Klarner-Rado sequence
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
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
.
<lang algol68>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</lang>
- 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.
<lang applescript>-- 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()</lang>
- Output:
<lang applescript>"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"</lang>
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.
<lang applescript>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()</lang>
F#
<lang fsharp> // 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] </lang>
- 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
<lang freebasic>#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</lang>
- Output:
Same as Phix entry.
J
Implementation:<lang J>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
}}</lang>
Task examples (including stretch):<lang J> #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</lang>
Julia
<lang 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])
</lang>
- 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. <lang ruby>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])
</lang>
- Output:
same as above version.
Phix
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
... 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.
<lang pli>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</lang>
- 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
<lang 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] :,}')
</lang>
- 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
<lang python>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] :,}')
</lang>
- Output:
Same as previous version.
Raku
<lang perl6>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";</lang>
- 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. <lang vbnet>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</lang>
- 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
There's no actual sorting here. The Find class (and its binary search methods) just happen to be in Wren-sort. <lang ecmascript>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])
} </lang>
- 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)
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. <lang ecmascript>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
}</lang>
- Output:
Same as Version 1.