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Calkin-Wilf sequence

From Rosetta Code
Task
Calkin-Wilf sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The Calkin-Wilf sequence contains every nonnegative rational number exactly once.

It can be calculated recursively as follows:

       a1   =  1 
       an+1  =  1/(2⌊an⌋+1-an) for n > 1 


Task part 1
  • Show on this page terms 1 through 20 of the Calkin-Wilf sequence.


To avoid floating point error, you may want to use a rational number data type.


It is also possible, given a non-negative rational number, to determine where it appears in the sequence without calculating the sequence. The procedure is to get the continued fraction representation of the rational and use it as the run-length encoding of the binary representation of the term number, beginning from the end of the continued fraction. It only works if the number of terms in the continued fraction is odd- use either of the two equivalent representations to achieve this:

       [a0; a1, a2, ..., an]  =  [a0; a1, a2 ,..., an-1, 1] 


Example

The fraction   9/4   has odd continued fraction representation     2; 3, 1,     giving a binary representation of   100011,
which means   9/4   appears as the   35th   term of the sequence.


Task part 2
  • Find the position of the number   83116/51639   in the Calkin-Wilf sequence.


See also



11l[edit]

Translation of: Nim
T CalkinWilf
n = 1
d = 1
 
F ()()
V r = (.n, .d)
.n = 2 * (.n I/ .d) * .d + .d - .n
swap(&.n, &.d)
R r
 
print(‘The first 20 terms of the Calkwin-Wilf sequence are:’)
V cw = CalkinWilf()
[String] seq
L 20
V (n, d) = cw()
seq.append(I d == 1 {String(n)} E n‘/’d)
print(seq.join(‘, ’))
 
cw = CalkinWilf()
V index = 1
L cw() != (83116, 51639)
index++
print("\nThe element 83116/51639 is at position "index‘ in the sequence.’)
Output:
The first 20 terms of the Calkwin-Wilf sequence are:
1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4, 1/5, 5/4, 4/7, 7/3, 3/8

The element 83116/51639 is at position 123456789 in the sequence.

AppleScript[edit]

-- Return the first n terms of the sequence. Tree generation. Faster for this purpose.
on CalkinWilfSequence(n)
script o
property sequence : {{1, 1}} -- Initialised with the first term ({numerator, denominator}).
end script
 
-- Work through the growing sequence list, adding the two children of each term to the end and
-- converting each term to text representing the vulgar fraction. Stop adding children halfway through.
set halfway to n div 2
repeat with position from 1 to n
set {numerator, denominator} to item position of o's sequence
if (position ≤ halfway) then
tell numerator + denominator
set end of o's sequence to {numerator, it}
if ((position < halfway) or (position * 2 < n)) then set end of o's sequence to {it, denominator}
end tell
end if
set item position of o's sequence to (numerator as text) & "/" & denominator
end repeat
 
return o's sequence
end CalkinWilfSequence
 
-- Alternatively, return terms pos1 to pos2. Binary run-length encoding. Doesn't need to work from the beginning of the sequence.
on CalkinWilfSequence2(pos1, pos2)
script o
property sequence : {}
end script
 
repeat with position from pos1 to pos2
-- Build a continued fraction list from the binary run-length encoding of this position index.
-- There's no need to put the last value into the list as it's used immediately.
set continuedFraction to {}
set bitValue to 1
set runLength to 0
repeat until (position = 0)
if (position mod 2 = bitValue) then
set runLength to runLength + 1
else
set end of continuedFraction to runLength
set bitValue to (bitValue + 1) mod 2
set runLength to 1
end if
set position to position div 2
end repeat
-- Work out the numerator and denominator from the continued fraction and derive text representing the vulgar fraction.
set numerator to runLength
set denominator to 1
repeat with i from (count continuedFraction) to 1 by -1
tell numerator
set numerator to numerator * (item i of continuedFraction) + denominator
set denominator to it
end tell
end repeat
set end of o's sequence to (numerator as text) & "/" & denominator
end repeat
 
return o's sequence
end CalkinWilfSequence2
 
-- Return the sequence position of the term with the given numerator and denominator.
on CalkinWilfSequencePosition(numerator, denominator)
-- Build a continued fraction list from the input.
set continuedFraction to {}
repeat until (denominator is 0)
set end of continuedFraction to numerator div denominator
set {numerator, denominator} to {denominator, numerator mod denominator}
end repeat
-- If it has an even number of entries, convert to the equivalent odd number.
if ((count continuedFraction) mod 2 is 0) then
set last item of continuedFraction to (last item of continuedFraction) - 1
set end of continuedFraction to 1
end if
-- "Binary run-length decode" the entries to get the position index.
set position to 0
set bitValue to 1
repeat with i from (count continuedFraction) to 1 by -1
repeat (item i of continuedFraction) times
set position to position * 2 + bitValue
end repeat
set bitValue to (bitValue + 1) mod 2
end repeat
 
return position
end CalkinWilfSequencePosition
 
-- Task code:
local sequenceResult1, sequenceResult2, positionResult, output, astid
set sequenceResult1 to CalkinWilfSequence(20)
set sequenceResult2 to CalkinWilfSequence2(1, 20)
set positionResult to CalkinWilfSequencePosition(83116, 51639)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to ", "
set output to "First twenty terms of sequence using tree generation:" & (linefeed & sequenceResult1)
set output to output & (linefeed & "Ditto using binary run-length encoding:") & (linefeed & sequenceResult1)
set AppleScript's text item delimiters to astid
set output to output & (linefeed & "83116/51639 is term number " & positionResult)
return output
Output:
"First twenty terms of sequence using tree generation:
1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4/1, 1/5, 5/4, 4/7, 7/3, 3/8
Ditto using binary run-length encoding:
1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4/1, 1/5, 5/4, 4/7, 7/3, 3/8
83116/51639 is term number 123456789"

Arturo[edit]

n: new 1
d: new 1
calkinWilf: function [] .export:[n,d] [
n: (d - n) + 2 * (n/d) * d
tmp: d
d: n
n: tmp
return @[n d]
]
 
first20: [[1 1]] ++ map 1..19 => calkinWilf
print "The first 20 terms of the Calkwin-Wilf sequence are:"
print map first20 'f -> ~"|f\0|/|f\1|"
 
n: new 1
d: new 1
indx: new 1
 
target: [83116, 51639]
 
while ø [
inc 'indx
if target = calkinWilf -> break
]
 
print ""
print ["The element" ~"|target\0|/|target\1|" "is at position" indx "in the sequence."]
Output:
The first 20 terms of the Calkwin-Wilf sequence are:
1/1 1/2 2/1 1/3 3/2 2/3 3/1 1/4 4/3 3/5 5/2 2/5 5/3 3/4 4/1 1/5 5/4 4/7 7/3 3/8 

The element 83116/51639 is at position 123456789 in the sequence.

BQN[edit]

BQN does not have rational number arithmetic yet, so it is manually implemented.

Part 2 runs in ~150 secs on CBQN.

GCD and _while_ are idioms from BQNcrate.

GCD ← {m 𝕊⍟(0<m←𝕨|𝕩) 𝕨}
_while_ ← {𝔽⍟𝔾∘𝔽_𝕣_𝔾∘𝔽⍟𝔾𝕩}
Sim ← { # Simplify a fraction
x𝕊1: 𝕨‿1;
0𝕊y: 0‿𝕩;
⌊𝕨‿𝕩 ÷ 𝕨 GCD 𝕩
}
Add ← { # Add two fractions
0‿b 𝕊 𝕩: 𝕩;
𝕨 𝕊 0‿y: 𝕨;
a‿b 𝕊 x‿y:
((a×y)+x×b) Sim b×y
}
Next ← {n‿d: ⌽(2×⌊÷´n‿d)‿1 Add (d-n)‿d} # Next term
Cal ← {Next⍟𝕩 1‿1}
 
•Show Cal 1+↕20
 
•Show {
cnt‿fr:
⟨cnt+1,Next fr⟩
} _while_ {
cnt‿fr:
fr ≢ 83116‿51639
} ⟨1,1‿1⟩
⟨ ⟨ 1 2 ⟩ ⟨ 2 1 ⟩ ⟨ 1 3 ⟩ ⟨ 3 2 ⟩ ⟨ 2 3 ⟩ ⟨ 3 1 ⟩ ⟨ 1 4 ⟩ ⟨ 4 3 ⟩ ⟨ 3 5 ⟩ ⟨ 5 2 ⟩ ⟨ 2 5 ⟩ ⟨ 5 3 ⟩ ⟨ 3 4 ⟩ ⟨ 4 1 ⟩ ⟨ 1 5 ⟩ ⟨ 5 4 ⟩ ⟨ 4 7 ⟩ ⟨ 7 3 ⟩ ⟨ 3 8 ⟩ ⟨ 8 5 ⟩ ⟩
⟨ 123456789 ⟨ 83116 51639 ⟩ ⟩

You can try Part 1 here. Second part can and will hang your browser, so it is best to try locally on CBQN.

Bracmat[edit]

Translation of: Python
( 1:?a
& 0:?i
& whl
' ( 1+!i:<20:?i
& (2*div$(!a,1)+1+-1*!a)^-1:?a
& out$!a
)
& ( r2cf
= floor
. div$(!arg,1):?floor
& ( !floor:!arg
| !floor r2cf$((!arg+-1*!floor)^-1)
)
)
& ( get-term-num
= ans dig pwr
. (0,1,1):(?ans,?dig,?pwr)
& r2cf$!arg:?n
& map
$ ( (
=
. whl
' ( !arg+-1:~<0:?arg
& !dig*!pwr+!ans:?ans
& 2*!pwr:?pwr
)
& 1+-1*!dig:?dig
)
. !n
)
& !ans
)
& out$(get-term-num$83116/51639)
);
Output:
1/2
2
1/3
3/2
2/3
3
1/4
4/3
3/5
5/2
2/5
5/3
3/4
4
1/5
5/4
4/7
7/3
3/8
123456789

C++[edit]

Library: Boost
#include <iostream>
#include <vector>
#include <boost/rational.hpp>
 
using rational = boost::rational<unsigned long>;
 
unsigned long floor(const rational& r) {
return r.numerator()/r.denominator();
}
 
rational calkin_wilf_next(const rational& term) {
return 1UL/(2UL * floor(term) + 1UL - term);
}
 
std::vector<unsigned long> continued_fraction(const rational& r) {
unsigned long a = r.numerator();
unsigned long b = r.denominator();
std::vector<unsigned long> result;
do {
result.push_back(a/b);
unsigned long c = a;
a = b;
b = c % b;
} while (a != 1);
if (result.size() > 0 && result.size() % 2 == 0) {
--result.back();
result.push_back(1);
}
return result;
}
 
unsigned long term_number(const rational& r) {
unsigned long result = 0;
unsigned long d = 1;
unsigned long p = 0;
for (unsigned long n : continued_fraction(r)) {
for (unsigned long i = 0; i < n; ++i, ++p)
result |= (d << p);
d = !d;
}
return result;
}
 
int main() {
rational term = 1;
std::cout << "First 20 terms of the Calkin-Wilf sequence are:\n";
for (int i = 1; i <= 20; ++i) {
std::cout << std::setw(2) << i << ": " << term << '\n';
term = calkin_wilf_next(term);
}
rational r(83116, 51639);
std::cout << r << " is the " << term_number(r) << "th term of the sequence.\n";
}
Output:
First 20 terms of the Calkin-Wilf sequence are:
 1: 1/1
 2: 1/2
 3: 2/1
 4: 1/3
 5: 3/2
 6: 2/3
 7: 3/1
 8: 1/4
 9: 4/3
10: 3/5
11: 5/2
12: 2/5
13: 5/3
14: 3/4
15: 4/1
16: 1/5
17: 5/4
18: 4/7
19: 7/3
20: 3/8
83116/51639 is the 123456789th term of the sequence.

EDSAC order code[edit]

Find first n terms[edit]

Translation of: Pascal
 
[For Rosetta Code. EDSAC program, Initial Orders 2.
Prints the first 20 terms of the Calkin-Wilf sequence.
Uses term{n} to calculate term{n + 1}.]
 
[Print subroutine for non-negative 17-bit integers.
Parameters: 0F = integer to be printed (not preserved)
1F = character for leading zero (preserved)
Workspace: 4F, 5F. Even address; 40 locations]
T 56 K [define load address]
[email protected]@[email protected]@AFT5FT4FH38#@[email protected]
[email protected]@[email protected]
[email protected]@[email protected]@[email protected]
 
[Main routine]
T 100 K [define load address]
G K [set up relative addressing via @ (theta)]
[Constants]
[0] P 10 F [maximum index = 20, edit ad lib.]
[1] P D [constant 1]
[Teleprinter characters]
[2] # F [set figures mode]
[3] C F [colon (in figures mode)]
[4] X F [slash (in figures mode)]
[5] ! F [space]
[6] @ F [carriage return]
[7] & F [line feed]
[8] K 4096 F [null]
[Variables]
[9] P F [index]
[10] P F [a (where term = a/b)]
[11] P F [b]
[Enter with acc = 0]
[12] O 2 @ [set teleprinter to figures]
A 1 @ [acc := 1]
U 9 @ [index := 1]
U 10 @ [a := 1]
T 11 @ [b := 1 (and clear acc)]
E 34 @ [jump to print first term]
[Loop back here if not yet printed enough terms]
[18] A @ [restore index after test]
A 1 @ [add 1]
T 9 @ [update index]
[Calculate next term. New b := a + b - 2(a mod b).
Code below calculates c := (a mod b) - b, then new b := a - b - 2*c]
A 10 @ [acc := a]
[22] S 11 @ [subtract b]
E 22 @ [if acc >= 0, subtract again]
T F [result c < 0, store in 0F]
A 10 @ [acc := a]
S 11 @ [subtract b]
S F [subtract c]
S F [subtract c]
T F [new b = a - b - 2*c; store in 0F]
A 11 @ [acc := old b]
T 10 @ [copy to a]
A F [acc := new b]
T 11 @ [copy to b]
[Print index and a/b. Assume acc = 0 here.]
[34] A 5 @ [space to replace leading 0's]
T 1 F [pass to print subroutine]
A 9 @ [acc := index]
T F [pass to print subroutine]
[38] A 38 @ [for return from subroutine]
G 56 F [call subroutine, clears acc]
O 3 @ [print colon]
O 5 @ [print space]
A 8 @ [null to replace leading 0's]
T 1 F [pass to print subroutine]
[email protected] TF [email protected] G56F [email protected] [print a followed by slash]
[email protected] TF [email protected] G56F [email protected] [email protected] [print b followed by CR LF]
[Test whether enough terms have been printed]
A 9 @ [acc := index]
S @ [subtract maximum index]
G 18 @ [loop back if acc < 0]
[Exit]
O 8 @ [print null to flush teleprinter buffer]
Z F [stop]
E 12 Z [relative address of entry point]
P F [enter with acc = 0]
[end]
 
Output:
    1: 1/1
    2: 1/2
    3: 2/1
    4: 1/3
    5: 3/2
    6: 2/3
    7: 3/1
    8: 1/4
    9: 4/3
   10: 3/5
   11: 5/2
   12: 2/5
   13: 5/3
   14: 3/4
   15: 4/1
   16: 1/5
   17: 5/4
   18: 4/7
   19: 7/3
   20: 3/8

Find index of a given term[edit]

Translation of: Pascal
 
[For Rosetta Code. EDSAC program, Initial Orders 2.]
[Finds the index of a given rational in the Calkin-Wilf series.]
 
[Library subroutine R2: input of positive integers.
Runs during input of the program, and is then overwritten.
Allows integers to be written in decimal, rather than as "pseudo-orders".
See Wilkes, Williams & Gill, 1951 edn, pp. 96-97, 148.]
T 54 K [to access integers via C parameter]
P 110 F [where to load integers]
[email protected]@[email protected]@E13Z
T #C [tell R2 where to load integers]
[F after each integer except the last, and # after the last.]
83116F51639#
 
[Modified library subroutine P7.
Prints signed integer up to 10 digits, left-justified.
Input: Number to be printed is at 0D.
54 locations. Load at even address. Workspace 4D.]
 
T 56 K
[email protected]@[email protected]@[email protected]@[email protected]#@[email protected]
[email protected]@[email protected]@[email protected]@[email protected]
[email protected]@[email protected]@[email protected]@[email protected]
 
[Main routine.]
T 120 K [define load address (must be even)]
G K [set up relative addressing via @ (theta)]
 
[Put 35-bit values first, to ensure each is at an even address]
[Variables]
[0] P F P F [a]
[2] P F P F [b]
[4] P F P F [power of 2]
[6] P F P F [calculated index]
[Constants]
T8#Z PF T8Z [clears sandwich digit between 8 and 9]
[8] P D P F [35-bit constant 1]
[Teleprinter characters]
[10] # F [set figures mode]
[11] X F [slash (in figures mode)]
[12] K 2048 F [set letters mode]
[13] I F [letter I]
[14] R F [letter R]
[15]  ! F [space]
[16] @ F [carriage return]
[17] & F [line feed]
[18] K 4096 F [null char]
 
[Enter with acc = 0]
[19] A #C [acc := initial a]
T #@ [copy to variable]
A 2#C [acc := initial b]
T 2#@ [copy to variable]
[23] A 8#@ [acc := 1]
[24] T 4#@ [initialize power of 2]
T 6#@ [initialize index to 0]
[Loop]
[26] A #@ [acc := a]
[27] S 2#@ [subtract b]
[28] E 33 @ [jump if a >= b]
[Here if a < b]
T D [store a - b in 0D]
S D [negate]
T 2#@ [b := b - a]
E 40 @ [join common code]
[Here if a >= b]
[33] S 8#@ [acc = a - b; test for a = b]
G 45 @ [jump out of loop if so]
A 8#@ [restore a - b]
T #@ [a := a - b]
A 6#@ [acc := index]
A 4#@ [inc index by power of 2]
T 6#@
[Code common to both cases]
[40] A 4#@ [acc := power of 2]
L D [shift left]
G 76 @
T 4#@ [update power of 2]
E 26 @ [loop back]
[Exit from loop.]
[45] T D [dump acc to clear it]
A 6#@ [acc := index]
A 4#@ [add power of 2 ]
T 6#@ [store final value of index]
[Finished calcualting index, now do printing]
O 10 @ [set teleprinter to figures]
A #C [acc := initial a]
T D [to 0D for printing]
[52] A 52 @ [for return from subroutine]
G 56 F [call print subroutine, clears acc]
O 11 @ [print slash]
A 2#C [print initial b similarly]
T D
[57] A 57 @
G 56 F
O 12 @ [set teleprinter to letters and print ' IS AT ']
[email protected] [email protected] [email protected] [email protected] [email protected] [email protected] [email protected]
O 10 @ [set teleprinter to figures]
A 6#@ [acc := calculated index]
T D [send to print subroutine]
[70] A 70 @
G 56 F
[72] [email protected] [email protected] [print CR, LF]
O 18 @ [print null to flush teleprinter buffer]
Z F [stop]
[Here if power of 2 goes negative (accumulator overflow)]
[76] O 12 @ [set teleprinter to letters]
[email protected] [email protected] [email protected] [email protected] [email protected] [print'ERROR']
G 72 @ [jump to common exit]
E 19 Z [relative address of entry point]
P F [enter with acc = 0]
 
Output:
83116/51639 IS AT 123456789

F#[edit]

The Function[edit]

 
// Calkin Wilf Sequence. Nigel Galloway: January 9th., 2021
let cW=Seq.unfold(fun(n)->Some(n,seq{for n,g in n do yield (n,n+g); yield (n+g,g)}))(seq[(1,1)])|>Seq.concat
 

The Tasks[edit]

first 20
 
cW |> Seq.take 20 |> Seq.iter(fun(n,g)->printf "%d/%d " n g);printfn ""
 
Output:
1/1 1/2 2/1 1/3 3/2 2/3 3/1 1/4 4/3 3/5 5/2 2/5 5/3 3/4 4/1 1/5 5/4 4/7 7/3 3/8
Indexof 83116/51639
 
printfn "%d" (1+Seq.findIndex(fun n->n=(83116,51639)) cW)
 
Output:
123456789

Factor[edit]

Works with: Factor version 0.99 2020-08-14
USING: formatting io kernel lists lists.lazy math
math.continued-fractions math.functions math.parser prettyprint
sequences strings vectors ;
 
: next-cw ( x -- y ) [ floor dup + ] [ 1 swap - + recip ] bi ;
 
: calkin-wilf ( -- list ) 1 [ next-cw ] lfrom-by ;
 
: >continued-fraction ( x -- seq )
1vector [ dup last integer? ] [ dup next-approx ] until
dup length even? [ unclip-last 1 - suffix! 1 suffix! ] when ;
 
: cw-index ( x -- n )
>continued-fraction <reversed>
[ even? CHAR: 1 CHAR: 0 ? <string> ] map-index concat bin> ;
 
! Task
"First 20 terms of the Calkin-Wilf sequence:" print
20 calkin-wilf ltake [ pprint bl ] leach nl nl
 
83116/51639 cw-index "83116/51639 is at index %d.\n" printf
Output:
First 20 terms of the Calkin-Wilf sequence:
1 1/2 2 1/3 1+1/2 2/3 3 1/4 1+1/3 3/5 2+1/2 2/5 1+2/3 3/4 4 1/5 1+1/4 4/7 2+1/3 3/8 

83116/51639 is at index 123456789.

FreeBASIC[edit]

Uses the code from Greatest common divisor#FreeBASIC as an include.

#include "gcd.bas"
 
type rational
num as integer
den as integer
end type
 
dim shared as rational ONE, TWO
ONE.num = 1 : ONE.den = 1
TWO.num = 2 : TWO.den = 1
 
function simplify( byval a as rational ) as rational
dim as uinteger g = gcd( a.num, a.den )
a.num /= g : a.den /= g
if a.den < 0 then
a.den = -a.den
a.num = -a.num
end if
return a
end function
 
operator + ( a as rational, b as rational ) as rational
dim as rational ret
ret.num = a.num * b.den + b.num*a.den
ret.den = a.den * b.den
return simplify(ret)
end operator
 
operator - ( a as rational, b as rational ) as rational
dim as rational ret
ret.num = a.num * b.den - b.num*a.den
ret.den = a.den * b.den
return simplify(ret)
end operator
 
operator * ( a as rational, b as rational ) as rational
dim as rational ret
ret.num = a.num * b.num
ret.den = a.den * b.den
return simplify(ret)
end operator
 
operator / ( a as rational, b as rational ) as rational
dim as rational ret
ret.num = a.num * b.den
ret.den = a.den * b.num
return simplify(ret)
end operator
 
function floor( a as rational ) as rational
dim as rational ret
ret.den = 1
ret.num = a.num \ a.den
return ret
end function
 
function cw_nextterm( q as rational ) as rational
dim as rational ret = (TWO*floor(q))
ret = ret + ONE : ret = ret - q
return ONE / ret
end function
 
function frac_to_int( byval a as rational ) as uinteger
redim as uinteger cfrac(-1)
dim as integer lt = -1, ones = 1, ret = 0
do
lt += 1
redim preserve as uinteger cfrac(0 to lt)
cfrac(lt) = floor(a).num
a = a - floor(a) : a = ONE / a
loop until a.num = 0 or a.den = 0
if lt mod 2 = 1 and cfrac(lt) = 1 then
lt -= 1
cfrac(lt)+=1
redim preserve as uinteger cfrac(0 to lt)
end if
if lt mod 2 = 1 and cfrac(lt) > 1 then
cfrac(lt) -= 1
lt += 1
redim preserve as uinteger cfrac(0 to lt)
cfrac(lt) = 1
end if
for i as integer = lt to 0 step -1
for j as integer = 1 to cfrac(i)
ret *= 2
if ones = 1 then ret += 1
next j
ones = 1 - ones
next i
return ret
end function
 
function disp_rational( a as rational ) as string
if a.den = 1 or a.num= 0 then return str(a.num)
return str(a.num)+"/"+str(a.den)
end function
 
dim as rational q
q.num = 1
q.den = 1
for i as integer = 1 to 20
print i, disp_rational(q)
q = cw_nextterm(q)
next i
 
q.num = 83116
q.den = 51639
print disp_rational(q)+" is the "+str(frac_to_int(q))+"th term."
Output:
 1            1
 2            1/2
 3            2
 4            1/3
 5            3/2
 6            2/3
 7            3
 8            1/4
 9            4/3
 10           3/5
 11           5/2
 12           2/5
 13           5/3
 14           3/4
 15           4
 16           1/5
 17           5/4
 18           4/7
 19           7/3
 20           3/8
83116/51639 is the 123456789th term.

Fōrmulæ[edit]

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

In this page you can see the program(s) related to this task and their results.

Go[edit]

Translation of: Wren

Go just has arbitrary precision rational numbers which we use here whilst assuming the numbers needed for this task can be represented exactly by the 64 bit built-in types.

package main
 
import (
"fmt"
"math"
"math/big"
"strconv"
"strings"
)
 
func calkinWilf(n int) []*big.Rat {
cw := make([]*big.Rat, n+1)
cw[0] = big.NewRat(1, 1)
one := big.NewRat(1, 1)
two := big.NewRat(2, 1)
for i := 1; i < n; i++ {
t := new(big.Rat).Set(cw[i-1])
f, _ := t.Float64()
f = math.Floor(f)
t.SetFloat64(f)
t.Mul(t, two)
t.Sub(t, cw[i-1])
t.Add(t, one)
t.Inv(t)
cw[i] = new(big.Rat).Set(t)
}
return cw
}
 
func toContinued(r *big.Rat) []int {
a := r.Num().Int64()
b := r.Denom().Int64()
var res []int
for {
res = append(res, int(a/b))
t := a % b
a, b = b, t
if a == 1 {
break
}
}
le := len(res)
if le%2 == 0 { // ensure always odd
res[le-1]--
res = append(res, 1)
}
return res
}
 
func getTermNumber(cf []int) int {
b := ""
d := "1"
for _, n := range cf {
b = strings.Repeat(d, n) + b
if d == "1" {
d = "0"
} else {
d = "1"
}
}
i, _ := strconv.ParseInt(b, 2, 64)
return int(i)
}
 
func commatize(n int) string {
s := fmt.Sprintf("%d", n)
if n < 0 {
s = s[1:]
}
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
if n >= 0 {
return s
}
return "-" + s
}
 
func main() {
cw := calkinWilf(20)
fmt.Println("The first 20 terms of the Calkin-Wilf sequnence are:")
for i := 1; i <= 20; i++ {
fmt.Printf("%2d: %s\n", i, cw[i-1].RatString())
}
fmt.Println()
r := big.NewRat(83116, 51639)
cf := toContinued(r)
tn := getTermNumber(cf)
fmt.Printf("%s is the %sth term of the sequence.\n", r.RatString(), commatize(tn))
}
Output:
The first 20 terms of the Calkin-Wilf sequnence are:
 1: 1
 2: 1/2
 3: 2
 4: 1/3
 5: 3/2
 6: 2/3
 7: 3
 8: 1/4
 9: 4/3
10: 3/5
11: 5/2
12: 2/5
13: 5/3
14: 3/4
15: 4
16: 1/5
17: 5/4
18: 4/7
19: 7/3
20: 3/8

83116/51639 is the 123,456,789th term of the sequence.

Haskell[edit]

import Control.Monad (forM_)
import Data.Bool (bool)
import Data.List.NonEmpty (NonEmpty, fromList, toList, unfoldr)
import Text.Printf (printf)
 
-- The infinite Calkin-Wilf sequence, a(n), starting with a(1) = 1.
calkinWilfs :: [Rational]
calkinWilfs = iterate (recip . succ . ((-) =<< (2 *) . fromIntegral . floor)) 1
 
-- The index into the Calkin-Wilf sequence of a given rational number, starting
-- with 1 at index 1.
calkinWilfIdx :: Rational -> Integer
calkinWilfIdx = rld . cfo
 
-- A continued fraction representation of a given rational number, guaranteed
-- to have an odd length.
cfo :: Rational -> NonEmpty Int
cfo = oddLen . cf
 
-- The canonical (i.e. shortest) continued fraction representation of a given
-- rational number.
cf :: Rational -> NonEmpty Int
cf = unfoldr step
where
step r =
case properFraction r of
(n, 1) -> (succ n, Nothing)
(n, 0) -> (n, Nothing)
(n, f) -> (n, Just (recip f))
 
-- Ensure a continued fraction has an odd length.
oddLen :: NonEmpty Int -> NonEmpty Int
oddLen = fromList . go . toList
where
go [x, y] = [x, pred y, 1]
go (x:y:zs) = x : y : go zs
go xs = xs
 
-- Run-length decode a continued fraction.
rld :: NonEmpty Int -> Integer
rld = snd . foldr step (True, 0)
where
step i (b, n) =
let p = 2 ^ i
in (not b, n * p + bool 0 (pred p) b)
 
main :: IO ()
main = do
forM_ (take 20 $ zip [1 :: Int ..] calkinWilfs) $
\(i, r) -> printf "%2d  %s\n" i (show r)
let r = 83116 / 51639
printf
"\n%s is at index %d of the Calkin-Wilf sequence.\n"
(show r)
(calkinWilfIdx r)
Output:
 1  1 % 1
 2  1 % 2
 3  2 % 1
 4  1 % 3
 5  3 % 2
 6  2 % 3
 7  3 % 1
 8  1 % 4
 9  4 % 3
10  3 % 5
11  5 % 2
12  2 % 5
13  5 % 3
14  3 % 4
15  4 % 1
16  1 % 5
17  5 % 4
18  4 % 7
19  7 % 3
20  3 % 8

83116 % 51639 is at index 123456789 of the Calkin-Wilf sequence.

J[edit]

   cw_next_term^:(<20)1x
1 1r2 2 1r3 3r2 2r3 3 1r4 4r3 3r5 5r2 2r5 5r3 3r4 4 1r5 5r4 4r7 7r3 3r8

   (,. index_cw_term&>) 3r4 53r37 83116r51639
        3r4        14
      53r37      1081
83116r51639 123456789

given definitions

 
cw_next_term=: [: % +:@<. + -.
 
ccf =: compute_continued_fraction=: 3 :0
if. 0 -: y do.
, 0
else.
result=. i. 0
remainder=. % y
whilst. remainder do.
remainder=. % remainder
integer_part=. <. remainder
remainder=. remainder - integer_part
result=. result , integer_part
end.
end.
)
 
molcf =: make_odd_length_continued_fraction=: (}: , 1 ,~ <:@{:)^:(0 -: 2 | #)
 
NB. base 2 @ reverse @ the cf's representation copies of 1 0 1 0 ...
index_cw_term=: #[email protected]|[email protected](# 1 0 $~ #)@[email protected]
 

Julia[edit]

Translation of: Wren
function calkin_wilf(n)
cw = zeros(Rational, n + 1)
for i in 2:n + 1
t = Int(floor(cw[i - 1])) * 2 - cw[i - 1] + 1
cw[i] = 1 // t
end
return cw[2:end]
end
 
function continued(r::Rational)
a, b = r.num, r.den
res = []
while true
push!(res, Int(floor(a / b)))
a, b = b, a % b
a == 1 && break
end
return res
end
 
function term_number(cf)
b, d = "", "1"
for n in cf
b = d^n * b
d = (d == "1") ? "0" : "1"
end
return parse(Int, b, base=2)
end
 
const cw = calkin_wilf(20)
println("The first 20 terms of the Calkin-Wilf sequence are: $cw")
 
const r = 83116 // 51639
const cf = continued(r)
const tn = term_number(cf)
println("$r is the $tn-th term of the sequence.")
 
Output:
The first 20 terms of the Calkin-Wilf sequence are: Rational[1//1, 1//2, 2//1, 1//3, 3//2, 2//3, 3//1, 1//4, 4//3, 3//5, 5//2, 2//5, 5//3, 3//4, 4//1, 1//5, 5//4, 4//7, 7//3, 3//8]
83116//51639 is the 123456789-th term of the sequence.

Little Man Computer[edit]

Runs in a home-made simulator, which is mostly compatible with Peter Higginson's online simulator. Only, for better control of the output format, I've added an instruction OTX (extended output). To run the code in PH's simulator, replace OTX and its parameter with OUT and no parameter.

Find first n terms[edit]

Translation of: Pascal
 
// Little Man Computer, for Rosetta Code.
// Displays terms of Calkin-Wilf sequence up to the given index.
// The chosen algorithm calculates the i-th term directly from i
// (i.e. not using any previous terms).
input INP // get number of terms from user
BRZ exit // exit if 0
STA max_i // store maximum index
LDA c1 // index := 1
next_i STA i
// Write index followed by '->'
OTX 3 // non-standard: minimum width 3, no new line
LDA asc_hy
OTC
LDA asc_gt
OTC
// Find greatest power of 2 not exceeding i,
// and count the number of binary digits in i.
LDA c1
STA pwr2
loop2 STA nrDigits
LDA i
SUB pwr2
SUB pwr2
BRP double
BRA part2 // jump out if next power of 2 would exceed i
double LDA pwr2
ADD pwr2
STA pwr2
LDA nrDigits
ADD c1
BRA loop2
// The nth term a/b is calculated from the binary digits of i.
// The leading 1 is not used.
part2 LDA c1
STA a // a := 1
STA b // b := 1
LDA i
SUB pwr2
STA diff
// Pre-decrement count, since leading 1 is not used
dec_ct LDA nrDigits // count down the number of digits
SUB c1
BRZ output // if all digits done, output the result
STA nrDigits
// We now want to compare diff with pwr2/2.
// Since division is awkward in LMC, we compare 2*diff with pwr2.
LDA diff // diff := 2*diff
ADD diff
STA diff
SUB pwr2 // is diff >= pwr2 ?
BRP digit_1 // binary digit is 1 if yes, 0 if no
// If binary digit is 0 then set b := a + b
LDA a
ADD b
STA b
BRA dec_ct
// If binary digit is 1 then update diff and set a := a + b
digit_1 STA diff
LDA a
ADD b
STA a
BRA dec_ct
// Now have nth term a/b. Write it to the output.
output LDA a // write a
OTX 1 // non-standard: minimum width 1; no new line
LDA asc_sl // write slash
OTC
LDA b // write b
OTX 11 // non-standard: minimum width 1; add new line
LDA i // have we done maximum i yet?
SUB max_i
BRZ exit // if yes, exit
LDA i // if no, increment i and loop back
ADD c1
BRA next_i
exit HLT
// Constants
c1 DAT 1
asc_hy DAT 45
asc_gt DAT 62
asc_sl DAT 47
// Variables
i DAT
max_i DAT
pwr2 DAT
nrDigits DAT
diff DAT
a DAT
b DAT
// end
 
Output:
  1->1/1
  2->1/2
  3->2/1
  4->1/3
  5->3/2
  6->2/3
  7->3/1
  8->1/4
  9->4/3
 10->3/5
 11->5/2
 12->2/5
 13->5/3
 14->3/4
 15->4/1
 16->1/5
 17->5/4
 18->4/7
 19->7/3
 20->3/8

Find index of a given term[edit]

Translation of: Pascal

The numbers in part 2 of the task are too large for LMC, so the demo program just confirms the example, that 9/4 is the 35th term.

 
// Little Man Computer, for Rosetta Code.
// Calkin-Wilf sequence: displays index of term entered by user.
INP // get numerator from user
BRZ exit // exit if 0
STA num
STA a // initialize a := numerator
INP // get denominator from user
BRZ exit // exit if 0
STA den
STA b // initialize b := denominator
LDA c0 // initialize index := 0
STA index
LDA c1 // initialize power of 2 := 1
STA pwr2
// Build binary digits of the index
loop LDA a // is a = b yet?
SUB b
BRZ break // if yes, break out of loop
BRP a_gt_b // jump if a > b
// If a < b then b := b - a, binary digit is 0
LDA b
SUB a
STA b
BRA double
// If a > b then a := a - b, binary digit is 1
a_gt_b STA a
LDA index
ADD pwr2
STA index
// In either case, on to next power of 2
double LDA pwr2
ADD pwr2
STA pwr2
BRA loop
// Out of loop, add leading binary digit 1
break LDA index
ADD pwr2
STA index
// Output the result
LDA num
OTX 1 // non-standard: minimum width = 1, no new line
LDA asc_sl
OTC
LDA den
OTX 1
LDA asc_lt // write '<-' after fraction
OTC
LDA asc_hy
OTC
LDA index
OTX 11 // non-standard: minimum width = 1, add new line
exit HLT
// Constants
c0 DAT 0
c1 DAT 1
asc_sl DAT 47
asc_lt DAT 60
asc_hy DAT 45
// Variables
num DAT
den DAT
a DAT
b DAT
pwr2 DAT
index DAT
// end
 
Output:
9/4<-35


Mathematica / Wolfram Language[edit]

ClearAll[a]
a[1] = 1;
a[n_?(GreaterThan[1])] := a[n] = 1/(2 Floor[a[n - 1]] + 1 - a[n - 1])
a /@ Range[20]
 
ClearAll[a]
a = 1;
n = 1;
Dynamic[n]
done = False;
While[! done,
a = 1/(2 Floor[a] + 1 - a);
n++;
If[a == 83116/51639,
Print[n];
Break[];
]
]
Output:
{1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4, 1/5, 5/4, 4/7, 7/3, 3/8}
123456789

Nim[edit]

We ignored the standard module “rationals” which is slow and have rather defined a fraction as a tuple of two 32 bits unsigned integers (slightly faster than 64 bits signed integers and sufficient for this task). Moreover, we didn’t do operations on fractions and computed directly the numerator and denominator values at each step. The fractions built this way are irreducible (which avoids to compute a GCD which is a slow operation). With these optimizations, the program runs in less than 1.3 s on our laptop.

type Fraction = tuple[num, den: uint32]
 
iterator calkinWilf(): Fraction =
## Yield the successive values of the sequence.
var n, d = 1u32
yield (n, d)
while true:
n = 2 * (n div d) * d + d - n
swap n, d
yield (n, d)
 
proc `$`(fract: Fraction): string =
## Return the representation of a fraction.
$fract.num & '/' & $fract.den
 
func `==`(a, b: Fraction): bool {.inline.} =
## Compare two fractions. Slightly faster than comparison of tuples.
a.num == b.num and a.den == b.den
 
when isMainModule:
 
echo "The first 20 terms of the Calkwin-Wilf sequence are:"
var count = 0
for an in calkinWilf():
inc count
stdout.write $an & ' '
if count == 20: break
stdout.write '\n'
 
const Target: Fraction = (83116u32, 51639u32)
var index = 0
for an in calkinWilf():
inc index
if an == Target: break
echo "\nThe element ", $Target, " is at position ", $index, " in the sequence."
Output:
The first 20 terms of the Calkwin-Wilf sequence are:
1/1 1/2 2/1 1/3 3/2 2/3 3/1 1/4 4/3 3/5 5/2 2/5 5/3 3/4 4/1 1/5 5/4 4/7 7/3 3/8 

The element 83116/51639 is at position 123456789 in the sequence.

Pascal[edit]

These programs were written in Free Pascal, using the Lazarus IDE and the Free Pascal compiler version 3.2.0. They are based on the Wikipedia article "Calkin-Wilf tree", rather than the algorithms in the task description.

 
program CWTerms;
 
{-------------------------------------------------------------------------------
FreePascal command-line program.
Calculates the Calkin-Wilf sequence up to the specified maximum index,
where the first term 1/1 has index 1.
Command line format is: CWTerms <max_index>
 
The program demonstrates 3 algorithms for calculating the sequence:
(1) Calculate term[2n] and term[2n + 1] from term[n]
(2) Calculate term[n + 1] from term[n]
(3) Calculate term[n] directly from n, without using other terms
Algorithm 1 is called first, and stores the terms in an array.
Then the program calls Algorithms 2 and 3, and checks that they agree
with Algorithm 1.
-------------------------------------------------------------------------------}

 
uses SysUtils;
 
type TRational = record
Num, Den : integer;
end;
 
var
terms : array of TRational;
max_index, k : integer;
 
// Routine to calculate array of terms up the the maiximum index
procedure CalcTerms_algo_1();
var
j, k : integer;
begin
SetLength( terms, max_index + 1);
j := 1; // index to earlier term, from which current term is calculated
k := 1; // index to current term
terms[1].Num := 1;
terms[1].Den := 1;
while (k < max_index) do begin
inc(k);
if (k and 1) = 0 then begin // or could write "if not Odd(k)"
terms[k].Num := terms[j].Num;
terms[k].Den := terms[j].Num + terms[j].Den;
end
else begin
terms[k].Num := terms[j].Num + terms[j].Den;
terms[k].Den := terms[j].Den;
inc(j);
end;
end;
end;
 
// Method to get each term from the preceding term.
// a/b --> b/(a + b - 2(a mod b));
function CheckTerms_algo_2() : boolean;
var
index, a, b, temp : integer;
begin
result := true;
index := 1;
a := 1;
b := 1;
while (index <= max_index) do begin
if (a <> terms[index].Num) or (b <> terms[index].Den) then
result := false;
temp := a + b - 2*(a mod b);
a := b;
b := temp;
inc( index)
end;
end;
 
// Mathod to calcualte each term from its index, without using other terms.
function CheckTerms_algo_3() : boolean;
var
index, a, b, pwr2, idiv2 : integer;
begin
result := true;
for index := 1 to max_index do begin
 
idiv2 := index div 2;
pwr2 := 1;
while (pwr2 <= idiv2) do pwr2 := pwr2 shl 1;
a := 1;
b := 1;
while (pwr2 > 1) do begin
pwr2 := pwr2 shr 1;
if (pwr2 and index) = 0 then
inc( b, a)
else
inc( a, b);
end;
if (a <> terms[index].Num) or (b <> terms[index].Den) then
result := false;
end;
end;
 
begin
// Read and validate maximum index
max_index := SysUtils.StrToIntDef( paramStr(1), -1); // -1 if not an integer
if (max_index <= 0) then begin
WriteLn( 'Maximum index must be a positive integer');
exit;
end;
 
// Calculate terms by algo 1, then check that algos 2 and 3 agree.
CalcTerms_algo_1();
if not CheckTerms_algo_2() then begin
WriteLn( 'Algorithm 2 failed');
exit;
end;
if not CheckTerms_algo_3() then begin
WriteLn( 'Algorithm 3 failed');
exit;
end;
 
// Display the terms
for k := 1 to max_index do
with terms[k] do
WriteLn( SysUtils.Format( '%8d: %d/%d', [k, Num, Den]));
end.
 
Output:
       1: 1/1
       2: 1/2
       3: 2/1
       4: 1/3
       5: 3/2
       6: 2/3
       7: 3/1
       8: 1/4
       9: 4/3
      10: 3/5
      11: 5/2
      12: 2/5
      13: 5/3
      14: 3/4
      15: 4/1
      16: 1/5
      17: 5/4
      18: 4/7
      19: 7/3
      20: 3/8
 
program CWIndex;
 
{-------------------------------------------------------------------------------
FreePascal command-line program.
Calculates index of a rational number in the Calkin-Wilf sequence,
where the first term 1/1 has index 1.
Command line format is
CWIndex <numerator> <denominator>
e.g. for the Rosetta Code example
CWIndex 83116 51639
-------------------------------------------------------------------------------}

 
uses SysUtils;
 
var
num, den : integer;
a, b : integer;
pwr2, index : qword; // 64-bit unsiged
begin
// Read and validate input.
num := SysUtils.StrToIntDef( paramStr(1), -1); // return -1 if not an integer
den := SysUtils.StrToIntDef( paramStr(2), -1);
if (num <= 0) or (den <= 0) then begin
WriteLn( 'Numerator and denominator must be positive integers');
exit;
end;
 
// Input OK, calculate and display index of num/den
// The index may overflow 64 bits, so turn on overflow detection
{$Q+}
a := num;
b := den;
pwr2 := 1;
index := 0;
try
while (a <> b) do begin
if (a < b) then
dec( b, a)
else begin
dec( a, b);
inc( index, pwr2);
end;
pwr2 := 2*pwr2;
end;
inc( index, pwr2);
WriteLn( SysUtils.Format( 'Index of %d/%d is %u', [num, den, index]));
except
WriteLn( 'Index is too large for 64 bits');
end;
end.
 
Output:
Index of 83116/51639 is 123456789

Perl[edit]

Translation of: Raku
Library: ntheory
use strict;
use warnings;
use feature qw(say state);
 
use ntheory 'fromdigits';
use List::Lazy 'lazy_list';
use Math::AnyNum ':overload';
 
my $calkin_wilf = lazy_list { state @cw = 1; push @cw, 1 / ( (2 * int $cw[0]) + 1 - $cw[0] ); shift @cw };
 
sub r2cf {
my($num, $den) = @_;
my($n, @cf);
my $f = sub { return unless $den;
my $q = int($num/$den);
($num, $den) = ($den, $num - $q*$den);
$q;
};
push @cf, $n while defined($n = $f->());
reverse @cf;
}
 
sub r2cw {
my($num, $den) = @_;
my $bits;
my @f = r2cf($num, $den);
$bits .= ($_%2 ? 0 : 1) x $f[$_] for 0..$#f;
fromdigits($bits, 2);
}
 
say 'First twenty terms of the Calkin-Wilf sequence:';
printf "%s ", $calkin_wilf->next() for 1..20;
say "\n\n83116/51639 is at index: " . r2cw(83116,51639);
Output:
First twenty terms of the Calkin-Wilf sequence:
1 1/2 2 1/3 3/2 2/3 3 1/4 4/3 3/5 5/2 2/5 5/3 3/4 4 1/5 5/4 4/7 7/3 3/8

83116/51639 is at index: 123456789

Phix[edit]

with javascript_semantics
requires("1.0.0")   -- (new even() builtin)

function calkin_wilf(integer len)
    sequence cw = repeat(0,len)
    integer n=0, d=1
    for i=1 to len do
        {n,d} = {d,(floor(n/d)*2+1)*d-n}
        cw[i] = {n,d}
    end for
    return cw
end function

function odd_length(sequence cf)
    -- replace even length continued fraction with odd length equivalent
--  if remainder(length(cf),2)=0 then
    if even(length(cf)) then
        cf[$] -= 1
        cf &= 1
    end if
    return cf
end function
 
function to_continued_fraction(sequence r)
    integer {a,b} = r
    sequence cf = {}
    while true do
        cf &= floor(a/b)
        {a, b} = {b, remainder(a,b)}
        if a=1 then exit end if
    end while
    cf = odd_length(cf)
    return cf
end function
 
function get_term_number(sequence cf)
    sequence b = {}
    integer d = 1
    for i=1 to length(cf) do
        b &= repeat(d,cf[i])
        d = 1-d
    end for
    integer t = bits_to_int(b)
    return t
end function
 
-- additional verification methods (2 of)
function i_to_cf(integer i)
--  sequence b = trim_tail(int_to_bits(i,32),0)&2,
    sequence b = int_to_bits(i)&2,
             cf = iff(b[1]=0?{0}:{})
    while length(b)>1 do
        for j=2 to length(b) do
            if b[j]!=b[1] then
                cf &= j-1
                b = b[j..$]
                exit
            end if
        end for
    end while
    cf = odd_length(cf)
    return cf
end function
 
function cf2r(sequence cf)
    integer n=0, d=1
    for i=length(cf) to 2 by -1 do
        {n,d} = {d,n+d*cf[i]}
    end for
    return {n+cf[1]*d,d}
end function
 
function prettyr(sequence r)
    integer {n,d} = r
    return iff(d=1?sprintf("%d",n):sprintf("%d/%d",{n,d}))
end function
 
sequence cw = calkin_wilf(20)
printf(1,"The first 20 terms of the Calkin-Wilf sequence are:\n")
for i=1 to 20 do
    string s = prettyr(cw[i]),
           r = prettyr(cf2r(i_to_cf(i)))
    integer t = get_term_number(to_continued_fraction(cw[i]))
    printf(1,"%2d: %-4s [==> %2d: %-3s]\n", {i, s, t, r})
end for
printf(1,"\n")
sequence r = {83116,51639}
sequence cf = to_continued_fraction(r)
integer tn = get_term_number(cf)
printf(1,"%d/%d is the %,d%s term of the sequence.\n", r&{tn,ord(tn)})
Output:
The first 20 terms of the Calkin-Wilf sequence are:
 1: 1    [==>  1: 1  ]
 2: 1/2  [==>  2: 1/2]
 3: 2    [==>  3: 2  ]
 4: 1/3  [==>  4: 1/3]
 5: 3/2  [==>  5: 3/2]
 6: 2/3  [==>  6: 2/3]
 7: 3    [==>  7: 3  ]
 8: 1/4  [==>  8: 1/4]
 9: 4/3  [==>  9: 4/3]
10: 3/5  [==> 10: 3/5]
11: 5/2  [==> 11: 5/2]
12: 2/5  [==> 12: 2/5]
13: 5/3  [==> 13: 5/3]
14: 3/4  [==> 14: 3/4]
15: 4    [==> 15: 4  ]
16: 1/5  [==> 16: 1/5]
17: 5/4  [==> 17: 5/4]
18: 4/7  [==> 18: 4/7]
19: 7/3  [==> 19: 7/3]
20: 3/8  [==> 20: 3/8]

83116/51639 is the 123,456,789th term of the sequence.

Prolog[edit]

 
% John Devou: 26-Nov-2021
 
% g(N,X):- consecutively generate in X the first N elements of the Calkin-Wilf sequence
 
g(N,[A/B|_]-_,A/B):- N > 0.
g(N,[A/B|Ls]-[A/C,C/B|Ys],X):- N > 1, M is N-1, C is A+B, g(M,Ls-Ys,X).
g(N,X):- g(N,[1/1|Ls]-Ls,X).
 
% t(A/B,X):- generate in X the index of A/B in the Calkin-Wilf sequence
 
t(A/1,S,C,X):- X is C*(2**(A-1+S)-S).
t(A/B,S,C,X):- B > 1, divmod(A,B,M,N), T is 1-S, D is C*2**M, t(B/N,T,D,Y), X is Y + S*C*(2**M-1).
t(A/B,X):- t(A/B,1,1,X), !.
 
Output:
?- findall(X, g(20,X), L), write(L).
[1/1,1/2,2/1,1/3,3/2,2/3,3/1,1/4,4/3,3/5,5/2,2/5,5/3,3/4,4/1,1/5,5/4,4/7,7/3,3/8]
L = [1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, ... / ...|...].

?- t(83116/51639,X).
X = 123456789.

Python[edit]

from fractions import Fraction
from math import floor
from itertools import islice, groupby
 
 
def cw():
a = Fraction(1)
while True:
yield a
a = 1 / (2 * floor(a) + 1 - a)
 
def r2cf(rational):
num, den = rational.numerator, rational.denominator
while den:
num, (digit, den) = den, divmod(num, den)
yield digit
 
def get_term_num(rational):
ans, dig, pwr = 0, 1, 0
for n in r2cf(rational):
for _ in range(n):
ans |= dig << pwr
pwr += 1
dig ^= 1
return ans
 
 
if __name__ == '__main__':
print('TERMS 1..20: ', ', '.join(str(x) for x in islice(cw(), 20)))
x = Fraction(83116, 51639)
print(f"\n{x} is the {get_term_num(x):_}'th term.")
Output:
TERMS 1..20:  1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4, 1/5, 5/4, 4/7, 7/3, 3/8

83116/51639 is the 123_456_789'th term.

Raku[edit]

In Raku, arrays are indexed from 0. The Calkin-Wilf sequence does not have a term defined at 0.

This implementation includes a bogus undefined value at position 0, having the bogus first term shifts the indices up by one, making the ordinal position and index match. Useful due to how reversibility function works.

my @calkin-wilf = Any, 1, {1 / (.Int × 2 + 1 - $_)}*;
 
# Rational to Calkin-Wilf index
sub r2cw (Rat $rat) { :2( join '', flat (flat (1,0) xx *) Zxx reverse r2cf $rat ) }
 
# The task
 
say "First twenty terms of the Calkin-Wilf sequence: ",
@calkin-wilf[1..20]».&prat.join: ', ';
 
say "\n99991st through 100000th: ",
(my @tests = @calkin-wilf[99_991 .. 100_000])».&prat.join: ', ';
 
say "\nCheck reversibility: ", @tests».Rat».&r2cw.join: ', ';
 
say "\n83116/51639 is at index: ", r2cw 83116/51639;
 
 
# Helper subs
sub r2cf (Rat $rat is copy) { # Rational to continued fraction
gather loop {
$rat -= take $rat.floor;
last if !$rat;
$rat = 1 / $rat;
}
}
 
sub prat ($num) { # pretty Rat
return $num unless $num ~~ Rat|FatRat;
return $num.numerator if $num.denominator == 1;
$num.nude.join: '/';
}
Output:
First twenty terms of the Calkin-Wilf sequence: 1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4, 1/5, 5/4, 4/7, 7/3, 3/8

99991st through 100000th: 1085/303, 303/1036, 1036/733, 733/1163, 1163/430, 430/987, 987/557, 557/684, 684/127, 127/713

Check reversibility: 99991, 99992, 99993, 99994, 99995, 99996, 99997, 99998, 99999, 100000

83116/51639 is at index: 123456789


REXX[edit]

The meat of this REXX program was provided by Paul Kislanko.

/*REXX pgm finds the Nth value of the  Calkin─Wilf  sequence (which will be a fraction),*/
/*────────────────────── or finds which sequence number contains a specified fraction). */
numeric digits 2000 /*be able to handle ginormic integers. */
parse arg LO HI te . /*obtain optional arguments from the CL*/
if LO=='' | LO=="," then LO= 1 /*Not specified? Then use the default.*/
if HI=='' | HI=="," then HI= 20 /* " " " " " " */
if te=='' | te=="," then te= '/' /* " " " " " " */
if datatype(LO, 'W') then call CW_terms /*Is LO numeric? Then show some terms.*/
if pos('/', te)>0 then call CW_frac te /*Does TE have a / ? Then find term #*/
exit 0
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
th: parse arg th; return word('th st nd rd', 1+(th//10) *(th//100%10\==1) *(th//10<4))
/*──────────────────────────────────────────────────────────────────────────────────────*/
CW_frac: procedure; parse arg p '/' q .; say
if q=='' then do; p= 83116; q= 51639; end
n= rle2dec( frac2cf(p q) ); @CWS= 'the Calkin─Wilf sequence'
say 'for ' p"/"q', the element number for' @CWS "is: " commas(n)th(n)
if length(n)<10 then return
say; say 'The above number has ' commas(length(n)) " decimal digits."
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
CW_term: procedure; parse arg z; dd= 1; nn= 0
do z
parse value dd dd*(2*(nn%dd)+1)-nn with nn dd
end /*z*/
return nn'/'dd
/*──────────────────────────────────────────────────────────────────────────────────────*/
CW_terms: $=; if LO\==0 then do j=LO to HI; $= $ CW_term(j)','
end /*j*/
if $=='' then return
say 'Calkin─Wilf sequence terms for ' commas(LO) " ──► " commas(HI) ' are:'
say strip( strip($), 'T', ",")
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
frac2cf: procedure; parse arg p q; if q=='' then return p; cf= p % q; m= q
p= p - cf*q; n= p; if p==0 then return cf
do k=1 until n==0; @.k= m % n
m= m - @.k * n; parse value n m with m n /*swap N M*/
end /*k*/
/*for inverse Calkin─Wilf, K must be even.*/
if k//2 then do; @.k= @.k - 1; k= k + 1; @.k= 1; end
do k=1 for k; cf= cf @.k; end /*k*/
return cf
/*──────────────────────────────────────────────────────────────────────────────────────*/
rle2dec: procedure; parse arg f1 rle; obin= copies(1, f1)
do until rle==''; parse var rle f0 f1 rle
obin= copies(1, f1)copies(0, f0)obin
end /*until*/
return x2d( b2x(obin) ) /*RLE2DEC: Run Length Encoding ──► decimal*/
output   when using the default inputs:
Calkin─Wilf sequence terms for  1  ──►  20  are:
1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4/1, 1/5, 5/4, 4/7, 7/3, 3/8

for  83116/51639,  the element number for the Calkin─Wilf sequence is:  123,456,789th

Ruby[edit]

Translation of: Python
cw = Enumerator.new do |y|
y << a = 1.to_r
loop { y << a = 1/(2*a.floor + 1 - a) }
end
 
def term_num(rat)
num, den, res, pwr, dig = rat.numerator, rat.denominator, 0, 0, 1
while den > 0
num, (digit, den) = den, num.divmod(den)
digit.times do
res |= dig << pwr
pwr += 1
end
dig ^= 1
end
res
end
 
puts cw.take(20).join(", ")
puts term_num (83116/51639r)
 
1/1, 1/2, 2/1, 1/3, 3/2, 2/3, 3/1, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4/1, 1/5, 5/4, 4/7, 7/3, 3/8
123456789

Rust[edit]

// [dependencies]
// num = "0.3"
 
use num::rational::Rational;
 
fn calkin_wilf_next(term: &Rational) -> Rational {
Rational::from_integer(1) / (Rational::from_integer(2) * term.floor() + 1 - term)
}
 
fn continued_fraction(r: &Rational) -> Vec<isize> {
let mut a = *r.numer();
let mut b = *r.denom();
let mut result = Vec::new();
loop {
let (q, r) = num::integer::div_rem(a, b);
result.push(q);
a = b;
b = r;
if a == 1 {
break;
}
}
let len = result.len();
if len != 0 && len % 2 == 0 {
result[len - 1] -= 1;
result.push(1);
}
result
}
 
fn term_number(r: &Rational) -> usize {
let mut result: usize = 0;
let mut d: usize = 1;
let mut p: usize = 0;
for n in continued_fraction(r) {
for _ in 0..n {
result |= d << p;
p += 1;
}
d ^= 1;
}
result
}
 
fn main() {
println!("First 20 terms of the Calkin-Wilf sequence are:");
let mut term = Rational::from_integer(1);
for i in 1..=20 {
println!("{:2}: {}", i, term);
term = calkin_wilf_next(&term);
}
let r = Rational::new(83116, 51639);
println!("{} is the {}th term of the sequence.", r, term_number(&r));
}
Output:
First 20 terms of the Calkin-Wilf sequence are:
 1: 1
 2: 1/2
 3: 2
 4: 1/3
 5: 3/2
 6: 2/3
 7: 3
 8: 1/4
 9: 4/3
10: 3/5
11: 5/2
12: 2/5
13: 5/3
14: 3/4
15: 4
16: 1/5
17: 5/4
18: 4/7
19: 7/3
20: 3/8
83116/51639 is the 123456789th term of the sequence.

Sidef[edit]

func calkin_wilf(n) is cached {
return 1 if (n == 1)
1/(2*floor(__FUNC__(n-1)) + 1 - __FUNC__(n-1))
}
 
func r2cw(r) {
 
var cfrac = r.as_cfrac
cfrac.len.is_odd || return nil
 
Num(cfrac.flip.map_kv {|k,v| (k.is_odd ? '0' : '1') * v }.join, 2)
}
 
with (20) {|n|
say "First #{n} terms of the Calkin-Wilf sequence:"
say calkin_wilf.map(1..n)
}
 
with (83116/51639) {|r|
say ("\n#{r.as_rat} is at index: ", r2cw(r))
}
Output:
First 20 terms of the Calkin-Wilf sequence:
[1, 1/2, 2, 1/3, 3/2, 2/3, 3, 1/4, 4/3, 3/5, 5/2, 2/5, 5/3, 3/4, 4, 1/5, 5/4, 4/7, 7/3, 3/8]

83116/51639 is at index: 123456789

Wren[edit]

Library: Wren-rat
Library: Wren-fmt
import "/rat" for Rat
import "/fmt" for Fmt, Conv
 
var calkinWilf = Fn.new { |n|
var cw = List.filled(n, null)
cw[0] = Rat.one
for (i in 1...n) {
var t = cw[i-1].floor * 2 - cw[i-1] + 1
cw[i] = Rat.one / t
}
return cw
}
 
var toContinued = Fn.new { |r|
var a = r.num
var b = r.den
var res = []
while (true) {
res.add((a/b).floor)
var t = a % b
a = b
b = t
if (a == 1) break
}
if (res.count%2 == 0) { // ensure always odd
res[-1] = res[-1] - 1
res.add(1)
}
return res
}
 
var getTermNumber = Fn.new { |cf|
var b = ""
var d = "1"
for (n in cf) {
b = (d * n) + b
d = (d == "1") ? "0" : "1"
}
return Conv.atoi(b, 2)
}
 
var cw = calkinWilf.call(20)
System.print("The first 20 terms of the Calkin-Wilf sequence are:")
Rat.showAsInt = true
for (i in 1..20) Fmt.print("$2d: $s", i, cw[i-1])
System.print()
var r = Rat.new(83116, 51639)
var cf = toContinued.call(r)
var tn = getTermNumber.call(cf)
Fmt.print("$s is the $,r term of the sequence.", r, tn)
Output:
The first 20 terms of the Calkin-Wilf sequence are:
 1: 1
 2: 1/2
 3: 2
 4: 1/3
 5: 3/2
 6: 2/3
 7: 3
 8: 1/4
 9: 4/3
10: 3/5
11: 5/2
12: 2/5
13: 5/3
14: 3/4
15: 4
16: 1/5
17: 5/4
18: 4/7
19: 7/3
20: 3/8

83116/51639 is the 123,456,789th term of the sequence.