Klarner-Rado sequence

Klarner-Rado sequence is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Klarner-Rado sequences are a class of similar sequences that were studied by the mathematicians David Klarner and Richard Rado.

The most well known is defined as the thinnest strictly ascending sequence K which starts 1, then, for each element n, it will also contain somewhere in the sequence, 2 × n + 1 and 3 × n + 1.

So, the sequence K starts with 1.

Set n equal to the first element 1; the sequence will also contain 2 × n + 1 and 3 × n + 1, or 3 and 4.

Set n equal to the next element: 3, somewhere in the sequence it will contain 2 × n + 1 and 3 × n + 1, or 7 and 10.

Continue setting n equal to each element in turn to add to the sequence.

Task

• Find and display the first one hundred elements of the sequence.
• Find and display the one thousandth and ten thousandth elements of the sequence.

Preferably without needing to find an over abundance and sorting.

Stretch

• Find and display the one hundred thousandth and one millionth elements of the sequence.

See also

• OEIS:A002977 - Klarner-Rado sequence: a(1) = 1; subsequent terms are defined by the rule that if m is present so are 2m+1 and 3m+1

ALGOL 68

Generates the sequence in order. Note that to run this with ALGOL 68G under Windows (and probably Linux), a large heap size must be specified on the command line, e.g.: -heap 1024M.

BEGIN # find elements of the Klarner-Rado sequence                           #
      #    - if n is an element, so are 2n + 1 and 3n + 1                    #
    INT max element = 100 000 000;
    [ 0 : max element ]BOOL kr; FOR i FROM LWB kr TO UPB kr DO kr[ i ] := FALSE OD;
    INT n21      := 3;                    # next 2n+1 value                  #
    INT n31      := 4;                    # next 3n+1 value                  #
    INT p2       := 1;                    # the n for the next 2n+1 value    #
    INT p3       := 1;                    # the n for the next 3n+1 value    #
    kr[ 1 ]      := TRUE;
    INT kr count := 0;
    INT max count = 1 000 000;
    FOR i WHILE kr count < max count DO
        IF i = n21 THEN                   # found the next 2n+1 value        #
            IF kr[ p2 ] THEN
                kr[ i ] := TRUE
            FI;
            p2  +:= 1;
            n21 +:= 2
        FI;
        IF i = n31 THEN                   # found the next 3n+1 value        #
            IF kr[ p3 ] THEN
                kr[ i ] := TRUE
            FI;
            p3  +:= 1;
            n31 +:= 3
        FI;
        IF kr[ i ] THEN
            kr count +:= 1;
            IF   kr count <= 100 THEN
                print( ( " ", whole( i, -3 ) ) );
                IF kr count MOD 20 = 0 THEN print( ( newline ) ) FI
            ELIF kr count =     1 000 THEN
                print( ( "The     thousandth element is ", whole( i, -8 ), newline ) )
            ELIF kr count =    10 000 THEN
                print( ( "The ten thousandth element is ", whole( i, -8 ), newline ) )
            ELIF kr count =   100 000 THEN
                print( ( "The 100-thousandth element is ", whole( i, -8 ), newline ) )
            ELIF kr count = 1 000 000 THEN
                print( ( "The      millionth element is ", whole( i, -8 ), newline ) )
            FI
        FI
    OD
END

Output:

   1   3   4   7   9  10  13  15  19  21  22  27  28  31  39  40  43  45  46  55
  57  58  63  64  67  79  81  82  85  87  91  93  94 111 115 117 118 121 127 129
 130 135 136 139 159 163 165 166 171 172 175 183 187 189 190 193 202 223 231 235
 237 238 243 244 247 255 256 259 261 262 271 273 274 279 280 283 319 327 331 333
 334 343 345 346 351 352 355 364 367 375 379 381 382 387 388 391 405 406 409 418
The     thousandth element is     8487
The ten thousandth element is   157653
The 100-thousandth element is  2911581
The      millionth element is 54381285

AppleScript

One way to test numbers for membership of the sequence is to feed them to a recursive handler which determines whether or not there's a Klarner-Rado route from them down to 0. It makes finding the elements in order simple, but takes about five and a half minutes to get to the millionth one.

-- Is n in the Klarner-Rado sequence?
-- Fully recursive:
(* on isKlarnerRado(n)
    set n to n - 1
    return ((n = 0) or ((n mod 2 = 0) and (isKlarnerRado(n div 2))) or ((n mod 3 = 0) and (isKlarnerRado(n div 3))))
end isKlarnerRado *)

-- Optimised with tail call elimination. 90 seconds faster than the above in this script.
on isKlarnerRado(n)
    set n to n - 1
    repeat
        if ((n = 0) or ((n mod 3 = 0) and (isKlarnerRado(n div 3)))) then
            return true
        else if (n mod 2 = 0) then
            set n to n div 2 - 1
        else
            return false
        end if
    end repeat
end isKlarnerRado

on join(lst, delim)
    set astid to AppleScript's text item delimiters
    set AppleScript's text item delimiters to delim
    set txt to lst as text
    set AppleScript's text item delimiters to astid
    return txt
end join

on task()
    set output to {"First 100 elements:"}
    set n to 0
    set |count| to 0
    set K to {}
    repeat until (|count| = 100)
        set n to n + 1
        if (isKlarnerRado(n)) then
            set K's end to n
            set |count| to |count| + 1
        end if
    end repeat
    repeat with i from 1 to 100 by 20
        set output's end to join(K's items i thru (i + 19), "  ")
    end repeat
    
    repeat with this in {{1000, "1,000th element: "}, {10000, "10,000th element: "}, {100000, "100,000th element: "}, ¬
        {1000000, "1,000,000th element: "}}
        set {target, spiel} to this
        repeat until (|count| = target)
            set n to n + 1
            if (isKlarnerRado(n)) then set |count| to |count| + 1
        end repeat
        set output's end to spiel & n
    end repeat
    
    return join(output, linefeed)
end task

task()

Output:

"First 100 elements:
1  3  4  7  9  10  13  15  19  21  22  27  28  31  39  40  43  45  46  55
57  58  63  64  67  79  81  82  85  87  91  93  94  111  115  117  118  121  127  129
130  135  136  139  159  163  165  166  171  172  175  183  187  189  190  193  202  223  231  235
237  238  243  244  247  255  256  259  261  262  271  273  274  279  280  283  319  327  331  333
334  343  345  346  351  352  355  364  367  375  379  381  382  387  388  391  405  406  409  418
1,000th element: 8487
10,000th element: 157653
100,000th element: 2911581
1,000,000th element: 54381285"

Another way is to produce a list with Klarner-Rado elements in the slots indexed by those numbers and 'missing values' in the other slots. If a number being tested is exactly divisible by 2 or by 3, and the slot whose index is the result of the division contains a number instead of 'missing value', the tested number plus 1 is a Klarner-Rado element and should go into the slot it indexes. The list will contain vastly more 'missing values' than Klarner-Rado elements and it — or portions of it — ideally need to exist before the sequence elements are inserted. So while an overabundance and sorting of sequence elements isn't needed, an overabundance of 'missing values' is! The script below carries out the task in about 75 seconds on my current machine and produces the same output as above.

on KlarnerRadoSequence(target)
    -- To find a million KR numbers with this method nominally needs a list with at least 54,381,285
    -- slots. But the number isn't known in advance, "growing" a list to the required length would
    -- take forever, and accessing its individual items could also take a while. Instead, K will
    -- here contain reasonably sized and quickly produced sublists which are added as needed.
    -- The 1-based referencing calculations will be ((index - 1) div sublistLen + 1) for the sublist
    -- and ((index - 1) mod sublistLen + 1) for the slot within it.
    set sublistLen to 10000
    script o
        property spare : makeList(sublistLen, missing value)
        property K : {spare's items}
    end script
    
    -- Insert the initial 1, start the KR counter at 1, start the number-to-test variable at 2.
    set {o's K's beginning's 1st item, |count|, n} to {1, 1, 2}
    -- Test the first, second, third, and fifth of every six consecutive numbers starting at 2.
    -- These are known to be divisible by 2 or by 3 and the fourth and sixth known not to be.
    -- If the item at index (n div 2) or index (n div 3) isn't 'missing value', it's a number,
    -- so insert (n + 1) at index (n + 1).
    if (|count| < target) then ¬
        repeat -- Per increase of n by 6.
            -- The first of the six numbers in this range is divisible by 2.
            -- Precalculate (index - 1) for index (n div 2) to reduce code clutter and halve calculation time.
            set pre to n div 2 - 1
            if (o's K's item (pre div sublistLen + 1)'s item (pre mod sublistLen + 1) is missing value) then
            else
                -- Insert (n + 1) at index (n + 1). The 'n's in the reference calculations are for ((n + 1) - 1)!
                set o's K's item (n div sublistLen + 1)'s item (n mod sublistLen + 1) to n + 1
                set |count| to |count| + 1
                if (|count| = target) then exit repeat
            end if
            -- The second number of the six is divisible by 3.
            set n to n + 1
            set pre to n div 3 - 1
            if (o's K's item (pre div sublistLen + 1)'s item (pre mod sublistLen + 1) is missing value) then
            else
                set o's K's item (n div sublistLen + 1)'s item (n mod sublistLen + 1) to n + 1
                set |count| to |count| + 1
                if (|count| = target) then exit repeat
            end if
            -- The third is divisible by 2.
            set n to n + 1
            set pre to n div 2 - 1
            if (o's K's item (pre div sublistLen + 1)'s item (pre mod sublistLen + 1) is missing value) then
            else
                set o's K's item (n div sublistLen + 1)'s item (n mod sublistLen + 1) to n + 1
                set |count| to |count| + 1
                if (|count| = target) then exit repeat
            end if
            -- The fifth is divisible by both 2 and 3.
            set n to n + 2
            set pre2 to n div 2 - 1
            set pre3 to n div 3 - 1
            if ((o's K's item (pre2 div sublistLen + 1)'s item (pre2 mod sublistLen + 1) is missing value) and ¬
                (o's K's item (pre3 div sublistLen + 1)'s item (pre3 mod sublistLen + 1) is missing value)) then
            else
                set o's K's item (n div sublistLen + 1)'s item (n mod sublistLen + 1) to n + 1
                set |count| to |count| + 1
                if (|count| = target) then exit repeat
            end if
            
            -- Advance to the first number of the next six.
            set n to n + 2
            -- If another sublist is about to be needed, append one to the end of K.
            if ((n + 6) mod sublistLen < 6) then copy o's spare to o's K's end
        end repeat
    
    -- Once the target's reached, replace the sublists with lists containing just the numbers.
    set sublistCount to (count o's K)
    repeat with i from 1 to sublistCount
        set o's K's item i to o's K's item i's numbers
    end repeat
    -- Concatenate the number lists to leave K as a single list of numbers.
    repeat while (sublistCount > 1)
        set o's spare to o's K
        set o's K to {}
        repeat with i from 2 to sublistCount by 2
            set end of o's K to o's spare's item (i - 1) & o's spare's item i
        end repeat
        if (i < sublistCount) then set o's K's last item to o's K's end & o's spare's end
        set sublistCount to sublistCount div 2
    end repeat
    set o's K to o's K's beginning
    
    return o's K
end KlarnerRadoSequence

on makeList(len, content)
    script o
        property lst : {}
    end script
    if (len > 0) then
        set o's lst's end to content
        set |count| to 1
        repeat until (|count| + |count| > len)
            set o's lst to o's lst & o's lst
            set |count| to |count| + |count|
        end repeat
        if (|count| < len) then set o's lst to o's lst & o's lst's items 1 thru (len - |count|)
    end if
    
    return o's lst
end makeList

on join(lst, delim)
    set astid to AppleScript's text item delimiters
    set AppleScript's text item delimiters to delim
    set txt to lst as text
    set AppleScript's text item delimiters to astid
    return txt
end join

on task()
    script o
        property K : KlarnerRadoSequence(1000000)
    end script
    set output to {"First 100 elements:"}
    repeat with i from 1 to 100 by 20
        set output's end to join(o's K's items i thru (i + 19), "  ")
    end repeat
    set output's end to "1,000th element: " & o's K's item 1000
    set output's end to "10,000th element: " & o's K's item 10000
    set output's end to "100,000th element: " & o's K's item 100000
    set output's end to "1,000,000th element: " & o's K's item 1000000
    
    return join(output, linefeed)
end task

task()

F#

// Klarner-Rado sequence. Nigel Galloway: August 19th., 20
let kr()=Seq.unfold(fun(n,g)->Some(g|>Seq.filter((>)n)|>Seq.sort,(n*2L+1L,seq[for n in g do yield n; yield n*2L+1L; yield n*3L+1L]|>Seq.filter((<=)n)|>Seq.distinct)))(3L,seq[1L])|>Seq.concat
let n=kr()|>Seq.take 1000000|>Array.ofSeq in n|>Array.take 100|>Array.iter(printf "%d "); printfn "\nkr[999] is %d\nkr[9999] is %d\nkr[99999] is %d\nkr[999999] is %d" n.[999] n.[9999] n.[99999] n.[999999]

Output:

1 3 4 7 9 10 13 15 19 21 22 27 28 31 39 40 43 45 46 55 57 58 63 64 67 79 81 82 85 87 91 93 94 111 115 117 118 121 127 129 130 135 136 139 159 163 165 166 171 172 175 183 187 189 190 193 202 223 231 235 237 238 243 244 247 255 256 259 261 262 271 273 274 279 280 283 319 327 331 333 334 343 345 346 351 352 355 364 367 375 379 381 382 387 388 391 405 406 409 418
kr[999] is 8487
kr[9999] is 157653
kr[99999] is 2911581
kr[999999] is 54381285

FreeBASIC

Translation of: Phix

#include "string.bi"

Dim As Integer limit = 1e8
Dim As Boolean kr(limit)
kr(1) = True
Dim As Integer n21 = 3, n31 = 4, p2 = 1, p3 = 1, count = 0, np10 = 1000
Print "First 100 Klarner-Rado sequence numbers:"
For i As Integer = 1 To limit
    If i = n21 Then
        If kr(p2) Then kr(i) = True
        p2  += 1
        n21 += 2
    End If
    If i = n31 Then
        If kr(p3) Then kr(i) = True
        p3  += 1
        n31 += 3
    End If
    If kr(i) Then
        count += 1
        If count <= 100 Then
            Print Using "####"; i;
            If(count Mod 20) = 0 Then Print
        Elseif count = np10 Then
            Print "The "; Format(count, "#,###,###"); "th Klarner-Rado number is "; Format(i, "#,###,###")
            np10 *= 10
        End If
    End If
Next i
Sleep

Output:

Same as Phix entry.

Haskell

import Data.List (intercalate)
import Data.List.Ordered (union)
import Data.List.Split (chunksOf)
import Text.Printf (printf)

----------------------- KLARNER-RADO ---------------------

klarnerRado :: (Num a, Ord a, Enum a) => [a]
klarnerRado =
  1 :
  union
    (succ . (* 2) <$> klarnerRado)
    (succ . (* 3) <$> klarnerRado)

--------------------------- TEST -------------------------
main :: IO ()
main = do
  putStrLn "First one hundred elements of the sequence:\n"
  mapM_
    putStrLn
    ( intercalate "  "
        <$> chunksOf
          10
          (printf "%3d" <$> take 100 klarnerRado)
    )

  putStrLn "\nKth and 10Kth elements of the sequence:\n"
  mapM_
    ( putStrLn
        . (<*>)
          (flip (printf "%5dth %s %7d") " ->")
          ((klarnerRado !!) . pred)
    )
    [1000, 10000]

Output:

First one hundred elements of the sequence:

  1    3    4    7    9   10   13   15   19   21
 22   27   28   31   39   40   43   45   46   55
 57   58   63   64   67   79   81   82   85   87
 91   93   94  111  115  117  118  121  127  129
130  135  136  139  159  163  165  166  171  172
175  183  187  189  190  193  202  223  231  235
237  238  243  244  247  255  256  259  261  262
271  273  274  279  280  283  319  327  331  333
334  343  345  346  351  352  355  364  367  375
379  381  382  387  388  391  405  406  409  418

Kth and 10Kth elements of the sequence:

 1000th  ->    8487
10000th  ->  157653

J

Implementation:

krsieve=: {{ 
  for_i. i.<.-:#b=. (1+y){.0 1 do.
    if. i{b do. b=. 1 ((#~ y&>)1+2 3*i)} b end.
  end.
  I.b
}}

Task examples (including stretch):

   #kr7e7=: krsieve 7e7
1215307
   100{.kr7e7
1 3 4 7 9 10 13 15 19 21 22 27 28 31 39 40 43 45 46 55 57 58 63 64 67 79 81 82 85 87 91 93 94 111 115 117 118 121 127 129 130 135 136 139 159 163 165 166 171 172 175 183 187 189 190 193 202 223 231 235 237 238 243 244 247 255 256 259 261 262 271 273 274 279 280 283 319 327 331 333 334 343 345 346 351 352 355 364 367 375 379 381 382 387 388 391 405 406 409 418
   (1e3-1){kr7e7
8487
   (1e4-1){kr7e7
157653
   (1e5-1){kr7e7
2911581
   (1e6-1){kr7e7
54381285

Julia

using Formatting

function KlarnerRado(N)
    K = [1]
    for i in 1:N
        j = K[i]
        firstadd, secondadd = 2j + 1, 3j + 1
        if firstadd < K[end]
            pos = findlast(<(firstadd), K) + 1
            K[pos] != firstadd && insert!(K, pos, firstadd)
        elseif K[end] != firstadd
            push!(K, firstadd)
        end
        if secondadd < K[end]
            pos = findlast(<(secondadd), K) + 1
            K[pos] != secondadd && insert!(K, pos, secondadd)
        elseif K[end] != secondadd
            push!(K, secondadd)
        end
    end
    return K
end

kr1m = KlarnerRado(1_000_000)

println("First 100 Klarner-Rado numbers:")
foreach(p -> print(rpad(p[2], 4), p[1] % 20 == 0 ? "\n" : ""), enumerate(kr1m[1:100]))
foreach(n -> println("The ", format(n, commas=true), "th Klarner-Rado number is ",
   format(kr1m[n], commas=true)), [1000, 10000, 100000, 1000000])

Output:

First 100 Klarner-Rado numbers:
1   3   4   7   9   10  13  15  19  21  22  27  28  31  39  40  43  45  46  55  
57  58  63  64  67  79  81  82  85  87  91  93  94  111 115 117 118 121 127 129
130 135 136 139 159 163 165 166 171 172 175 183 187 189 190 193 202 223 231 235
237 238 243 244 247 255 256 259 261 262 271 273 274 279 280 283 319 327 331 333 
334 343 345 346 351 352 355 364 367 375 379 381 382 387 388 391 405 406 409 418
The 1,000th Klarner-Rado number is 8,487
The 10,000th Klarner-Rado number is 157,653
The 100,000th Klarner-Rado number is 2,911,581
The 1,000,000th Klarner-Rado number is 54,381,285

Faster version

Probably does get an overabundance, but no sorting. `falses()` is implemented as a bit vector, so huge arrays are not needed. From ALGOL.

using Formatting

function KlamerRado(N)
    kr = falses(100 * N)
    kr[1] = true
    for i in 1:30N
        if kr[i]
            kr[2i + 1] = true
            kr[3i + 1] = true
        end
    end
    return [i for i in eachindex(kr) if kr[i]]
end

kr1m = KlamerRado(1000000)

println("First 100 Klarner-Rado numbers:")
foreach(p -> print(rpad(p[2], 4), p[1] % 20 == 0 ? "\n" : ""), enumerate(kr1m[1:100]))
foreach(n -> println("The ", format(n, commas=true), "th Klarner-Rado number is ",
   format(kr1m[n], commas=true)), [1000, 10000, 100000, 1000000])

Output:

same as above version.

Phix

Translation of: ALGOL_68

with javascript_semantics
constant limit = iff(platform()=JS?10_000_000:100_000_000)
sequence kr = repeat(false,limit); kr[1] = true
integer n21 = 3, n31 = 4, p2 = 1, p3 = 1, count = 0, np10 = 1_000
for i=1 to limit do
    if i = n21 then
        if kr[p2] then
            kr[i] := true
        end if
        p2  += 1
        n21 += 2
    end if
    if i = n31 then
        if kr[p3] then
            kr[i] := true
        end if
        p3  += 1
        n31 += 3
    end if
    if kr[i] then
        count += 1
        if count <= 100 then
            printf(1,"%4d%s",{i,iff(mod(count,20)=0?"\n":"")})
        elsif count = np10 then
            printf(1,"The %,d%s Klarner-Rado number is %,d\n",{count,ord(count),i})
            np10 *= 10
        end if
    end if
end for

Output:

   1   3   4   7   9  10  13  15  19  21  22  27  28  31  39  40  43  45  46  55
  57  58  63  64  67  79  81  82  85  87  91  93  94 111 115 117 118 121 127 129
 130 135 136 139 159 163 165 166 171 172 175 183 187 189 190 193 202 223 231 235
 237 238 243 244 247 255 256 259 261 262 271 273 274 279 280 283 319 327 331 333
 334 343 345 346 351 352 355 364 367 375 379 381 382 387 388 391 405 406 409 418
The 1,000th Klarner-Rado number is 8,487
The 10,000th Klarner-Rado number is 157,653
The 100,000th Klarner-Rado number is 2,911,581
The 1,000,000th Klarner-Rado number is 54,381,285

Unfortunately JavaScript can't quite cope/runs out of memory with a 100 million element sieve, so it only goes to the 10^5th under p2js.

PL/M

Works with: 8080 PL/M Compiler

... under CP/M (or an emulator)

As PL/M only handles unsigned 8 and 16 bit integers, this only finds the first 1000 elements. This is based on the Algol 68 sample, but as with the VB.NET sample, uses a "bit vector" (here an array of bytes) - as suggested by the Julia sample.

100H: /* EIND ELEMENTS OF THE KLARNER-RADO SEQUENCE                         */
      /*    - IF N IS AN ELEMENT, SO ARE 2N+1 AND 3N+!, 1 IS AN ELEMENT     */

   /* CP/M SYSTEM CALL AND I/O ROUTINES                                     */
   BDOS:      PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
   PR$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C );  END;
   PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S );  END;
   PR$NL:     PROCEDURE;   CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
   PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
      V = N;
      W = LAST( N$STR );
      N$STR( W ) = '$';
      N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      DO WHILE( ( V := V / 10 ) > 0 );
         N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      END;
      CALL PR$STRING( .N$STR( W ) );
   END PR$NUMBER;

   /* TASK                                                                  */

   DECLARE MAX$COUNT LITERALLY '1$000';

   DECLARE BIT ( 8 )BYTE INITIAL( 128, 1, 2, 4, 8, 16, 32, 64 );

   DECLARE ( N21, N31, P2, P3, KR$COUNT, I, I$3 ) ADDRESS;
   DECLARE KR ( 1251 )BYTE; DO I = 0 TO LAST( KR ); KR( I ) = 0; END;
   KR( 0 ) = BIT( 1 );

   KR$COUNT = 0; N21 = 3; N31 = 4; P2, P3 = 1;
   I = 0;
   DO WHILE( KR$COUNT < MAX$COUNT );
      I = I + 1;
      I$3 = SHR( I, 3 );
      IF I = N21 THEN DO;                         /* I IS 2N+1 WHERE N = P2 */
         IF ( KR( SHR( P2, 3 ) ) AND BIT( P2 AND 7 ) ) <> 0 THEN DO;
            KR( I$3 ) = KR( I$3 ) OR BIT( I AND 7 );
         END;
         P2  = P2  + 1;
         N21 = N21 + 2;
      END;
      IF I = N31 THEN DO;                         /* I IS 3N+1 WHERE N = P3 */
         IF ( KR( SHR( P3, 3 ) ) AND BIT( P3 AND 7 ) ) <> 0 THEN DO;
            KR( I$3 ) = KR( I$3 ) OR BIT( I AND 7 );
         END;
         P3  = P3  + 1;
         N31 = N31 + 3;
      END;
      IF ( KR( I$3 ) AND BIT( I AND 7 ) ) <> 0 THEN DO;
         KR$COUNT = KR$COUNT + 1;
         IF KR$COUNT <= 100 THEN DO;
            CALL PR$CHAR( ' ' );
            IF I <  10 THEN CALL PR$CHAR( ' ' );
            IF I < 100 THEN CALL PR$CHAR( ' ' );
            CALL PR$NUMBER( I );
            IF KR$COUNT MOD 20 = 0 THEN CALL PR$NL;
            END;
         ELSE IF KR$COUNT = MAX$COUNT THEN DO;
            CALL PR$STRING( .'ELEMENT $' );
            CALL PR$NUMBER( MAX$COUNT );
            CALL PR$STRING( .' IS: $' );
            CALL PR$NUMBER( I );
         END;
      END;
   END;

EOF

Output:

   1   3   4   7   9  10  13  15  19  21  22  27  28  31  39  40  43  45  46  55
  57  58  63  64  67  79  81  82  85  87  91  93  94 111 115 117 118 121 127 129
 130 135 136 139 159 163 165 166 171 172 175 183 187 189 190 193 202 223 231 235
 237 238 243 244 247 255 256 259 261 262 271 273 274 279 280 283 319 327 331 333
 334 343 345 346 351 352 355 364 367 375 379 381 382 387 388 391 405 406 409 418
ELEMENT 1000 IS: 8487

Python

def KlarnerRado(N):
    K = [1]
    for i in range(N):
        j = K[i]
        firstadd, secondadd = 2 * j + 1, 3 * j + 1
        if firstadd < K[-1]:
            for pos in range(len(K)-1, 1, -1):
                if K[pos] < firstadd < K[pos + 1]:
                    K.insert(pos + 1, firstadd)
                    break
        elif firstadd > K[-1]:
            K.append(firstadd)
        if secondadd < K[-1]:
            for pos in range(len(K)-1, 1, -1):
                if K[pos] < secondadd < K[pos + 1]:
                    K.insert(pos + 1, secondadd)
                    break
        elif secondadd > K[-1]:
            K.append(secondadd)

    return K

kr1m = KlarnerRado(100_000)

print('First 100 Klarner-Rado sequence numbers:')
for idx, v in enumerate(kr1m[:100]):
    print(f'{v: 4}', end='\n' if (idx + 1) % 20 == 0 else '')
for n in [1000, 10_000, 100_000]:
    print(f'The {n :,}th Klarner-Rado number is {kr1m[n-1] :,}')

Output:

First 100 Klarner-Rado sequence numbers:
   1   3   4   7   9  10  13  15  19  21  22  27  28  31  39  40  43  45  46  55
  57  58  63  64  67  79  81  82  85  87  91  93  94 111 115 117 118 121 127 129
 130 135 136 139 159 163 165 166 171 172 175 183 187 189 190 193 202 223 231 235
 237 238 243 244 247 255 256 259 261 262 271 273 274 279 280 283 319 327 331 333
 334 343 345 346 351 352 355 364 367 375 379 381 382 387 388 391 405 406 409 418
The 1,000th Klarner-Rado number is 8,487
The 10,000th Klarner-Rado number is 157,653
The 100,000th Klarner-Rado number is 2,911,581

faster version

from numpy import ndarray

def KlarnerRado(N):
    kr = ndarray(100 * N, dtype=bool)
    kr[1] = True
    for i in range(30 * N):
        if kr[i]:
            kr[2 * i + 1] = True
            kr[3 * i + 1] = True

    return [i for i in range(100 * N) if kr[i]]
 
kr1m = KlarnerRado(1_000_000)
 
print('First 100 Klarner-Rado sequence numbers:')
for idx, v in enumerate(kr1m[:100]):
    print(f'{v: 4}', end='\n' if (idx + 1) % 20 == 0 else '')
for n in [1000, 10_000, 100_000, 1_000_000]:
    print(f'The {n :,}th Klarner-Rado number is {kr1m[n-1] :,}')

Output:

Same as previous version.

Raku

sub Klarner-Rado ($n) {
    my @klarner-rado = 1;
    my @next = x2, x3;

    loop {
        @klarner-rado.push: my $min = @next.min;
        @next[0] = x2 if @next[0] == $min;
        @next[1] = x3 if @next[1] == $min;
        last if +@klarner-rado > $n.max;
    }

    sub x2 { state $i = 0; @klarner-rado[$i++] × 2 + 1 }
    sub x3 { state $i = 0; @klarner-rado[$i++] × 3 + 1 }

    @klarner-rado[|$n]
}

use Lingua::EN::Numbers;

put "First 100 elements of Klarner-Rado sequence:";
put Klarner-Rado(^100).batch(10)».fmt("%3g").join: "\n";
put '';
put (($_».Int».&ordinal».tc »~» ' element: ').List Z~ |(List.new: (Klarner-Rado ($_ »-» 1))».&comma)
    given $(1e0, 1e1, 1e2, 1e3, 1e4, 1e5, 1e6)).join: "\n";

Output:

First 100 elements of Klarner-Rado sequence:
  1   3   4   7   9  10  13  15  19  21
 22  27  28  31  39  40  43  45  46  55
 57  58  63  64  67  79  81  82  85  87
 91  93  94 111 115 117 118 121 127 129
130 135 136 139 159 163 165 166 171 172
175 183 187 189 190 193 202 223 231 235
237 238 243 244 247 255 256 259 261 262
271 273 274 279 280 283 319 327 331 333
334 343 345 346 351 352 355 364 367 375
379 381 382 387 388 391 405 406 409 418

First element: 1
Tenth element: 21
One hundredth element: 418
One thousandth element: 8,487
Ten thousandth element: 157,653
One hundred thousandth element: 2,911,581
One millionth element: 54,381,285

Visual Basic .NET

Based on the ALGOL 68 sample, but using a "bit vector" (array of integers), as suggested by the Julia sample. Simplifies printing "the powers of ten" elements, as in the Phix sample.

Option Strict On
Option Explicit On

Imports System.IO

Module KlarnerRado

    Private Const bitsWidth As Integer = 31

    Private bitMask() As Integer = _
        New Integer(){ 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 _
                     , 2048, 4096, 8192, 16384, 32768, 65536, 131072 _
                     , 262144, 524288, 1048576, 2097152, 4194304, 8388608 _
                     , 16777216, 33554432, 67108864, 134217728, 268435456 _
                     , 536870912, 1073741824 _
                     }

    Private Const maxElement As Integer = 1100000000


    Private Function BitSet(bit As Integer, v As Integer) As Boolean
        Return (v And bitMask(bit - 1)) <> 0
    End Function

    Private Function SetBit(ByVal bit As Integer, ByVal b As Integer) As Integer
        Return b Or bitMask(bit - 1)
    End Function

    Public Sub Main
        Dim  kr(maxElement \ bitsWidth) As Integer

        For i As Integer = 0 To kr.Count() - 1
            kr(i) = 0
        Next  i

        Dim krCount As Integer =  0    ' count of sequende elements
        Dim n21 As Integer = 3         ' next 2n+1 value
        Dim n31 As Integer = 4         ' next 3n+1 value
        Dim p10 As Integer = 1000      ' next power of ten to show the count/element at
        Dim iBit As Integer = 0        ' i Mod bitsWidth  - used to set the ith bit
        Dim iOverBw As Integer = 0     ' i \ bitsWidth    - used to set the ith bit
        '   p2                         ' the n for the 2n+1 value
        Dim p2Bit As Integer = 1       ' p2 Mod bitsWidth  - used to set the p2th bit
        Dim p2OverBw As Integer = 0    ' p2 \ bitsWidth    - used to set the p2th bit
        '   p3                         ' the n for the 3n+1 value
        Dim p3Bit As Integer = 1       ' p3 Mod bitsWidth  - used to set the p3th bit
        Dim p3OverBw As Integer = 0    ' p3 \ bitsWidth    - used to set the p3th bit

        kr(0) = SetBit(1, kr(0))
        Dim kri As Boolean = True
        Dim lastI As Integer = 0
        For i As Integer = 1 To  maxElement
            iBit += 1
            If iBit > bitsWidth Then
                iBit = 1
                iOverBw += 1
            End If
            If i = n21 Then            ' found the next 2n+1 value
                If BitSet(p2Bit, kr(p2OverBw)) Then
                    kri = True
                End If
                p2Bit += 1
                If p2Bit > bitsWidth Then
                    p2Bit = 1
                    p2OverBw += 1
                End If
                n21 += 2
            End If
            If i = n31 Then            ' found the next 3n+1 value
                If BitSet(p3Bit, kr(p3OverBw)) Then
                    kri = True
                End If
                p3Bit += 1
                If p3Bit > bitsWidth Then
                    p3Bit = 1
                    p3OverBw += 1
                End If
                n31 += 3
            End If
            If kri Then
                lastI = i
                kr(iOverBw) = SetBit(iBit, kr(iOverBw))
                krCount += 1
                If krCount <= 100 Then
                    Console.Out.Write(" " & i.ToString().PadLeft(3))
                    If krCount Mod 20 = 0 Then
                        Console.Out.WriteLine()
                    End If
                ElseIf krCount = p10 Then
                    Console.Out.WriteLine("Element " & p10.ToString().PadLeft(10) & " is " & i.ToString().PadLeft(10))
                    p10 *= 10
                End If
                kri = False
            End If
        Next  i
        Console.Out.WriteLine("Element " & krCount.ToString().PadLeft(10) & " is " & lastI.ToString().PadLeft(10))

    End Sub

End Module

Output:

   1   3   4   7   9  10  13  15  19  21  22  27  28  31  39  40  43  45  46  55
  57  58  63  64  67  79  81  82  85  87  91  93  94 111 115 117 118 121 127 129
 130 135 136 139 159 163 165 166 171 172 175 183 187 189 190 193 202 223 231 235
 237 238 243 244 247 255 256 259 261 262 271 273 274 279 280 283 319 327 331 333
 334 343 345 346 351 352 355 364 367 375 379 381 382 387 388 391 405 406 409 418
Element       1000 is       8487
Element      10000 is     157653
Element     100000 is    2911581
Element    1000000 is   54381285
Element   10000000 is 1031926801
Element   10543878 is 1099640002

Wren

Version 1

Library: Wren-sort

Library: Wren-fmt

There's no actual sorting here. The Find class (and its binary search methods) just happen to be in Wren-sort.

import "./sort" for Find
import "./fmt" for Fmt

var klarnerRado = Fn.new { |limit|
    var kr = [1]
    for (i in 0...limit) {
        var n = kr[i]
        for (e in [2*n + 1, 3*n + 1]) {
            if (e > kr[-1]) {
                kr.add(e)
            } else {
                var ix = Find.nearest(kr, e)  // binary search
                if (kr[ix] != e) kr.insert(ix, e)
            }
        }
    }
    return kr[0...limit]
}

var kr = klarnerRado.call(1e6)
System.print("First 100 elements of Klarner-Rado sequence:")
Fmt.tprint("$3d ", kr[0..99], 10)
System.print()
var limits = [1, 10, 1e2, 1e3, 1e4, 1e5, 1e6]
for (limit in limits) {
    Fmt.print("The $,r element: $,d", limit, kr[limit-1])
}

Output:

First 100 elements of Klarner-Rado sequence:
  1    3    4    7    9   10   13   15   19   21  
 22   27   28   31   39   40   43   45   46   55  
 57   58   63   64   67   79   81   82   85   87  
 91   93   94  111  115  117  118  121  127  129  
130  135  136  139  159  163  165  166  171  172  
175  183  187  189  190  193  202  223  231  235  
237  238  243  244  247  255  256  259  261  262  
271  273  274  279  280  283  319  327  331  333  
334  343  345  346  351  352  355  364  367  375  
379  381  382  387  388  391  405  406  409  418  

The 1st element: 1
The 10th element: 21
The 100th element: 418
The 1,000th element: 8,487
The 10,000th element: 157,653
The 100,000th element: 2,911,581
The 1,000,000th element: 54,381,285

Version 2 (faster)

Translation of: ALGOL 68

Library: Wren-array

Takes around 13.8 seconds to find the millionth element compared to 48 seconds for the first version.

As in the case of some other examples, uses a bit array to save memory. However, whilst the BitArray class crams 32 bools into a 64 bit float (Wren doesn't have integers as such), it is significantly slower to index than the native List class. If the latter is used instead, the time comes down to 5.3 seconds.

Although not shown here, if the size of the BitArray is increased to 1.1 billion and 'max' to 1e7, then the 10 millionth element (1,031,926,801) will eventually be found but takes 4 minutes 50 seconds to do so.

import "./array" for BitArray
import "./fmt" for Fmt

var kr = BitArray.new(1e8)
var first100 = List.filled(100, 0)
var pow10 = {}
kr[1] = true
var n21 = 3
var n31 = 4
var p2 = 1
var p3 = 1
var count = 0
var max = 1e6
var i = 1
var limit = 1
while (count < max) {
    if (i == n21) {
        if (kr[p2]) kr[i] = true
        p2 = p2 + 1
        n21 = n21 + 2
    }
    if (i == n31) {
        if (kr[p3]) kr[i] = true
        p3 = p3 + 1
        n31 = n31 + 3
    }
    if (kr[i]) {
        count = count + 1
        if (count <= 100) first100[count-1] = i
        if (count == limit) {
            pow10[count] = i
            if (limit == max) break
            limit = 10 * limit
        }
    }
    i = i + 1
}

System.print("First 100 elements of Klarner-Rado sequence:")
Fmt.tprint("$3d ", first100, 10)
System.print()
limit = 1
while (limit <= max) {
    Fmt.print("The $,r element: $,d", limit, pow10[limit])
    limit = 10 * limit
}

Output:

Same as Version 1.