ludic_keep(T),
ludic(T2,[T[1]|Ludic0],Ludic).
% which elements to keep
ludic_keep([]) = [].
ludic_keep([H|List]) = Ludic =>
ludic_keep(H,1,List,[],Ludic).
ludic_keep(_H,_C,[],Ludic0,Ludic) ?=>
Ludic = Ludic0.reverse().
ludic_keep(H,C,[H1|T],Ludic0,Ludic) =>
(
C mod H > 0 ->
ludic_keep(H,C+1,T,[H1|Ludic0],Ludic)
;
ludic_keep(H,C+1,T,Ludic0,Ludic)
).Imperative approach
ludic2(N) = Ludic =>
A = 1..N,
Ludic = [1],
A := delete(A, 1),
while(A.length > 0)
T = A[1],
Ludic := Ludic ++ [T],
A := delete(A,T),
A := [A[J] : J in 1..A.length, J mod T > 0]
end.Test
The recursive variant is about 10 times faster than the imperative.
go =>
time(check(ludic)),
time(check(ludic2)),
nl.
check(LudicFunc) =>
println(ludicFunc=LudicFunc),
Ludic1000 = apply(LudicFunc,1000),
% first 25
println(first_25=Ludic1000[1..25]),
% below 1000
println(num_below_1000=Ludic1000.length),
% 2000..2005
Ludic22000 = apply(LudicFunc,22000),
println(len_22000=Ludic22000.length),
println(ludic_2000_2005=[Ludic22000[I] : I in 2000..2005]),
% Triplets
Ludic2500 = apply(LudicFunc,2500),
Triplets=[[N,N+2,N+6] : N in 1..Ludic2500.length,
membchk(N,Ludic2500),
membchk(N+2,Ludic2500),
membchk(N+6,Ludic2500)],
foreach(Triplet in Triplets)
println(Triplet)
end,
nl.- Output:
ludicFunc = 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] num_below_1000 = 142 len_22000 = 2042 ludic_2000_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] CPU time 0.288 seconds. ludicFunc = ludic2 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] num_below_1000 = 142 len_22000 = 2042 ludic_2000_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] CPU time 2.835 seconds.
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)- 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
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;
put edit ('(', L(1), L(3), L(5), ') ' ) (A, 3 F(4), A);
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: ( 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/SQL
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..2005''th 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;
/
- 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
PowerShell
# Start with a pool large enough to meet the requirements
$Pool = [System.Collections.ArrayList]( 2..22000 )
# Start with 1, because it's grandfathered in
$Ludic = @( 1 )
# While the size of the pool is still larger than the next Ludic number...
While ( $Pool.Count -gt $Pool[0] )
{
# Add the next Ludic number to the list
$Ludic += $Pool[0]
# Remove from the pool all entries whose index is a multiple of the next Ludic number
[math]::Truncate( ( $Pool.Count - 1 )/ $Pool[0])..0 | ForEach { $Pool.RemoveAt( $_ * $Pool[0] ) }
}
# Add the rest of the numbers in the pool to the list of Ludic numbers
$Ludic += $Pool.ToArray()
# Display the first 25 Ludic numbers
$Ludic[0..24] -join ", "
''
# Display the count of all Ludic numbers under 1000
$Ludic.Where{ $_ -le 1000 }.Count
''
# Display the 2000th through the 2005th Ludic number
$Ludic[1999..2004] -join ", "
''
# Display all Ludic triplets less than 250
$TripletStart = $Ludic.Where{ $_ -lt 244 -and ( $_ + 2 ) -in $Ludic -and ( $_ + 6 ) -in $Ludic }
$TripletStart.ForEach{ $_, ( $_ + 2 ), ( $_ + 6 ) -join ", " }
- Output:
1, 2, 3, 5, 7, 11, 13, 17, 23, 25, 29, 37, 41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97, 107 142 21475, 21481, 21487, 21493, 21503, 21511 1, 3, 7 5, 7, 11 11, 13, 17 23, 25, 29 41, 43, 47 173, 175, 179 221, 223, 227 233, 235, 239
Prolog
Simple, straightforward implementation
% John Devou: 26-Nov-2021
d(_,_,[],[]).
d(N,N,[_|Xs],Rs):- d(N,1,Xs,Rs).
d(N,M,[X|Xs],[X|Rs]):- M < N, M_ is M+1, d(N,M_,Xs,Rs).
l([],[]).
l([X|Xs],[X|Rs]):- d(X,1,Xs,Ys), l(Ys,Rs).
% g(N,L):- generate in L a list with Ludic numbers up to N
g(N,[1|X]):- numlist(2,N,L), l(L,X).
s(0,Xs,[],Xs).
s(N,[X|Xs],[X|Ls],Rs):- N > 0, M is N-1, s(M,Xs,Ls,Rs).
t([X,Y,Z|_],[X,Y,Z]):- Y =:= X+2, Z =:= X+6.
t([_,Y,Z|Xs],R):- t([Y,Z|Xs],R).
% tasks
t1:- g(500,L), s(25,L,X,_), write(X), !.
t2:- g(1000,L), length(L,X), write(X), !.
t3:- g(22000,L), s(1999,L,_,R), s(6,R,X,_), write(X), !.
t4:- g(249,L), findall(A, t(L,A), X), write(X), !.
- Output:
?- t1. [1,2,3,5,7,11,13,17,23,25,29,37,41,43,47,53,61,67,71,77,83,89,91,97,107] true. ?- t2. 142 true. ?- t3. [21475,21481,21487,21493,21503,21511] true. ?- t4. [[5,7,11],[11,13,17],[23,25,29],[41,43,47],[173,175,179],[221,223,227],[233,235,239]] true.
PureBasic
EnableExplicit
If Not OpenConsole() : End 1 : EndIf
#NMAX=25000
Dim ludic.b(#NMAX)
FillMemory(@ludic(0),SizeOf(Byte)*#NMAX,#True,#PB_Byte)
Define.i i,j,n,c1,c2
Define r1$,r2$,r3$,r4$
For i=2 To Int(#NMAX/2)
If ludic(i)
n=0
For j=i+1 To #NMAX
If ludic(j) : n+1 : EndIf
If n=i : ludic(j)=#False : n=0 : EndIf
Next j
EndIf
Next i
n=0 : c1=0 : c2=0
For i=1 To #NMAX
If ludic(i) And n<25 : n+1 : r1$+Str(i)+" " : EndIf
If i<=1000 : c1+Bool(ludic(i)) : EndIf
c2+Bool(ludic(i))
If c2>=2000 And c2<=2005 And ludic(i) : r3$+Str(i)+" " : EndIf
If i<243 And ludic(i) And ludic(i+2) And ludic(i+6)
r4$+"["+Str(i)+" "+Str(i+2)+" "+Str(i+6)+"] "
EndIf
Next
r2$=Str(c1)
PrintN("First 25 Ludic numbers: " +r1$)
PrintN("Ludic numbers below 1000: " +r2$)
PrintN("Ludic numbers 2000 to 2005: " +r3$)
PrintN("Ludic Triplets below 250: " +r4$)
Input()
End
- 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]
Python
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 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))
- 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.
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.
Python: lazy streaming generator
The following streaming version of function ludic also returns ludic numbers until reaching system limits.
Instead of creating a bounded table and deleting elements, it uses the insight that after each iteration the remaining numbers are shuffled left, modifying their indices in a regular way. Reversing this process tracks the k'th ludic number in the final list back to its position in the initial list of integers, and hence determines its value without any need to build the table. The only storage requirement is for the list of numbers found so far.
Based on the similar algorithm for lucky numbers at https://oeis.org/A000959/a000959.txt.
Function triplets wraps ludic and uses a similar stream-filtering approach to find triplets.
def ludic():
yield 1
ludics = []
while True:
k = 0
for j in reversed(ludics):
k = (k*j)//(j-1) + 1
ludics.append(k+2)
yield k+2
def triplets():
a, b, c, d = 0, 0, 0, 0
for k in ludic():
if k-4 in (b, c, d) and k-6 in (a, b, c):
yield k-6, k-4, k
a, b, c, d = b, c, d, k
first_25 = [k for i, k in zip(range(25), gen_ludic())]
print(f'First 25 ludic numbers: {first_25}')
count = 0
for k in gen_ludic():
if k > 1000:
break
count += 1
print(f'Number of ludic numbers <= 1000: {count}')
it = iter(gen_ludic())
for i in range(1999):
next(it)
ludic2000 = [next(it) for i in range(6)]
print(f'Ludic numbers 2000..2005: {ludic2000}')
print('Ludic triplets < 250:')
for a, b, c in triplets():
if c >= 250:
break
print(f'[{a}, {b}, {c}]')
- 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] Number of ludic numbers <= 1000: 142 Ludic numbers 2000..2005: [21475, 21481, 21487, 21493, 21503, 21511] 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]
Quackery
[ 0 over of
swap times
[ i 1+ swap i poke ]
1 split
[ dup 0 peek
rot over join unrot
over size over > while
1 - temp put
[] swap
[ behead drop
temp share split
dip join
dup [] = until ]
drop temp release
again ]
drop behead drop join ] is ludic ( n --> [ )
999 ludic
say "First 25 ludic numbers: "
dup 25 split drop echo
cr cr
say "There are "
size echo
say " ludic numbers less than 1000."
cr cr
25000 ludic
say "Ludic numbers 2000 to 2005: "
1999 split nip 6 split drop echo- 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 less than 1000. Ludic numbers 2000 to 2005: [ 21475 21481 21487 21493 21503 21511 ]
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
Raku
(formerly 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.sort -> $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>)
REXX
/*REXX program gens/shows (a range of) ludic numbers, or a count when a range is used.*/
parse arg N count bot top triples . /*obtain optional arguments from the CL*/
if N=='' | N=="," then N= 25 /*Not specified? Then use the default.*/
if count=='' | count=="," then count= 1000 /* " " " " " " */
if bot=='' | bot=="," then bot= 2000 /* " " " " " " */
if top=='' | top=="," then top= 2005 /* " " " " " " */
if triples=='' | triples=="," then triples= 249 /* " " " " " " */
#= 0 /*the number of ludic numbers (so far).*/
$= ludic( max(N, count, bot, top, triples) ) /*generate enough ludic nums*/
say 'The first ' N " ludic numbers: " subword($,1,25) /*display 1st N ludic nums*/
do j=1 until word($, j) > count /*search up to a specific #.*/
end /*j*/
say
say "There are " j - 1 ' ludic numbers that are ≤ ' count
say
say "The " bot '───►' top ' (inclusive) ludic numbers are: ' subword($, bot)
@= /*list of ludic triples found (so far).*/
do j=1 for words($)
_= word($, j) /*it is known that ludic _ exists. */
if _>=triples then leave /*only process up to a specific number.*/
if wordpos(_+2, $)==0 | wordpos(_+6, $)==0 then iterate /*Not triple? Skip it.*/
#= # + 1 /*bump the triple counter. */
@= @ '◄'_ _+2 _+6"► " /*append the newly found triple ──► @ */
end /*j*/
say
if @=='' then say 'From 1──►'triples", no triples found."
else say 'From 1──►'triples", " # ' triples found:' @
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
ludic: procedure; parse arg m,,@; $= 1 2 /*$≡ludic numbers superset; @≡sequence*/
do j=3 by 2 to m*15; @= @ j /*construct an initial list of numbers.*/
end /*j*/
@= @' '; n= words(@) /*append a blank to the number sequence*/
do while n\==0; f= word(@, 1) /*examine the first word in the @ list.*/
$= $ f /*add the word to the $ list. */
do d=1 by f while d<=n; n= n-1 /*use 1st number, elide all occurrences*/
@= changestr(' 'word(@, d)" ", @, ' . ') /*cross─out a number in @ */
end /*d*/ /* [↑] done eliding the "1st" number. */
@= translate(@, , .) /*change dots to blanks; count numbers.*/
end /*while*/ /* [↑] done eliding ludic numbers. */
return subword($, 1, m) /*return a range of ludic numbers. */
Some older REXXes don't have a changestr BIF, so one is included here ──► CHANGESTR.REX.
output using the default values 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 that are ≤ 1000 The 2000 ───► 2005 (inclusive) 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►
Ring
# Project : Ludic numbers
ludic = list(300)
resludic = []
nr = 1
for n = 1 to len(ludic)
ludic[n] = n+1
next
see "the first 25 Ludic numbers are:" + nl
ludicnumbers(ludic)
showarray(resludic)
see nl
see "Ludic numbers below 1000: "
ludic = list(3000)
resludic = []
for n = 1 to len(ludic)
ludic[n] = n+1
next
ludicnumbers(ludic)
showarray2(resludic)
see nr
see nl + nl
see "the 2000..2005th Ludic numbers are:" + nl
ludic = list(60000)
resludic = []
for n = 1 to len(ludic)
ludic[n] = n+1
next
ludicnumbers(ludic)
showarray3(resludic)
func ludicnumbers(ludic)
for n = 1 to len(ludic)
delludic = []
for m = 1 to len(ludic) step ludic[1]
add(delludic, m)
next
add(resludic, ludic[1])
for p = len(delludic) to 1 step -1
del(ludic, delludic[p])
next
next
func showarray(vect)
see "[1, "
svect = ""
for n = 1 to 24
svect = svect + vect[n] + ", "
next
svect = left(svect, len(svect) - 2)
see svect
see "]" + nl
func showarray2(vect)
for n = 1 to len(vect)
if vect[n] <= 1000
nr = nr + 1
ok
next
return nr
func showarray3(vect)
see "["
svect = ""
for n = 1999 to 2004
svect = svect + vect[n] + ", "
next
svect = left(svect, len(svect) - 2)
see svect
see "]" + nlOutput:
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] Ludic numbers below 1000: 142 the 2000..2005th ludic numbers are: [21475, 21481, 21487, 21493, 21503, 21511]
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]]
Rust
const ARRAY_MAX: usize = 25_000;
const LUDIC_MAX: usize = 2100;
/// Calculates and returns the first `LUDIC_MAX` Ludic numbers.
///
/// Needs a sufficiently large `ARRAY_MAX`.
fn ludic_numbers() -> Vec<usize> {
// The first two Ludic numbers
let mut numbers = vec![1, 2];
// We start the array with an immediate first removal to reduce memory usage by
// collecting only odd numbers.
numbers.extend((3..ARRAY_MAX).step_by(2));
// We keep the correct Ludic numbers in place, removing the incorrect ones.
for ludic_idx in 2..LUDIC_MAX {
let next_ludic = numbers[ludic_idx];
// We remove incorrect numbers by counting the indices after the correct numbers.
// We start from zero and keep until we reach the potentially incorrect numbers.
// Then we keep only those not divisible by the `next_ludic`.
let mut idx = 0;
numbers.retain(|_| {
let keep = idx <= ludic_idx || (idx - ludic_idx) % next_ludic != 0;
idx += 1;
keep
});
}
numbers
}
fn main() {
let ludic_numbers = ludic_numbers();
print!("First 25: ");
print_n_ludics(&ludic_numbers, 25);
println!();
print!("Number of Ludics below 1000: ");
print_num_ludics_upto(&ludic_numbers, 1000);
println!();
print!("Ludics from 2000 to 2005: ");
print_ludics_from_to(&ludic_numbers, 2000, 2005);
println!();
println!("Triplets below 250: ");
print_triplets_until(&ludic_numbers, 250);
}
/// Prints the first `n` Ludic numbers.
fn print_n_ludics(x: &[usize], n: usize) {
println!("{:?}", &x[..n]);
}
/// Calculates how many Ludic numbers are below `max_num`.
fn print_num_ludics_upto(x: &[usize], max_num: usize) {
let num = x.iter().take_while(|&&i| i < max_num).count();
println!("{}", num);
}
/// Prints Ludic numbers between two numbers.
fn print_ludics_from_to(x: &[usize], from: usize, to: usize) {
println!("{:?}", &x[from - 1..to - 1]);
}
/// Calculates triplets until a certain Ludic number.
fn triplets_below(ludics: &[usize], limit: usize) -> Vec<(usize, usize, usize)> {
ludics
.iter()
.enumerate()
.take_while(|&(_, &num)| num < limit)
.filter_map(|(idx, &number)| {
let triplet_2 = number + 2;
let triplet_3 = number + 6;
// Search for the other two triplet numbers. We know they are larger than
// `number` so we can give the searches lower bounds of `idx + 1` and
// `idx + 2`. We also know that the `n + 2` number can only ever be two
// numbers away from the previous and the `n + 6` number can only be four
// away (because we removed some in between). Short circuiting and doing
// the check more likely to fail first are also useful.
let is_triplet = ludics[idx + 1..idx + 3].binary_search(&triplet_2).is_ok()
&& ludics[idx + 2..idx + 5].binary_search(&triplet_3).is_ok();
if is_triplet {
Some((number, triplet_2, triplet_3))
} else {
None
}
})
.collect()
}
/// Prints triplets until a certain Ludic number.
fn print_triplets_until(ludics: &[usize], limit: usize) {
for (number, triplet_2, triplet_3) in triplets_below(ludics, limit) {
println!("{} {} {}", number, triplet_2, triplet_3);
}
}
- 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] Number of Ludics below 1000: 142 Ludics from 2000 to 2005: [21475, 21481, 21487, 21493, 21503] 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
Scala
In this example, we define a function to drop every nth element from a list and use it to build a lazily evaluated list of all Ludic numbers. We then generate a lazy list of triplets and filter for the triplets of Ludic numbers.
object Ludic {
def main(args: Array[String]): Unit = {
println(
s"""|First 25 Ludic Numbers: ${ludic.take(25).mkString(", ")}
|Ludic Numbers <= 1000: ${ludic.takeWhile(_ <= 1000).size}
|2000-2005th Ludic Numbers: ${ludic.slice(1999, 2005).mkString(", ")}
|Triplets <= 250: ${triplets.takeWhile(_._3 <= 250).mkString(", ")}""".stripMargin)
}
def dropByN[T](lst: LazyList[T], len: Int): LazyList[T] = lst.grouped(len).flatMap(_.init).to(LazyList)
def ludic: LazyList[Int] = 1 #:: LazyList.unfold(LazyList.from(2)){case n +: ns => Some((n, dropByN(ns, n)))}
def triplets: LazyList[(Int, Int, Int)] = LazyList.from(1).map(n => (n, n + 2, n + 6)).filter{case (a, b, c) => Seq(a, b, c).forall(ludic.takeWhile(_ <= c).contains)}
}
- 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 <= 1000: 142 2000-2005th Ludic Numbers: 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)
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)
SequenceL
import <Utilities/Set.sl>;
ludic(v(1), result(1)) :=
let
n := head(v);
filtered[i] := v[i] when (i-1) mod n /= 0;
in
result when size(v) < 1 else
ludic(filtered, result ++ [n]);
count : int(1) * int * int -> int;
count(v(1), top, index) :=
index-1 when v[index] > top else
count(v, top, index + 1);
main() :=
let
ludics := ludic(2...100000, [1]);
ludics250 := ludics[1 ... count(ludics, 250, 1)];
triplets[i] := [i, i+2, i+6] when elementOf(i+2, ludics250) and elementOf(i+6, ludics250)
foreach i within ludics250;
in
"First 25:\n" ++ toString(ludics[1...25]) ++
"\n\nLudics below 1000:\n" ++ toString(count(ludics, 1000, 1)) ++
"\n\nLudic 2000 to 2005:\n" ++ toString(ludics[2000...2005]) ++
"\n\nTriples below 250:\n" ++ toString(triplets) ;- 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]]
Sidef
func ludics_upto(nmax=100000) {
Enumerator({ |collect|
collect(1)
var arr = @(2..nmax)
while (arr) {
collect(var n = arr[0])
arr.range.by(n).each {|i| arr[i] = nil}
arr.compact!
}
})
}
func ludics_first(n) {
ludics_upto(n * n.log2).first(n)
}
say("First 25 Ludic numbers: ", ludics_first(25).join(' '))
say("Ludics below 1000: ", ludics_upto(1000).len)
say("Ludic numbers 2000 to 2005: ", ludics_first(2005).last(6).join(' '))
var a = ludics_upto(250).to_a
say("Ludic triples below 250: ", a.grep{|x| a.contains_all([x+2, x+6]) } \
.map {|x| '(' + [x, x+2, x+6].join(' ') + ')' } \
.join(' '))
- 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)
Standard ML
open List;
fun Ludi [] = []
| Ludi (T as h::L) =
let
fun next (h:: L ) =
let
val nw = #2 (ListPair.unzip (filter (fn (a,b) => a mod #2 h <> 0) L) )
in
ListPair.zip ( List.tabulate(List.length nw,fn i=>i) ,nw)
end
in
h :: Ludi ( next T)
end;
val ludics = 1:: (#2 (ListPair.unzip(Ludi (ListPair.zip ( List.tabulate(25000,fn i=>i),tabulate (25000,fn i=>i+2)) )) ));
app ((fn e => print (e^" ")) o Int.toString ) (take (ludics,25));
length (filter (fn e=> e <= 1000) ludics);
drop (take (ludics,2005),1999);
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 val it = () : unit - val it = 142 : int - val it = [21475,21481,21487,21493,21503,21511] : int list
Tcl
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)
VBScript
Set list = CreateObject("System.Collections.Arraylist")
Set ludic = CreateObject("System.Collections.Arraylist")
'populate the list
For i = 1 To 25000
list.Add i
Next
'set 1 as the first ludic number
ludic.Add list(0)
list.RemoveAt(0)
'variable to count ludic numbers <= 1000
up_to_1k = 1
'determine the succeeding ludic numbers
For j = 2 To 2005
If list.Count > 0 Then
If list(0) <= 1000 Then
up_to_1k = up_to_1k + 1
End If
ludic.Add list(0)
Else
Exit For
End If
increment = list(0) - 1
n = 0
Do While n <= list.Count - 1
list.RemoveAt(n)
n = n + increment
Loop
Next
'the first 25 ludics
WScript.StdOut.WriteLine "First 25 Ludic Numbers:"
For k = 0 To 24
If k < 24 Then
WScript.StdOut.Write ludic(k) & ", "
Else
WScript.StdOut.Write ludic(k)
End If
Next
WScript.StdOut.WriteBlankLines(2)
'the number of ludics up to 1000
WScript.StdOut.WriteLine "Ludics up to 1000: "
WScript.StdOut.WriteLine up_to_1k
WScript.StdOut.WriteBlankLines(1)
'2000th - 2005th ludics
WScript.StdOut.WriteLine "The 2000th - 2005th Ludic Numbers:"
For k = 1999 To 2004
If k < 2004 Then
WScript.StdOut.Write ludic(k) & ", "
Else
WScript.StdOut.Write ludic(k)
End If
Next
WScript.StdOut.WriteBlankLines(2)
'triplets up to 250: x, x+2, and x+6
WScript.StdOut.WriteLine "Ludic Triplets up to 250: "
triplets = ""
k = 0
Do While ludic(k) + 6 <= 250
x2 = ludic(k) + 2
x6 = ludic(k) + 6
If ludic.IndexOf(x2,1) > 0 And ludic.IndexOf(x6,1) > 0 Then
triplets = triplets & ludic(k) & ", " & x2 & ", " & x6 & vbCrLf
End If
k = k + 1
Loop
WScript.StdOut.WriteLine triplets
- 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 up to 1000: 142 The 2000th - 2005th Ludic Numbers: 21475, 21481, 21487, 21493, 21503, 21511 Ludic 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
V (Vlang)
const max_i32 = 1<<31 - 1 // i.e. math.MaxInt32
// 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.
fn ludic(nn int, m int) []u32 {
mut n := nn
mut max := m
if max > 0 && n < 0 {
n = max_i32
}
if n < 1 {
return []
}
if max < 0 {
max = max_i32
}
mut sieve := []u32{len: 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, u32(3); i < sieve.len; 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++ {
mut l := int(sieve[k])
if l >= max {
n = k
break
}
mut i := l
l--
// last is the last valid index
mut last := k + i - 1
for j := k + i + 1; j < sieve.len; 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 < sieve.len-1 {
sieve = sieve[..last+1]
}
}
}
if n > sieve.len {
panic("program error") // should never happen
}
return sieve[..n]
}
fn has(x []u32, v u32) bool {
for i := 0; i < x.len && x[i] <= v; i++ {
if x[i] == v {
return true
}
}
return false
}
fn main() {
// ludic() is so quick we just call it repeatedly
println("First 25: ${ludic(25, -1)}")
println("Numner of ludics below 1000: ${ludic(-1, 1000).len}")
println("ludic 2000 to 2005: ${ludic(2005, -1)[1999..]}")
print("Tripples below 250:")
x := ludic(-1, 250)
for i, v in x[..x.len-2] {
if has(x[i+1..], v+2) && has(x[i+2..], v+6) {
print(", ($v ${v+2} ${v+6})")
}
}
println('')
}- 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)
Wren
import "./fmt" for Fmt
var ludic = Fn.new { |n, max|
var maxInt32 = 2.pow(31) - 1
if (max > 0 && n < 0) n = maxInt32
if (n < 1) return []
if (max < 0) max = maxInt32
var sieve = List.filled(10760, 0)
sieve[0] = 1
sieve[1] = 2
if (n > 2) {
var j = 3
for (i in 2...sieve.count) {
sieve[i] = j
j = j + 2
}
for (k in 2...n) {
var l = sieve[k]
if (l >= max) {
n = k
break
}
var i = l
l = l - 1
var last = k + i - 1
var j = k + i + 1
while (j < sieve.count) {
last = k + i
sieve[last] = sieve[j]
if (i%l == 0) j = j + 1
i = i + 1
j = j + 1
}
if (last < sieve.count-1) sieve = sieve[0..last]
}
}
if (n > sieve.count) Fiber.abort("Program error.")
return sieve[0...n]
}
var has = Fn.new { |x, v|
var i = 0
while (i < x.count && x[i] <= v) {
if (x[i] == v) return true
i = i + 1
}
return false
}
System.print("First 25: %(ludic.call(25, -1))")
System.print("Number of Ludics below 1000: %(ludic.call(-1, 1000).count)")
System.print("Ludics 2000 to 2005: %(ludic.call(2005, -1)[1999..-1])")
System.print("Triplets below 250:")
var x = ludic.call(-1, 250)
var i = 0
var triples = []
for (v in x.take(x.count-2)) {
if (has.call(x[i+1..-1], v+2) && has.call(x[i+2..-1], v+6)) {
triples.add([v, v+2, v+6])
}
i = i + 1
}
System.print(triples)
- 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] Number of Ludics below 1000: 142 Ludics 2000 to 2005: [21475, 21481, 21487, 21493, 21503, 21511] 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]]
XPL0
def Size = 25000;
char Sieve(1+Size);
int I, Count, Ludic;
[\Make sieve for Ludic numbers
for I:= 1 to Size do Sieve(I):= true;
Ludic:= 2;
loop [I:= Ludic; Count:= Ludic;
loop [repeat I:= I+1;
if I > Size then quit;
until Sieve(I);
Count:= Count-1;
if Count = 0 then
[Sieve(I):= false;
Count:= Ludic;
];
];
repeat Ludic:= Ludic+1;
if Ludic > Size then quit;
until Sieve(Ludic);
];
\Show first 25 Ludic numbers
Count:= 0; I:= 1;
loop [if Sieve(I) then
[IntOut(0, I); ChOut(0, ^ );
Count:= Count+1;
if Count >= 25 then quit;
];
I:= I+1;
];
CrLf(0);
\Show how many Ludic numbers are <= 1000
Count:= 0;
for I:= 1 to 1000 do
if Sieve(I) then Count:= Count+1;
IntOut(0, Count);
CrLf(0);
\Show Ludic numbers from 2000 to 2005
Count:= 0; I:= 1;
loop [if Sieve(I) then
[Count:= Count+1;
if Count >= 2000 then
[IntOut(0, I); ChOut(0, ^ )];
if Count >= 2005 then quit;
];
I:= I+1;
];
CrLf(0);
\Show triplets of Ludic numbers < 250
for I:= 1 to 250-1-6 do
if Sieve(I) & Sieve(I+2) & Sieve(I+6) then
[ChOut(0, ^();
IntOut(0, I); ChOut(0, ^ );
IntOut(0, I+2); ChOut(0, ^ );
IntOut(0, I+6); Text(0, ") ");
];
CrLf(0);
]- 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)
zkl
This solution builds an iterator with filters, one for each Ludic number, each extending the previous filter. A "master" iterator sits at the top and provides the interface. When the next Ludic number is requested, the next odd number sent down the list of filters and if it makes to the end, it is the next Ludic number. A new filter is then attached [to the iterator] with a starting index of 1 and which indexes to strike.
fcn dropNth(n,seq){
seq.tweak(fcn(n,skipper,idx){ if(0==idx.inc()%skipper) Void.Skip else n }
.fp1(n,Ref(1))) // skip every nth number of previous sequence
}
fcn ludic{ //-->Walker
Walker(fcn(rw){ w:=rw.value; n:=w.next(); rw.set(dropNth(n,w)); n }
.fp(Ref([3..*,2]))) // odd numbers starting at 3
.push(1,2); // first two Ludic numbers
}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();- 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))
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