# Zeckendorf number representation

Zeckendorf number representation
You are encouraged to solve this task according to the task description, using any language you may know.

Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.

Recall that the first six distinct Fibonacci numbers are: 1, 2, 3, 5, 8, 13.

The decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.

10100 is not the only way to make 11 from the Fibonacci numbers however; 0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that no two consecutive Fibonacci numbers can be used which leads to the former unique solution.

Generate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.

The intention in this task to find the Zeckendorf form of an arbitrary integer. The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task.

Also see

## 360 Assembly

Translation of: BBC BASIC
*        Zeckendorf number representation  04/04/2017ZECKEN   CSECT         USING  ZECKEN,R13         base register         B      72(R15)            skip savearea         DC     17F'0'             savearea         STM    R14,R12,12(R13)    save previous context         ST     R13,4(R15)         link backward         ST     R15,8(R13)         link forward         LR     R13,R15            set addressability         LA     R6,0               i=0       DO WHILE=(C,R6,LE,=A(20))   do i=0 to 20         MVC    PG,=CL80'xx : '      init buffer         LA     R10,PG               pgi=0         XDECO  R6,XDEC              i         MVC    0(2,R10),XDEC+10     output i         LA     R10,5(R10)           pgi+=5         MVC    FIB,=A(1)            fib(1)=1         MVC    FIB+4,=A(2)          fib(2)=2         LA     R7,2                 j=2         LR     R1,R7                j         SLA    R1,2                 @fib(j)       DO WHILE=(C,R6,GT,FIB-4(R1)   do while fib(j)<i         LA     R7,1(R7)               j++         LR     R1,R7                  j         SLA    R1,2                   ~         L      R2,FIB-8(R1)           fib(j-1)         A      R2,FIB-12(R1)          fib(j-2)         ST     R2,FIB-4(R1)           fib(j)=fib(j-1)+fib(j-2)         LR     R1,R7                  j         SLA    R1,2                   @fib(j)       ENDDO    ,                    enddo j         LR     R8,R6                k=i         MVI    BB,X'00'             bb=false       DO WHILE=(C,R7,GE,=A(1))      do j=j to 1 by -1         LR     R1,R7                  j         SLA    R1,2                   ~       IF C,R8,GE,FIB-4(R1) THEN       if fib(j)<=k then         MVI    BB,X'01'                 bb=true         MVC    0(1,R10),=C'1'           output '1'         LA     R10,1(R10)               pgi+=1         LR     R1,R7                    j         SLA    R1,2                     ~         S      R8,FIB-4(R1)             k=k-fib(j)       ELSE     ,                      else       IF CLI,BB,EQ,X'01' THEN           if bb then         MVC    0(1,R10),=C'0'             output '0'         LA     R10,1(R10)                 pgi+=1       ENDIF    ,                        endif       ENDIF    ,                      endif         BCTR   R7,0                   j--       ENDDO    ,                    enddo j       IF CLI,BB,NE,X'01' THEN       if not bb then         MVC    0(1,R10),=C'0'         output '0'       ENDIF    ,                    endif         XPRNT  PG,L'PG              print buffer         LA     R6,1(R6)             i++       ENDDO    ,                  enddo i         L      R13,4(0,R13)       restore previous savearea pointer         LM     R14,R12,12(R13)    restore previous context         XR     R15,R15            rc=0         BR     R14                exitFIB      DS     32F                Fibonnacci tableBB       DS     X                  flagPG       DS     CL80               bufferXDEC     DS     CL12               temp         YREGS         END    ZECKEN
Output:
 0 : 0
1 : 1
2 : 10
3 : 100
4 : 101
5 : 1000
6 : 1001
7 : 1010
8 : 10000
9 : 10001
10 : 10010
11 : 10100
12 : 10101
13 : 100000
14 : 100001
15 : 100010
16 : 100100
17 : 100101
18 : 101000
19 : 101001
20 : 101010


with Ada.Text_IO, Ada.Strings.Unbounded; procedure Print_Zeck is    function Zeck_Increment(Z: String) return String is   begin      if Z="" then 	 return "1";      elsif Z(Z'Last) = '1' then	 return Zeck_Increment(Z(Z'First .. Z'Last-1)) & '0';      elsif Z(Z'Last-1) = '0' then	 return Z(Z'First .. Z'Last-1) & '1';      else -- Z has at least two digits and ends with "10"	 return Zeck_Increment(Z(Z'First .. Z'Last-2)) & "00";      end if;   end Zeck_Increment;    use Ada.Strings.Unbounded;   Current: Unbounded_String := Null_Unbounded_String; begin   for I in 1 .. 20 loop      Current := To_Unbounded_String(Zeck_Increment(To_String(Current)));      Ada.Text_IO.Put(To_String(Current) & " ");   end loop;   end Print_Zeck;
Output:
1 10 100 101 1000 1001 1010 10000 10001 10010 10100 10101 100000 100001 100010 100100 100101 101000 101001 101010

## ALGOL 68

# print some Zeckendorf number representations                             # # We handle 32-bit numbers, the maximum fibonacci number that can fit in a ## 32 bit number is F(45)                                                   # # build a table of 32-bit fibonacci numbers                                #[ 45 ]INT fibonacci;fibonacci[ 1 ] := 1;fibonacci[ 2 ] := 2;FOR i FROM 3 TO UPB fibonacci DO fibonacci[ i ] := fibonacci[ i - 1 ] + fibonacci[ i - 2 ] OD; # returns the Zeckendorf representation of n or "?" if one cannot be found #PROC to zeckendorf = ( INT n )STRING:     IF n = 0 THEN        "0"     ELSE        STRING result := "";        INT    f pos  := UPB fibonacci;        INT    rest   := ABS n;        # find the first non-zero Zeckendorf digit                        #        WHILE f pos > LWB fibonacci AND rest < fibonacci[ f pos ] DO            f pos -:= 1        OD;        # if we found a digit, build the representation                   #        IF f pos >= LWB fibonacci THEN            # have a digit                                                #            BOOL skip digit := FALSE;            WHILE f pos >= LWB fibonacci DO                IF   rest <= 0 THEN                    result    +:= "0"                ELIF skip digit THEN                    # we used the previous digit                          #                    skip digit := FALSE;                    result    +:= "0"                ELIF rest < fibonacci[ f pos ] THEN                    # can't use the digit at f pos                        #                    skip digit := FALSE;                    result    +:= "0"                ELSE                    # can use this digit                                  #                    skip digit := TRUE;                    result    +:= "1";                    rest      -:= fibonacci[ f pos ]                FI;                f pos -:= 1            OD        FI;        IF rest = 0 THEN            # found a representation                                      #            result        ELSE            # can't find a representation                                 #            "?"        FI     FI; # to zeckendorf # FOR i FROM 0 TO 20 DO    print( ( whole( i, -3 ), " ", to zeckendorf( i ), newline ) )OD
Output:
  0 0
1 1
2 10
3 100
4 101
5 1000
6 1001
7 1010
8 10000
9 10001
10 10010
11 10100
12 10101
13 100000
14 100001
15 100010
16 100100
17 100101
18 101000
19 101001
20 101010

## AppleScript

Translation of: JavaScript
(mapAccumuL example)
-- ZECKENDORF NUMBERS -------------------------------------------------------- -- zeckendorf :: Int -> Stringon zeckendorf(n)    script f        on |λ|(n, x)            if n < x then                [n, 0]            else                [n - x, 1]            end if        end |λ|    end script     if n = 0 then        {0} as string    else        item 2 of mapAccumL(f, n, |reverse|(just of tailMay(fibUntil(n)))) as string    end ifend zeckendorf -- fibUntil :: Int -> [Int]on fibUntil(n)    set xs to {}    set limit to n     script atLimit        property ceiling : limit        on |λ|(x)            (item 2 of x) > (atLimit's ceiling)        end |λ|    end script     script nextPair        property series : xs        on |λ|([a, b])            set nextPair's series to nextPair's series & b            [b, a + b]        end |λ|    end script     |until|(atLimit, nextPair, {0, 1})    return nextPair's seriesend fibUntil -- TEST ----------------------------------------------------------------------on run     intercalate(linefeed, ¬        map(zeckendorf, enumFromTo(0, 20))) end run -- GENERIC FUNCTIONS --------------------------------------------------------- -- enumFromTo :: Int -> Int -> [Int]on enumFromTo(m, n)    if m > n then        set d to -1    else        set d to 1    end if    set lst to {}    repeat with i from m to n by d        set end of lst to i    end repeat    return lstend enumFromTo -- foldl :: (a -> b -> a) -> a -> [b] -> aon foldl(f, startValue, xs)    tell mReturn(f)        set v to startValue        set lng to length of xs        repeat with i from 1 to lng            set v to |λ|(v, item i of xs, i, xs)        end repeat        return v    end tellend foldl -- 'The mapAccumL function behaves like a combination of map and foldl; -- it applies a function to each element of a list, passing an -- accumulating parameter from left to right, and returning a final -- value of this accumulator together with the new list.' (see Hoogle) -- mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])on mapAccumL(f, acc, xs)    script        on |λ|(a, x)            tell mReturn(f) to set pair to |λ|(item 1 of a, x)            [item 1 of pair, (item 2 of a) & item 2 of pair]        end |λ|    end script     foldl(result, [acc, []], xs)end mapAccumL -- map :: (a -> b) -> [a] -> [b]on map(f, xs)    tell mReturn(f)        set lng to length of xs        set lst to {}        repeat with i from 1 to lng            set end of lst to |λ|(item i of xs, i, xs)        end repeat        return lst    end tellend map -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Scripton mReturn(f)    if class of f is script then        f    else        script            property |λ| : f        end script    end ifend mReturn -- intercalate :: Text -> [Text] -> Texton intercalate(strText, lstText)    set {dlm, my text item delimiters} to {my text item delimiters, strText}    set strJoined to lstText as text    set my text item delimiters to dlm    return strJoinedend intercalate -- reverse :: [a] -> [a]on |reverse|(xs)    if class of xs is text then        (reverse of characters of xs) as text    else        reverse of xs    end ifend |reverse| -- tailMay :: [a] -> Maybe [a]on tailMay(xs)    if length of xs > 1 then        {nothing:false, just:items 2 thru -1 of xs}    else        {nothing:true}    end ifend tailMay -- until :: (a -> Bool) -> (a -> a) -> a -> aon |until|(p, f, x)    set mp to mReturn(p)    set v to x     tell mReturn(f)        repeat until mp's |λ|(v)            set v to |λ|(v)        end repeat    end tell    return vend |until|
Output:
0
1
10
100
101
1000
1001
1010
10000
10001
10010
10100
10101
100000
100001
100010
100100
100101
101000
101001
101010

## AutoHotkey

Works with: AutoHotkey_L
Fib := NStepSequence(1, 2, 2, 20)Loop, 21 {	i := A_Index - 1	, Out .= i ":t", n := ""	Loop, % Fib.MaxIndex() {		x := Fib.MaxIndex() + 1 - A_Index		if (Fib[x] <= i)			n .= 1, i -= Fib[x]		else			n .= 0	}	Out .= (n ? LTrim(n, "0") : 0) "n"}MsgBox, % Out NStepSequence(v1, v2, n, k) {    a := [v1, v2]	Loop, % k - 2 {		a[j := A_Index + 2] := 0		Loop, % j < n + 2 ? j - 1 : n			a[j] += a[j - A_Index]	}	return, a}
NStepSequence()

Output:

0:	0
1:	1
2:	10
3:	100
4:	101
5:	1000
6:	1001
7:	1010
8:	10000
9:	10001
10:	10010
11:	10100
12:	10101
13:	100000
14:	100001
15:	100010
16:	100100
17:	100101
18:	101000
19:	101001
20:	101010

## AutoIt

 For $i = 0 To 20 ConsoleWrite($i &": "& Zeckendorf($i)&@CRLF)Next Func Zeckendorf($int, $Fibarray = "") If Not IsArray($Fibarray) Then $Fibarray = Fibonacci($int)	Local $ret = "" For$i = UBound($Fibarray) - 1 To 1 Step -1 If$Fibarray[$i] >$int And $ret = "" Then ContinueLoop ; dont use Leading Zeros If$Fibarray[$i] >$int Then			$ret &= "0" Else If StringRight($ret, 1) <>  "1" Then				$ret &= "1"$int -= $Fibarray[$i]			Else				$ret &= "0" EndIf EndIf Next If$ret = "" Then $ret = "0" Return$retEndFunc   ;==>Zeckendorf Func Fibonacci($max)$AList = ObjCreate("System.Collections.ArrayList")	$AList.add("0")$current = 0	While True		If $current > 1 Then$count = $AList.Count$current = $AList.Item($count - 1)			$current =$current + $AList.Item($count - 2)		Else			$current += 1 EndIf$AList.add($current) If$current > $max Then ExitLoop WEnd$Array = $AList.ToArray Return$ArrayEndFunc   ;==>Fibonacci

Output:

0: 0
1: 1
2: 10
3: 100
4: 101
5: 1000
6: 1001
7: 1010
8: 10000
9: 10001
10: 10010
11: 10100
12: 10101
13: 100000
14: 100001
15: 100010
16: 100100
17: 100101
18: 101000
19: 101001
20: 101010


## BBC BASIC

      FOR n% = 0 TO 20        PRINT n% RIGHT$(" " + FNzeckendorf(n%), 8) NEXT PRINT '"Checking numbers up to 10000..." FOR n% = 21 TO 10000 IF INSTR(FNzeckendorf(n%), "11") STOP NEXT PRINT "No Zeckendorf numbers contain consecutive 1's" END DEF FNzeckendorf(n%) LOCAL i%, o$, fib%() : DIM fib%(45)      fib%(0) = 1 : fib%(1) = 1 : i% = 1      REPEAT        i% += 1        fib%(i%) = fib%(i%-1) + fib%(i%-2)      UNTIL fib%(i%) > n%      REPEAT        i% -= 1        IF n% >= fib%(i%) THEN          o$+= "1" n% -= fib%(i%) ELSE o$ += "0"        ENDIF      UNTIL i% = 1      = o$ Output:  0 0 1 1 2 10 3 100 4 101 5 1000 6 1001 7 1010 8 10000 9 10001 10 10010 11 10100 12 10101 13 100000 14 100001 15 100010 16 100100 17 100101 18 101000 19 101001 20 101010 Checking numbers up to 10000... No Zeckendorf numbers contain consecutive 1's  ## Befunge The first number on the stack, 45*, specifies the range of values to display. However, the algorithm depends on a hardcoded list of Fibonacci values (currently just 10) so the theoretical maximum is 143. It's also constrained by the range of a Befunge data cell, which on many implementations will be as low as 127. 45*83p0>:::.0"0"vv53210p 39+!:,,9+<>858+37 *66g"7Y":v>3g#@_^ v\g39$<^8:+1,+5_5<>-:0\|v:-\g39_^#:<*:p39<>0\:!"0"+#^ ,#_^ Output: 0 0 1 1 2 10 3 100 4 101 5 1000 6 1001 7 1010 8 10000 9 10001 10 10010 11 10100 12 10101 13 100000 14 100001 15 100010 16 100100 17 100101 18 101000 19 101001 20 101010 ## C #include <stdio.h> typedef unsigned long long u64; #define FIB_INVALID (~(u64)0) u64 fib[] = { 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073ULL, 4807526976ULL, 7778742049ULL, 12586269025ULL, 20365011074ULL, 32951280099ULL, 53316291173ULL, 86267571272ULL, 139583862445ULL, 225851433717ULL, 365435296162ULL, 591286729879ULL, 956722026041ULL, 1548008755920ULL, 2504730781961ULL, 4052739537881ULL, 6557470319842ULL, 10610209857723ULL, 17167680177565ULL, 27777890035288ULL // this 65-th one is for range check}; u64 fibbinary(u64 n){ if (n >= fib[64]) return FIB_INVALID; u64 ret = 0; int i; for (i = 64; i--; ) if (n >= fib[i]) { ret |= 1ULL << i; n -= fib[i]; } return ret;} void bprint(u64 n, int width){ if (width > 64) width = 64; u64 b; for (b = 1ULL << (width - 1); b; b >>= 1) putchar(b == 1 && !n ? '0' : b > n ? ' ' : b & n ? '1' : '0'); putchar('\n');} int main(void){ int i; for (i = 0; i <= 20; i++) printf("%2d:", i), bprint(fibbinary(i), 8); return 0;} Output:  0: 0 1: 1 2: 10 3: 100 4: 101 5: 1000 6: 1001 7: 1010 8: 10000 9: 10001 10: 10010 11: 10100 12: 10101 13: 100000 14: 100001 15: 100010 16: 100100 17: 100101 18: 101000 19: 101001 20: 101010  ## C++ ### Using a C++11 User Defined Literal Works with: C++11 see Here for a further example using this class.  // For a class N which implements Zeckendorf numbers:// I define an increment operation ++()// I define a comparison operation <=(other N)// Nigel Galloway October 22nd., 2012#include <iostream>class N {private: int dVal = 0, dLen;public: N(char const* x = "0"){ int i = 0, q = 1; for (; x[i] > 0; i++); for (dLen = --i/2; i >= 0; i--) { dVal+=(x[i]-48)*q; q*=2; }} const N& operator++() { for (int i = 0;;i++) { if (dLen < i) dLen = i; switch ((dVal >> (i*2)) & 3) { case 0: dVal += (1 << (i*2)); return *this; case 1: dVal += (1 << (i*2)); if (((dVal >> ((i+1)*2)) & 1) != 1) return *this; case 2: dVal &= ~(3 << (i*2)); }}} const bool operator<=(const N& other) const {return dVal <= other.dVal;} friend std::ostream& operator<<(std::ostream&, const N&);};N operator "" N(char const* x) {return N(x);}std::ostream &operator<<(std::ostream &os, const N &G) { const static std::string dig[] {"00","01","10"}, dig1[] {"","1","10"}; if (G.dVal == 0) return os << "0"; os << dig1[(G.dVal >> (G.dLen*2)) & 3]; for (int i = G.dLen-1; i >= 0; i--) os << dig[(G.dVal >> (i*2)) & 3]; return os;}  I may now write:  int main(void) {//for (N G; G <= 101010N; ++G) std::cout << G << std::endl; // from zero to 101010M for (N G(101N); G <= 101010N; ++G) std::cout << G << std::endl; // from 101N to 101010N return 0;}  Which produces: Output: 101 1000 1001 1010 10000 10001 10010 10100 10101 100000 100001 100010 100100 100101 101000 101001 101010  ## C#  using System;using System.Collections.Generic;using System.Linq;using System.Text; namespace Zeckendorf{ class Program { private static uint Fibonacci(uint n) { if (n < 2) { return n; } else { return Fibonacci(n - 1) + Fibonacci(n - 2); } } private static string Zeckendorf(uint num) { IList<uint> fibonacciNumbers = new List<uint>(); uint fibPosition = 2; uint currentFibonaciNum = Fibonacci(fibPosition); do { fibonacciNumbers.Add(currentFibonaciNum); currentFibonaciNum = Fibonacci(++fibPosition); } while (currentFibonaciNum <= num); uint temp = num; StringBuilder output = new StringBuilder(); foreach (uint item in fibonacciNumbers.Reverse()) { if (item <= temp) { output.Append("1"); temp -= item; } else { output.Append("0"); } } return output.ToString(); } static void Main(string[] args) { for (uint i = 1; i <= 20; i++) { string zeckendorfRepresentation = Zeckendorf(i); Console.WriteLine(string.Format("{0} : {1}", i, zeckendorfRepresentation)); } Console.ReadKey(); } }}  Output: 1 : 1 2 : 10 3 : 100 4 : 101 5 : 1000 6 : 1001 7 : 1010 8 : 10000 9 : 10001 10 : 10010 11 : 10100 12 : 10101 13 : 100000 14 : 100001 15 : 100010 16 : 100100 17 : 100101 18 : 101000 19 : 101001 20 : 101010  ## Clojure (def fibs (lazy-cat [1 1] (map + fibs (rest fibs)))) (defn z [n] (if (zero? n) "0" (let [ps (->> fibs (take-while #(<= % n)) rest reverse) fz (fn [[s n] p] (if (>= n p) [(conj s 1) (- n p)] [(conj s 0) n]))] (->> ps (reduce fz [[] n]) first (apply str))))) (doseq [n (range 0 21)] (println n (z n))) ## Common Lisp Common Lisp's arbitrary precision integers should handle any positive input: (defun zeckendorf (n) "returns zeckendorf integer of n (see OEIS A003714)" (let ((fib '(2 1))) ;; extend Fibonacci sequence long enough (loop while (<= (car fib) n) do (push (+ (car fib) (cadr fib)) fib)) (loop with r = 0 for f in fib do (setf r (* 2 r)) (when (>= n f) (setf n (- n f)) (incf r)) finally (return r)))) ;;; task requirement(loop for i from 0 to 20 do (format t "~2D: ~2R~%" i (zeckendorf i)))  ;; Print Zeckendorf numbers upto 20.;; I have implemented this as a state machine.;; Nigel Galloway - October 13th., 2012;;(let ((fibz '(13 8 5 3 2 1))) (dotimes (G 21) (progn (format t "~S is " G) (let ((z 0) (ng G)) (dolist (N fibz) (if (> z 1) (progn (setq z 1) (format t "~S" 0)) (if (>= ng N) (progn (setq z 2) (setq ng (- ng N)) (format t "~S" 1)) (if (= z 1) (format t "~S" 0))))) (if (= z 0) (format t "~S~%" 0) (format t "~%"))))))  Output:  0 is 0 1 is 1 2 is 10 3 is 100 4 is 101 5 is 1000 6 is 1001 7 is 1010 8 is 10000 9 is 10001 10 is 10010 11 is 10100 12 is 10101 13 is 100000 14 is 100001 15 is 100010 16 is 100100 17 is 100101 18 is 101000 19 is 101001 20 is 101010  ## D Translation of: Haskell import std.stdio, std.range, std.algorithm, std.functional; void main() { size_t .max .iota .filter!q{ !(a & (a >> 1)) } .take(21) .binaryReverseArgs!writefln("%(%b\n%)");} Output: 0 1 10 100 101 1000 1001 1010 10000 10001 10010 10100 10101 100000 100001 100010 100100 100101 101000 101001 101010 Translation of: Go import std.stdio, std.typecons; int zeckendorf(in int n) pure nothrow { Tuple!(int,"remaining", int,"set") zr(in int fib0, in int fib1, in int n, in uint bit) pure nothrow { if (fib1 > n) return typeof(return)(n, 0); auto rs = zr(fib1, fib0 + fib1, n, bit + 1); if (fib1 <= rs.remaining) { rs.set |= 1 << bit; rs.remaining -= fib1; } return rs; } return zr(1, 1, n, 0)[1];} void main() { foreach (i; 0 .. 21) writefln("%2d: %6b", i, zeckendorf(i));} Output:  0: 0 1: 1 2: 10 3: 100 4: 101 5: 1000 6: 1001 7: 1010 8: 10000 9: 10001 10: 10010 11: 10100 12: 10101 13: 100000 14: 100001 15: 100010 16: 100100 17: 100101 18: 101000 19: 101001 20: 101010 Translation of: Tcl (Same output.) import std.stdio, std.algorithm, std.range; string zeckendorf(size_t n) { if (n == 0) return "0"; auto fibs = recurrence!q{a[n - 1] + a[n - 2]}(1, 2); string result; foreach_reverse (immutable f; fibs.until!(x => x > n).array) { result ~= (f <= n) ? '1' : '0'; if (f <= n) n -= f; } return result;} void main() { foreach (immutable i; 0 .. 21) writefln("%2d: %6s", i, i.zeckendorf);} ## EchoLisp We analytically find the first fibonacci(i) >= n, using the formula i = log((n* Φ) + 0.5) / log(Φ) .  ;; special fib's starting with 1 2 3 5 ...(define (fibonacci n) (+ (fibonacci (1- n)) (fibonacci (- n 2))))(remember 'fibonacci #(1 2)) (define-constant Φ (// (1+ (sqrt 5)) 2))(define-constant logΦ (log Φ));; find i : fib(i) >= n(define (iFib n) (floor (// (log (+ (* n Φ) 0.5)) logΦ))) ;; left trim zeroes(string-delimiter "")(define (zeck->string digits) (if (!= 0 (first digits)) (string-join digits "") (zeck->string (rest digits)))) (define (Zeck n) (cond (( < n 0) "no negative zeck") ((inexact? n) "no floating zeck") ((zero? n) "0") (else (zeck->string (for/list ((s (reverse (take fibonacci (iFib n))))) (if ( > s n) 0 (begin (-= n s) 1 )))))))  Output: (take Zeck 21) → (0 1 10 100 101 1000 1001 1010 10000 10001 10010 10100 10101 100000 100001 100010 100100 100101 101000 101001 101010) (Zeck 1000000000) → 1010000100100001010101000001000101000101001  ## Elena Translation of: C# ELENA 3.4 : import system'routines.import system'collections.import system'text.import extensions. extension op{ fibonacci [ if (self < 2) [ ^ self ]; [ ^ (self - 1) fibonacci + (self - 2) fibonacci. ]. ] zeckendorf [ var fibonacciNumbers := List<IntNumber>(10). int num := self. int fibPosition := 2. int currentFibonaciNum := fibPosition fibonacci. while (currentFibonaciNum <= num) [ fibonacciNumbers append:currentFibonaciNum. fibPosition := fibPosition + 1. currentFibonaciNum := fibPosition fibonacci. ]. auto output := TextBuilder new. int temp := num. fibonacciNumbers sequenceReverse; forEach(:item) [ if (item <= temp) [ output << "1". temp := temp - item. ]; [ output << "0". ]. ]. ^ output literal. ]} public program[ 1 to:20 do(:i) [ var zeckendorfRepresentation := i zeckendorf. console printFormatted("{0} : {1}",i,i zeckendorf); writeLine. ]. console readChar.] Output: 1 : 1 2 : 10 3 : 100 4 : 101 5 : 1000 6 : 1001 7 : 1010 8 : 10000 9 : 10001 10 : 10010 11 : 10100 12 : 10101 13 : 100000 14 : 100001 15 : 100010 16 : 100100 17 : 100101 18 : 101000 19 : 101001 20 : 101010 ## Elixir Translation of: Ruby Stream generator: defmodule Zeckendorf do def number do Stream.unfold(0, fn n -> zn_loop(n) end) end defp zn_loop(n) do bin = Integer.to_string(n, 2) if String.match?(bin, ~r/11/), do: zn_loop(n+1), else: {bin, n+1} endend Zeckendorf.number |> Enum.take(21) |> Enum.with_index|> Enum.each(fn {zn, i} -> IO.puts "#{i}: #{zn}" end) Output: 0: 0 1: 1 2: 10 3: 100 4: 101 5: 1000 6: 1001 7: 1010 8: 10000 9: 10001 10: 10010 11: 10100 12: 10101 13: 100000 14: 100001 15: 100010 16: 100100 17: 100101 18: 101000 19: 101001 20: 101010  Fibonacci numbers: defmodule Zeckendorf do def number(n) do fib_loop(n, [2,1]) |> Enum.reduce({"",n}, fn f,{dig,i} -> if f <= i, do: {dig<>"1", i-f}, else: {dig<>"0", i} end) |> elem(0) |> String.to_integer end defp fib_loop(n, fib) when n < hd(fib), do: fib defp fib_loop(n, [a,b|_]=fib), do: fib_loop(n, [a+b | fib])end for i <- 0..20, do: IO.puts "#{i}: #{Zeckendorf.number(i)}" same output ## F# let fib = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (1,2) let zeckendorf n = if n = 0 then ["0"] else let folder k state = let (n, z) = (fst state), (snd state) if n >= k then (n - k, "1" :: z) else (n, "0" :: z) let fb = fib |> Seq.takeWhile (fun i -> i<=n) |> Seq.toList snd (List.foldBack folder fb (n, [])) |> List.rev for i in 0 .. 20 do printfn "%2d: %8s" i (String.concat "" (zeckendorf i)) Output:  0: 0 1: 1 2: 10 3: 100 4: 101 5: 1000 6: 1001 7: 1010 8: 10000 9: 10001 10: 10010 11: 10100 12: 10101 13: 100000 14: 100001 15: 100010 16: 100100 17: 100101 18: 101000 19: 101001 20: 101010 ## Factor USING: formatting kernel locals make math sequences ; :: fib<= ( n -- seq ) 1 2 [ [ dup n <= ] [ 2dup + [ , ] 2dip ] while drop , ] { } make ; :: zeck ( n -- str ) 0 :> s! n fib<= <reversed> [ dup s + n <= [ s + s! 49 ] [ drop 48 ] if ] "" map-as ; 21 <iota> [ dup zeck "%2d: %6s\n" printf ] each Output:  0: 0 1: 1 2: 10 3: 100 4: 101 5: 1000 6: 1001 7: 1010 8: 10000 9: 10001 10: 10010 11: 10100 12: 10101 13: 100000 14: 100001 15: 100010 16: 100100 17: 100101 18: 101000 19: 101001 20: 101010  ## Forth : fib<= ( n -- n ) >r 0 1 BEGIN dup [email protected] <= WHILE tuck + REPEAT drop rdrop ; : z. ( n -- ) dup fib<= dup . - BEGIN ?dup WHILE dup fib<= dup [char] + emit space . - REPEAT ; : tab 9 emit ; : zeckendorf ( -- ) 21 0 DO cr i 2 .r tab i z. LOOP ; Output: zeckendorf 0 0 1 1 2 2 3 3 4 3 + 1 5 5 6 5 + 1 7 5 + 2 8 8 9 8 + 1 10 8 + 2 11 8 + 3 12 8 + 3 + 1 13 13 14 13 + 1 15 13 + 2 16 13 + 3 17 13 + 3 + 1 18 13 + 5 19 13 + 5 + 1 20 13 + 5 + 2 ok  ## Fortran The simplest representation of a number in the Zeckendorf manner is as a sequence of digits, such as are used in multi-precision arithmetic, and for this an array of integers will do. Rather excessively, as only two states are required and the default integer style is usually thirty-two bits these days. Some compilers allow the specification of one-byte integers, as in INTEGER*1 D(0:ZLAST) so that would be only an eight-fold excess. One could escalate to fiddling with individual bits within a number (as is done in Extensible_prime_generator#Fortran) and a 16-bit integer would be adequate for the specified tests, but Fortran syntax has not been extended to offer simple methods for manipulating bits such as D(7:7) to obtain the seventh bit of D. Instead one might use special library routines as supplied by F90 such as IBITS(D,7,1) for the same effect, though possibly at a cost in code size and execution time. Less storage may be saved through cramming bits than is consumed by the code needed to extract individual bits. Such values could then be displayed using theB format code. However, the source code would now be littered with the details of bit access rather than the form of the Zeckendorf procedure. An alternative lies in noting that only the sequences 00, 01, and 10 can appear (because 11 is unnecessary; see below), so a base three scheme could be used to represent the three such pairs of bits. But this still contains redundancies. Suppose a 01 value is somewhere in the sequence: then it may be followed only by 00 or 01, and likewise 10 be preceded only by 10 or 00; just two values, not three. Perhaps a still more cunning compaction scheme could be devised to take advantage of these details, or some other scheme concocted. For simplicity, no compaction will be attempted so the states of 0 and 1 will be represented by a simple integer devoted to that bit. The conversion scheme requires the values of the Fibonacci sequence, except not quite: the Fibonacci sequence starts F0 = 0, F1 = 1, F2 = 1, F3 = 2, F4 = 3, etc. but what is wanted is to start with the second 1, so F1 = 1, F2 = 2, F3 = 3, F4 = 5, etc. so this sequence has been named the Fib1nacci sequence to replace conceptual dissonance with lexical dissonance, and similarly with array F1B instead of FIB. Initial investigations show that F45 is the last before a thirty-two bit two's complement integer is overflowed, though systems offering INTEGER*8 could be pushed further. Initialising this table in array F1B could be done via specifying the relevant values (computed separately, even by hand) or by some banal initialisation loop that would be executed on the first invocation of any of the routines requiring those values, a tedious and annoying rigmarole to organise. More interesting are the facilities offered by the PARAMETER statement (introduced with F77), with the further possibility that constants, so defined, would be held safe from accidental change, nor would there be initialising code to execute at run time. Alas, the obvious approach (commented out in the source) using array F1B is rejected by the F90 compiler even though it does allow a value to be defined in terms of other defined values, as is demonstrated by the horde of simple variables following. Despite being a multi-pass compiler, the dependencies will not be unravelled if the statements appear out of order. Fortran does not include a standard pre-processor stage, unlike say pl/i where it is built-in to the language and uses much the same syntax as normal pl/i statements, so loops, IF-tests and so forth are available. By such means, the upper limit of 45 could be determined, the initial values calculated, and the array be defined with initial values, all at compile time. By declaring the horde of simple names to have the PRIVATE attribute, they will not litter the name space of routines invoking the module, but alas, they will still occupy their own storage space. Another possibility would be to use the EQUIVALENCE statement to have them placed within array F1B, but alas, as noted in 15_Puzzle_Game#Fortran, the compiler will not countenance PARAMETER statements for names engaged in such misbehviour. A pity. Still another possibility would be to take advantage of the formula for calculating the values of the Fibonacci series directly (with careful attention to the offsets needed for the Fib1nacci sequence), but this formula is rather intimidating: F(N) = ((1 + SQRT(5))**N - (1 - SQRT(5))**N)/(SQRT(5)*2**N) It can easily be coded as a Fortran function (and would have to be double precision because 32-bit floating-point arithmetic is not accurate enough for integer constants approaching 32 bits), but alas, the compiler does not allow itself to take the risk of invoking a user-written function in a PARAMETER statement, even if the compiler had itself compiled it. For, who knows what it might do? Given the array F1B the conversion from an integer to Zeckendorf digit sequence starts from the high-order end to find the highest F1B value not exceeding the number. There is a formula for this mentioned in the EchoLisp section, but it too is intimidating and the rounding of its result would also require checking. Rather than a linear search, a binary chop could be used, though at the cost of additional code. The location of the high-order digit is recorded on general principles, it being useful in formatting output for example. This requires a "first-time" test within the loop, that could be avoided if the conversion were to be done in two stages. The special feature of the conversion lies in noting that F1B(n + 1) = F1B(n) + F1B(n - 1), the defining feature of the Fibonacci sequence. Thus, when a 1-bit is found (say it is bit n), the next bit down must be a 0 and so the test for it may be skipped, by incrementing L This is because if it were not 0, then the bit above (bit n + 1) would have been turned on in the previous stage instead. Because of this adjustment, the controlling loop cannot be DO L = ZLAST,1,-1 to step down the entries in array F1B because modifications to the index variable of a DO-loop, if not rejected out-of-hand by the compiler, may have no effect on the execution of the loop. This is because the execution of a DO-loop is often controlled by an "iteration count", calculated on entry to the DO-loop, which is thereby unaffected by changes to the index variable, or indeed to the bounds and step size of the loop. Other implementations of a DO-loop will offer other behaviour. There being no equivalent in Fortran of Repeat ... until ... ; (whereby the test is at the end, and there is no initial test), a GO TO appears... The source uses F90 for its MODULE facility, in particular having array F1B available without having to mess about with additional parameters or COMMON statements. This also enables the specification of arrays with a lower bound other than one, which makes it easy to define the digit arrays to have a current length, stored in element zero. This sort of "string" facility is often restricted only to strings of characters, but the notion "string of <type>" is often useful. If in routines declared within a MODULE the size of an array parameter is declared via : there are secret additional parameters defining its size, accessible via special functions such as UBOUND so there is no need for an explicit parameter doing so as would be the case prior to F90. With F90 it is also possible to define a compound data type for the digit sequence, but a simple array seems more flexible. The pleasing name, "MODULE ZECKENDORF ARITHMETIC" causes some odd behaviour, even though Fortran source normally involves spaces having no significance outside text literals.  MODULE ZECKENDORF ARITHMETIC !Using the Fibonacci series, rather than powers of some base. INTEGER ZLAST !The standard 32-bit two's complement integers PARAMETER (ZLAST = 45) !only get so far, just as there's a limit to the highest power. INTEGER F1B(ZLAST) !I want the Fibonacci series, and, starting with its second one.c PARAMETER (F1B = (/1,2, !But alas, the compiler doesn't allowc 3 F1B(1) + F1B(2), !for this sort of carpet-unrollingc 4 F1B(2) + F1B(3), !initialisation sequence. INTEGER,PRIVATE:: F01,F02,F03,F04,F05,F06,F07,F08,F09,F10, !So, not bothering with F00, 1 F11,F12,F13,F14,F15,F16,F17,F18,F19,F20, !Prepare a horde of simple names, 2 F21,F22,F23,F24,F25,F26,F27,F28,F29,F30, !which can be initialised 3 F31,F32,F33,F34,F35,F36,F37,F38,F39,F40, !in a certain way, 4 F41,F42,F43,F44,F45 !without scaring the compiler. PARAMETER (F01 = 1, F02 = 2, F03 = F02 + F01, F04 = F03 + F02, !Thusly. 1 F05=F04+F03,F06=F05+F04,F07=F06+F05,F08=F07+F06,F09=F08+F07, !Typing all this 2 F10=F09+F08,F11=F10+F09,F12=F11+F10,F13=F12+F11,F14=F13+F12, !is an invitation 3 F15=F14+F13,F16=F15+F14,F17=F16+F15,F18=F17+F16,F19=F18+F17, !for mistypes. 4 F20=F19+F18,F21=F20+F19,F22=F21+F20,F23=F22+F21,F24=F23+F22, !So a regular layout 5 F25=F24+F23,F26=F25+F24,F27=F26+F25,F28=F27+F26,F29=F28+F27, !helps a little. 6 F30=F29+F28,F31=F30+F29,F32=F31+F30,F33=F32+F31,F34=F33+F32, !Otherwise, 7 F35=F34+F33,F36=F35+F34,F37=F36+F35,F38=F37+F36,F39=F38+F37, !devise a prog. 8 F40=F39+F38,F41=F40+F39,F42=F41+F40,F43=F42+F41,F44=F43+F42, !to generate these texts... 9 F45=F44+F43) !The next is 2971215073. Too big for 32-bit two's complement integers. PARAMETER (F1B = (/F01,F02,F03,F04,F05,F06,F07,F08,F09,F10, !And now, 1 F11, F12, F13, F14, F15, F16, F17, F18, F19, F20, !Here is the desired 2 F21, F22, F23, F24, F25, F26, F27, F28, F29, F30, !array of constants. 3 F31, F32, F33, F34, F35, F36, F37, F38, F39, F40, !And as such, possibly 4 F41, F42, F43, F44, F45/)) !protected from alteration. CONTAINS !After all that, here we go. SUBROUTINE ZECK(N,D) !Convert N to a "Zeckendorf" digit sequence.Counts upwards from digit one. D(i) ~ F1B(i). D(0) fingers the high-order digit. INTEGER N !The normal number, in the computer's base. INTEGER D(0:) !The digits, to be determined. INTEGER R !The remnant. INTEGER L !A finger, similar to the power of the base. IF (N.LT.0) STOP "ZECK! No negative numbers!" !I'm not thinking about them. R = N !Grab a copy that I can mess with. D = 0 !Scrub the lot in one go. L = ZLAST !As if starting with BASE**MAX, rather than BASE**0. 10 IF (R.GE.F1B(L)) THEN !Has the remnant sufficient for this digit? R = R - F1B(L) !Yes! Remove that amount. IF (D(0).EQ.0) THEN !Is this the first non-zero digit? IF (L.GT.UBOUND(D,DIM=1)) STOP "ZECK! Not enough digits!" !Yes. D(0) = L !Remember the location of the high-order digit. END IF !Two loops instead, to avoid repeated testing? D(L) = 1 !Place the digit, knowing a place awaits. L = L - 1 !Never need a ...11... sequence because F1B(i) + F1B(i+1) = F1B(i+2). END IF !So much for that digit "power". L = L - 1 !Down a digit. IF (L.GT.0 .AND. R.GT.0) GO TO 10 !Are we there yet? IF (N.EQ.0) D(0) = 1 !Zero has one digit. END SUBROUTINE ZECK !That was fun. INTEGER FUNCTION ZECKN(D) !Converts a "Zeckendorf" digit sequence to a number. INTEGER D(0:) !The digits. D(0) fingers the high-order digit. IF (D(0).LE.0) STOP "ZECKN! Empty number!" !General paranoia. IF (D(0).GT.ZLAST) STOP "ZECKN! Oversize number!" !I hate array bound hiccoughs. ZECKN = SUM(D(1:D(0))*F1B(1:D(0))) !This is what positional notation means. IF (ZECKN.LT.0) STOP "ZECKN! Integer overflow!" !Oh for IF OVERFLOW as in First Fortran. END FUNCTION ZECKN !Overflows by a small amount will produce a negative number. END MODULE ZECKENDORF ARITHMETIC !Odd stuff. PROGRAM POKE USE ZECKENDORF ARITHMETIC !Please. INTEGER ZD(0:ZLAST) !A scratchpad. INTEGER I,J,W CHARACTER*1 DIGIT(0:1) !Assistance for the output. PARAMETER (DIGIT = (/"0","1"/), W = 6) !This field width suffices.c WRITE (6,*) F1Bc WRITE (6,*) INT8(F1B(44)) + INT8(F1B(45)) WRITE (6,1) F1B(1:4),ZLAST,ZLAST,F1B(ZLAST),HUGE(I) !Show some provenance. 1 FORMAT ("Converts integers to their Zeckendorf digit string " 1 "using the Fib1nacci sequence (",4(I0,","), 2 " ...) as the equivalent of powers."/ 3 "At most, ",I0," digits because Fib1nacci(",I0,") = ",I0, 4 " and the integer limit is ",I0,".",//," N ZN") !Ends with a heading. DO I = 0,20 !Step through the specified range. CALL ZECK(I,ZD) !Convert I to ZD.c WRITE (6,2) I,ZD(ZD(0):1:-1) !Show digits from high-order to low.c 2 FORMAT (I3,1X,66I1) !Or, WRITE (6,2) I,(ZD(J), J = ZD(0),1,-1) WRITE (6,3) I,(" ",J = ZD(0) + 1,W),DIGIT(ZD(ZD(0):1:-1)) !Right-aligned in field width W. 3 FORMAT (I3,1X,66A1) !The digits appear as characters. IF (I.NE.ZECKN(ZD)) STOP "Huh?" !Should never happen... END DO !On to the next. END Output: shown aligned right for a more regular table. Producing leading spaces or digits required a conversion from a numerical digit to a character digit, so that all the output could use the A format code. Converts integers to their Zeckendorf digit string using the Fib1nacci sequence (1,2,3,5, ...) as the equivalent of powers. At most, 45 digits because Fib1nacci(45) = 1836311903 and the integer limit is 2147483647. N ZN 0 0 1 1 2 10 3 100 4 101 5 1000 6 1001 7 1010 8 10000 9 10001 10 10010 11 10100 12 10101 13 100000 14 100001 15 100010 16 100100 17 100101 18 101000 19 101001 20 101010  ## FreeBASIC ' version 17-10-2016' compile with: fbc -s console #Define max 92 ' max for Fibonacci number Dim Shared As ULongInt fib(max) fib(0) = 1fib(1) = 1 For x As Integer = 2 To max fib(x) = fib(x-1) + fib(x-2)Next Function num2zeck(n As Integer) As String If n < 0 Then Print "Error: no negative numbers allowed" Beep : Sleep 5000,1 : EndEnd If If n < 2 Then Return Str(n) Dim As String zeckendorf For x As Integer = max To 1 Step -1 If fib(x) <= n Then zeckendorf = zeckendorf + "1" n = n - fib(x) Else zeckendorf = zeckendorf + "0" End If Next return LTrim(zeckendorf, "0") ' get rid of leading zeroesEnd Function ' ------=< MAIN >=------ Dim As Integer x, eDim As String zeckendorfPrint "number zeckendorf" For x = 0 To 200000 zeckendorf = num2zeck(x) If x <= 20 Then Print x, zeckendorf ' check for two consecutive Fibonacci numbers If InStr(zeckendorf, "11") <> 0 Then Print " Error: two consecutive Fibonacci numbers "; x, zeckendorf e = e +1 End IfNext PrintIf e = 0 Then Print " No Zeckendorf numbers with two consecutive Fibonacci numbers found"Else Print e; " error(s) found"End If ' empty keyboard bufferWhile Inkey <> "" : WendPrint : Print "hit any key to end program"SleepEnd Output: number zeckendorf 0 0 1 1 2 10 3 100 4 101 5 1000 6 1001 7 1010 8 10000 9 10001 10 10010 11 10100 12 10101 13 100000 14 100001 15 100010 16 100100 17 100101 18 101000 19 101001 20 101010 No Zeckendorf numbers with two consecutive Fibonacci numbers found ## Go package main import "fmt" func main() { for i := 0; i <= 20; i++ { fmt.Printf("%2d %7b\n", i, zeckendorf(i)) }} func zeckendorf(n int) int { // initial arguments of fib0 = 1 and fib1 = 1 will produce // the Fibonacci sequence {1, 2, 3,..} on the stack as successive // values of fib1. _, set := zr(1, 1, n, 0) return set} func zr(fib0, fib1, n int, bit uint) (remaining, set int) { if fib1 > n { return n, 0 } // recurse. // construct sequence on the way in, construct ZR on the way out. remaining, set = zr(fib1, fib0+fib1, n, bit+1) if fib1 <= remaining { set |= 1 << bit remaining -= fib1 } return} Output:  0 0 1 1 2 10 3 100 4 101 5 1000 6 1001 7 1010 8 10000 9 10001 10 10010 11 10100 12 10101 13 100000 14 100001 15 100010 16 100100 17 100101 18 101000 19 101001 20 101010 ## Haskell Using "no consecutive 1s" rule: import Data.Bitsimport Numeric zeckendorf = map b filter ones [0..] where	ones :: Int -> Bool	ones x = 0 == x .&. (x shiftR 1)	b x = showIntAtBase 2 ("01"!!) x "" main = mapM_ putStrLn $take 21 zeckendorf which is the same as zeckendorf = "0":"1":[s++[d] | s <- tail zeckendorf, d <- "01", last s /= '1' || d /= '1'] main = mapM putStrLn$ take 21 zeckendorf

or a different way to generate the sequence:

import Numeric fib = 1 : 1 : zipWith (+) fib (tail fib)pow2 = iterate (2*) 1 zeckendorf = map b z where	z = 0:concat (zipWith f fib pow2)	f x y = map (y+) (take x z)	b x = showIntAtBase 2 ("01"!!) x "" main = mapM_ putStrLn $take 21 zeckendorf Creating a string for an individual number: import Data.List (mapAccumL) fib :: [Int]fib = 1 : 2 : zipWith (+) fib (tail fib) zeckendorf :: Int -> Stringzeckendorf 0 = "0"zeckendorf n = snd$ mapAccumL f n $reverse$ takeWhile (<= n) fib  where    f n x      | n < x = (n, '0')      | otherwise = (n - x, '1') main :: IO ()main = (putStrLn . unlines) $zeckendorf <$> [0 .. 20]
Output:
0
1
10
100
101
1000
1001
1010
10000
10001
10010
10100
10101
100000
100001
100010
100100
100101
101000
101001
101010


## J

Please enjoy our Zeckendorf essay.

 fib=: 3 : 0 " 0 mp=. +/ .* {.{: mp/ mp~^:(I.|.#:y) 2 2$0 1 1 1x) phi=: -:1+%:5 fi =: 3 : 'n - y<fib n=. 0>.(1=y)-~>.(phi^.%:5)+phi^.y' fsum=: 3 : 0 z=. 0$r=. y while. 3<r do.  m=. fib fi r  z=. z,m  r=. r-m end. z,r~(*r)+.0=y) Filter=: (#~)(:6) ' '&~:[email protected]:":@:#:@:#[email protected]:((|. fib 2+i.8) e. fsum)&.>i.3 7┌──────┬──────┬──────┬──────┬──────┬──────┬──────┐│0 │1 │10 │100 │101 │1000 │1001 │├──────┼──────┼──────┼──────┼──────┼──────┼──────┤│1010 │10000 │10001 │10010 │10100 │10101 │100000│├──────┼──────┼──────┼──────┼──────┼──────┼──────┤│100001│100010│100100│100101│101000│101001│101010│└──────┴──────┴──────┴──────┴──────┴──────┴──────┘  Explanation: fsum finds the canonical list of fibonacci terms which sum to its argument. fib finds the nth fibonacci term of the fibonacci sequence. This would be 0 1 1 2 3 5 8 13 21 34 55 89 ... but we ignore the first two values of that sequence for the purpose of this exercise. (|. fib 2+i.8) is 34 21 13 8 5 3 2 1. You can think of an eight bit Zeckendorf number such as 101010 as representing the inner product of its digits with (|. fib 2+i.8). Thus, we can find the relevant Zeckendorf bits by finding which which members of that sequence are in the result of fsum The rest is just formatting. (We convert from binary list to integer and then back to binary list, to eliminate leading zeros from the list. Then we convert to text and remove all the spaces. Since we arranged for each result to be in a box, the boxes will align giving us a somewhat readable presentation. ## Java Code: import java.util.*; class Zeckendorf{ public static String getZeckendorf(int n) { if (n == 0) return "0"; List<Integer> fibNumbers = new ArrayList<Integer>(); fibNumbers.add(1); int nextFib = 2; while (nextFib <= n) { fibNumbers.add(nextFib); nextFib += fibNumbers.get(fibNumbers.size() - 2); } StringBuilder sb = new StringBuilder(); for (int i = fibNumbers.size() - 1; i >= 0; i--) { int fibNumber = fibNumbers.get(i); sb.append((fibNumber <= n) ? "1" : "0"); if (fibNumber <= n) n -= fibNumber; } return sb.toString(); } public static void main(String[] args) { for (int i = 0; i <= 20; i++) System.out.println("Z(" + i + ")=" + getZeckendorf(i)); }} Output: Z(0)=0 Z(1)=1 Z(2)=10 Z(3)=100 Z(4)=101 Z(5)=1000 Z(6)=1001 Z(7)=1010 Z(8)=10000 Z(9)=10001 Z(10)=10010 Z(11)=10100 Z(12)=10101 Z(13)=100000 Z(14)=100001 Z(15)=100010 Z(16)=100100 Z(17)=100101 Z(18)=101000 Z(19)=101001 Z(20)=101010 ### Recursive Implementation Code: import java.util.ArrayList;import java.util.List; public class Zeckendorf { private List<Integer> getFibList(final int maxNum, final int n1, final int n2, final List<Integer> fibs){ if(n2 > maxNum) return fibs; fibs.add(n2); return getFibList(maxNum, n2, n1 + n2, fibs); } public String getZeckendorf(final int num) { if (num <= 0) return "0"; final List<Integer> fibs = getFibList(num, 1, 2, new ArrayList<Integer>(){{ add(1); }}); return getZeckString("", num, fibs.size() - 1, fibs); } private String getZeckString(final String zeck, final int num, final int index, final List<Integer> fibs){ final int curFib = fibs.get(index); final boolean placeZeck = num >= curFib; final String outString = placeZeck ? zeck + "1" : zeck + "0"; final int outNum = placeZeck ? num - curFib : num; if(index == 0) return outString; return getZeckString(outString, outNum, index - 1, fibs); } public static void main(final String[] args) { final Zeckendorf zeckendorf = new Zeckendorf(); for(int i =0; i <= 20; i++){ System.out.println("Z("+ i +"):\t" + zeckendorf.getZeckendorf(i)); } }} Output: Z(0): 0 Z(1): 1 Z(2): 10 Z(3): 100 Z(4): 101 Z(5): 1000 Z(6): 1001 Z(7): 1010 Z(8): 10000 Z(9): 10001 Z(10): 10010 Z(11): 10100 Z(12): 10101 Z(13): 100000 Z(14): 100001 Z(15): 100010 Z(16): 100100 Z(17): 100101 Z(18): 101000 Z(19): 101001 Z(20): 101010 ## JavaScript ### ES6 Translation of: Haskell (mapAccumuL example) (() => { 'use strict'; const main = () => unlines( map(n => concat(zeckendorf(n)), enumFromTo(0, 20) ) ); // zeckendorf :: Int -> String const zeckendorf = n => { const go = (n, x) => n < x ? ( Tuple(n, '0') ) : Tuple(n - x, '1') return 0 < n ? ( snd(mapAccumL( go, n, reverse(fibUntil(n)) )) ) : ['0']; }; // fibUntil :: Int -> [Int] const fibUntil = n => cons(1, takeWhile(x => n >= x, map(snd, iterateUntil( tpl => n <= fst(tpl), tpl => { const x = snd(tpl); return Tuple(x, x + fst(tpl)); }, Tuple(1, 2) )))); // GENERIC FUNCTIONS ---------------------------- // Tuple (,) :: a -> b -> (a, b) const Tuple = (a, b) => ({ type: 'Tuple', '0': a, '1': b, length: 2 }); // concat :: [[a]] -> [a] // concat :: [String] -> String const concat = xs => 0 < xs.length ? (() => { const unit = 'string' !== typeof xs[0] ? ( [] ) : ''; return unit.concat.apply(unit, xs); })() : []; // cons :: a -> [a] -> [a] const cons = (x, xs) => Array.isArray(xs) ? ( [x].concat(xs) ) : (x + xs); // enumFromTo :: Int -> Int -> [Int] const enumFromTo = (m, n) => m <= n ? iterateUntil( x => n <= x, x => 1 + x, m ) : []; // fst :: (a, b) -> a const fst = tpl => tpl[0]; // iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a] const iterateUntil = (p, f, x) => { const vs = [x]; let h = x; while (!p(h))(h = f(h), vs.push(h)); return vs; }; // map :: (a -> b) -> [a] -> [b] const map = (f, xs) => xs.map(f); // 'The mapAccumL function behaves like a combination of map and foldl; // it applies a function to each element of a list, passing an accumulating // parameter from left to right, and returning a final value of this // accumulator together with the new list.' (See Hoogle) // mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y]) const mapAccumL = (f, acc, xs) => xs.reduce((a, x, i) => { const pair = f(a[0], x, i); return Tuple(pair[0], a[1].concat(pair[1])); }, Tuple(acc, [])); // reverse :: [a] -> [a] const reverse = xs => 'string' !== typeof xs ? ( xs.slice(0).reverse() ) : xs.split('').reverse().join(''); // snd :: (a, b) -> b const snd = tpl => tpl[1]; // tail :: [a] -> [a] const tail = xs => 0 < xs.length ? xs.slice(1) : []; // takeWhile :: (a -> Bool) -> [a] -> [a] // takeWhile :: (Char -> Bool) -> String -> String const takeWhile = (p, xs) => { const lng = xs.length; return 0 < lng ? xs.slice( 0, until( i => i === lng || !p(xs[i]), i => 1 + i, 0 ) ) : []; }; // unlines :: [String] -> String const unlines = xs => xs.join('\n'); // until :: (a -> Bool) -> (a -> a) -> a -> a const until = (p, f, x) => { let v = x; while (!p(v)) v = f(v); return v; }; // MAIN --- return main();})(); Output: 0 1 10 100 101 1000 1001 1010 10000 10001 10010 10100 10101 100000 100001 100010 100100 100101 101000 101001 101010 ## jq Works with: jq version 1.4 def zeckendorf: # rfibs(n) returns an array of fibonnaci numbers up to n, # beginning with 1, 2, ..., in reverse order def rfibs(n): # input: [f(i-2), f(i-1)] [1,1] | [recurse( if .[1] >= n then empty else [.[1], add] end ) | .[1]] | reverse; . asn  # [n, rfibs, digit ]  | [$n, rfibs($n), "" ]  | [ recurse( .[0] as $n | .[1] as$f               | if ($f|length) == 0 then empty else$f[0] as $next | if$n >= $next then [ ($n - $next),$f[1:], "1"]		   else [ $n,$f[1:], "0"]                   end                 end )      | .[2] ]  | if .[1] == "0" then .[2:] else . end  # remove leading 0 if any  | join("") ;

Example:

range(0;21) | "\(.): \(zeckendorf)"
Output:
$jq -n -r -f zeckendorf.jq0: 1: 12: 103: 1004: 1015: 10006: 10017: 10108: 100009: 1000110: 1001011: 1010012: 1010113: 10000014: 10000115: 10001016: 10010017: 10010118: 10100019: 10100120: 101010 ## Julia Translation of: Python function zeck(n) n <= 0 && return 0 fib = [2,1]; while fib[1] < n unshift!(fib,sum(fib[1:2])) end dig = Int[]; for f in fib f <= n ? (push!(dig,1); n = n-f;) : push!(dig,0) end return dig[1] == 0 ? dig[2:end] : digend Output: julia> for x = 0:20 println(join(zeck(x))) end 0 1 10 100 101 1000 1001 1010 10000 10001 10010 10100 10101 100000 100001 100010 100100 100101 101000 101001 101010  ## Kotlin // version 1.0.6 const val LIMIT = 46 // to stay within range of signed 32 bit integerval fibs = fibonacci(LIMIT) fun fibonacci(n: Int): IntArray { if (n !in 2..LIMIT) throw IllegalArgumentException("n must be between 2 and$LIMIT")    val fibs = IntArray(n)    fibs[0] = 1    fibs[1] = 1    for (i in 2 until n) fibs[i] = fibs[i - 1] + fibs[i - 2]    return fibs} fun zeckendorf(n: Int): String {    if (n < 0) throw IllegalArgumentException("n must be non-negative")    if (n < 2) return n.toString()    var lastFibIndex = 1    for (i in 2..LIMIT)        if (fibs[i] > n) {            lastFibIndex = i - 1            break        }    var nn = n - fibs[lastFibIndex--]    val zr = StringBuilder("1")    for (i in lastFibIndex downTo 1)        if (fibs[i] <= nn) {            zr.append('1')            nn -= fibs[i]        } else {            zr.append('0')        }    return zr.toString()} fun main(args: Array<String>) {    println(" n   z")    for (i in 0..20) println("${"%2d".format(i)} :${zeckendorf(i)}")}
Output:
 n   z
0 : 0
1 : 1
2 : 10
3 : 100
4 : 101
5 : 1000
6 : 1001
7 : 1010
8 : 10000
9 : 10001
10 : 10010
11 : 10100
12 : 10101
13 : 100000
14 : 100001
15 : 100010
16 : 100100
17 : 100101
18 : 101000
19 : 101001
20 : 101010


## Lingo

Translation of: Lua
-- Return the distinct Fibonacci numbers not greater than 'n'on fibsUpTo (n)    fibList = []    last = 1    current = 1    repeat while current <= n        fibList.add(current)        nxt = last + current        last = current        current = nxt        end repeat    return fibListend -- Return the Zeckendorf representation of 'n'on zeckendorf (n)    fib = fibsUpTo(n)    zeck = ""    repeat with pos = fib.count down to 1        if n >= fib[pos] then            zeck = zeck & "1"            n = n - fib[pos]        else            zeck = zeck & "0"        end if    end repeat    if zeck = "" then return "0"    return zeckend
repeat with n = 0 to 20    put n & ": " & zeckendorf(n)end repeat
Output:
0: 0
1: 1
2: 10
3: 100
4: 101
5: 1000
6: 1001
7: 1010
8: 10000
9: 10001
10: 10010
11: 10100
12: 10101
13: 100000
14: 100001
15: 100010
16: 100100
17: 100101
18: 101000
19: 101001
20: 101010


## Logo

; return the (N+1)th Fibonacci number (1,2,3,5,8,13,...)to fib m  local "n  make "n sum :m 1  if [lessequal? :n 0] [output difference fib sum :n 2 fib sum :n 1]  global "_fib  if [not name? "_fib] [    make "_fib [1 1]  ]  local "length  make "length count :_fib  while [greater? :n :length] [    make "_fib (lput (sum (last :_fib) (last (butlast :_fib))) :_fib)    make "length sum :length 1  ]  output item :n :_fibend ; return the binary Zeckendorf representation of a nonnegative numberto zeckendorf n  if [less? :n 0] [(throw "error [Number must be nonnegative.])]  (local "i "f "result)  make "i :n  make "f fib :i  while [less? :f :n] [make "i sum :i 1 make "f fib :i]   make "result "||  while [greater? :i 0] [    ifelse [greaterequal? :n :f] [      make "result lput 1 :result      make "n difference :n :f    ] [      if [not empty? :result] [        make "result lput 0 :result      ]    ]    make "i difference :i 1    make "f fib :i  ]  if [equal? :result "||] [    make "result 0  ]  output :resultend type zeckendorf 0repeat 20 [  type word "| | zeckendorf repcount]print []bye
Output:
0 1 10 100 101 1000 1001 1010 10000 10001 10010 10100 10101 100000 100001 100010 100100 100101 101000 101001 101010

## Lua

-- Return the distinct Fibonacci numbers not greater than 'n'function fibsUpTo (n)    local fibList, last, current, nxt = {}, 1, 1    while current <= n do        table.insert(fibList, current)        nxt = last + current        last = current        current = nxt        end    return fibListend -- Return the Zeckendorf representation of 'n'function zeckendorf (n)    local fib, zeck = fibsUpTo(n), ""    for pos = #fib, 1, -1 do        if n >= fib[pos] then            zeck = zeck .. "1"            n = n - fib[pos]        else            zeck = zeck .. "0"        end    end    if zeck == "" then return "0" end    return zeckend -- Main procedureprint(" n\t| Zeckendorf(n)")print(string.rep("-", 23))for n = 0, 20 do    print(" " .. n, "| " .. zeckendorf(n))end
Output:
 n      | Zeckendorf(n)
-----------------------
0      | 0
1      | 1
2      | 10
3      | 100
4      | 101
5      | 1000
6      | 1001
7      | 1010
8      | 10000
9      | 10001
10     | 10010
11     | 10100
12     | 10101
13     | 100000
14     | 100001
15     | 100010
16     | 100100
17     | 100101
18     | 101000
19     | 101001
20     | 101010

## Mathematica

zeckendorf[0] = 0;zeckendorf[n_Integer] :=   10^(# - 1) + zeckendorf[n - Fibonacci[# + 1]] &@   LengthWhile[    Fibonacci /@      Range[2, [email protected][GoldenRatio, n [email protected]]], # <= n &];zeckendorf /@ Range[0, 20]
Output:
{0, 1, 10, 100, 101, 1000, 1001, 1010, 10000, 10001, 10010, 10100,
10101, 100000, 100001, 100010, 100100, 100101, 101000, 101001, 101010}

## Nim

Translation of: Python
import strutils proc z(n): string =  if n == 0: return "0"  var fib = @[2,1]  var n = n  while fib[0] < n: fib.insert(fib[0] + fib[1])  result = ""  for f in fib:    if f <= n:      result.add '1'      n -= f    else:      result.add '0'  if result[0] == '0':    result = result[1..result.high] for i in 0 .. 20:  echo align(i, 3)," ",align(z(i), 8) Output:  0 0 1 1 2 10 3 100 4 101 5 1000 6 1001 7 1010 8 10000 9 10001 10 10010 11 10100 12 10101 13 100000 14 100001 15 100010 16 100100 17 100101 18 101000 19 101001 20 101010 ## PARI/GP Z(n)=if(!n,print1(0));my(k=2);while(fibonacci(k)<=n,k++); forstep(i=k-1,2,-1,print1(if(fibonacci(i)<=n,n-=fibonacci(i);1,0)));printfor(n=0,20,Z(n)) 0 1 10 100 101 1000 1001 1010 10000 10001 10010 10100 10101 100000 100001 100010 100100 100101 101000 101001 101010 ## Perl my @fib; sub fib { myn = shift;  return 1 if $n < 2; return$fib[$n] //= fib($n-1)+fib($n-2);} sub zeckendorf { my$n = shift;  return "0" unless $n; my$i = 1;  $i++ while fib($i) <= $n; my$z = '';  while( --$i ) {$z .= "0", next if fib( $i ) >$n;    $z .= "1";$n -= fib( $i ); } return$z;} printf "%4d: %8s\n", $_, zeckendorf($_) for 0..20;
Output:
   0:        0
1:        1
2:       10
3:      100
4:      101
5:     1000
6:     1001
7:     1010
8:    10000
9:    10001
10:    10010
11:    10100
12:    10101
13:   100000
14:   100001
15:   100010
16:   100100
17:   100101
18:   101000
19:   101001
20:   101010


## Perl 6

Works with: rakudo version 2015.12
printf "%2d: %8s\n", $_, zeckendorf($_) for 0 .. 20; multi zeckendorf(0) { '0' }multi zeckendorf($n is copy) { constant FIBS = (1,2, *+* ... *).cache; [~] map {$n -= $_ if my$digit = $n >=$_;        +$digit; }, reverse FIBS ...^ * >$n;}
Output:
 0:        0
1:        1
2:       10
3:      100
4:      101
5:     1000
6:     1001
7:     1010
8:    10000
9:    10001
10:    10010
11:    10100
12:    10101
13:   100000
14:   100001
15:   100010
16:   100100
17:   100101
18:   101000
19:   101001
20:   101010

function zeckendorf(integer n)integer r = 0, csequence fib = {1,1}    while fib[$]<n do fib &= fib[$] + fib[$-1] end while for i=length(fib) to 2 by -1 do c = n>=fib[i] r += r+c n -= c*fib[i] end for return rend function for i=0 to 20 do printf(1,"%2d: %7b\n",{i,zeckendorf(i)})end for Output:  0: 0 1: 1 2: 10 3: 100 4: 101 5: 1000 6: 1001 7: 1010 8: 10000 9: 10001 10: 10010 11: 10100 12: 10101 13: 100000 14: 100001 15: 100010 16: 100100 17: 100101 18: 101000 19: 101001 20: 101010  ## PHP  <?php$m = 20; $F = array(1,1);while ($F[count($F)-1] <=$m)   $F[] =$F[count($F)-1] +$F[count($F)-2]; while ($n = $m--) { while ($F[count($F)-1] >$n) array_pop($F);$l = count($F)-1; print "$n: ";   while ($n) { if ($n >= $F[$l]) {         $n =$n - $F[$l];         print '1';      } else print '0';      --$l; } print str_repeat('0',$l);   print "\n";}?>
Output:
20: 101010
19: 101001
18: 101000
17: 100101
16: 100100
15: 100010
14: 100001
13: 100000
12: 10101
11: 10100
10: 10010
9: 10001
8: 10000
7: 1010
6: 1001
5: 1000
4: 101
3: 100
2: 10
1: 1


## PicoLisp

(de fib (N)   (let Fibs (1 1)      (while (>= N (+ (car Fibs) (cadr Fibs)))         (push 'Fibs (+ (car Fibs) (cadr Fibs))) )      (uniq Fibs) ) ) (de zecken1 (N)   (make      (for I (fib N)         (if (> I N)            (link 0)            (link 1)            (dec 'N I) ) ) ) ) (de zecken2 (N)   (make      (when (=0 N) (link 0))      (for I (fib N)         (when (<= I N)            (link I)            (dec 'N I) ) ) ) ) (for (N 0 (> 21 N) (inc N))   (tab (2 4 6 2 -10)      N       " -> "      (zecken1 N)      "  "      (glue " + " (zecken2 N)) ) )  (bye)
Output:
 0 ->      0  0
1 ->      1  1
2 ->     10  2
3 ->    100  3
4 ->    101  3 + 1
5 ->   1000  5
6 ->   1001  5 + 1
7 ->   1010  5 + 2
8 ->  10000  8
9 ->  10001  8 + 1
10 ->  10010  8 + 2
11 ->  10100  8 + 3
12 ->  10101  8 + 3 + 1
13 -> 100000  13
14 -> 100001  13 + 1
15 -> 100010  13 + 2
16 -> 100100  13 + 3
17 -> 100101  13 + 3 + 1
18 -> 101000  13 + 5
19 -> 101001  13 + 5 + 1
20 -> 101010  13 + 5 + 2

## Plain TeX

This code needs an etex engine.

\def\genfibolist#1{% #creates the fibo list which sum>=#1	\let\fibolist\empty\def\targetsum{#1}\def\fibosum{0}%	\genfibolistaux1,1\relax}\def\genfibolistaux#1,#2\relax{%	\ifnum\fibosum<\targetsum\relax		\edef\fibosum{\number\numexpr\fibosum+#2}%		\edef\fibolist{#2,\fibolist}%		\edef\tempfibo{\noexpand\genfibolistaux#2,\number\numexpr#1+#2\relax\relax}%		\expandafter\tempfibo	\fi}\def\zeckendorf#1{\expandafter\zeckendorfaux\fibolist,\relax#1\relax\relax0}\def\zeckendorfaux#1,#2\relax#3\relax#4\relax#5{%	\ifx\relax#2\relax		#4%	\else		\ifnum#3<#1			\edef\temp{#2\relax#3\relax#4\ifnum#5=1 0\fi\relax#5}%		\else			\edef\temp{#2\relax\number\numexpr#3-#1\relax\relax#41\relax1}%		\fi		\expandafter\expandafter\expandafter\zeckendorfaux\expandafter\temp	\fi}\newcount\ii\def\listzeckendorf#1{%	\genfibolist{#1}%	\ii=0	\loop		\ifnum\ii<#1		\advance\ii1		\number\ii: \zeckendorf\ii\endgraf	\repeat}\listzeckendorf{20}% any integer accepted\bye

pdf output looks like:

1: 1
2: 10
3: 100
4: 101
5: 1000
6: 1001
7: 1010
8: 10000
9: 10001
10: 10010
11: 10100
12: 10101
13: 100000
14: 100001
15: 100010
16: 100100
17: 100101
18: 101000
19: 101001
20: 101010


## PowerShell

Works with: PowerShell version 2
 function Get-ZeckendorfNumber ( $N ) { # Calculate relevant portation of Fibonacci series$Fib = @( 1, 1 )    While ( $Fib[-1] -lt$N ) { $Fib +=$Fib[-1] + $Fib[-2] } # Start with 0$ZeckendorfNumber = 0     #  For each number in the relevant portion of Fibonacci series    For ( $i =$Fib.Count - 1; $i -gt 0;$i-- )        {        #  If Fibonacci number is less than or equal to remainder of N        If ( $Fib[$i] -le $N ) { # Double Z number and add 1 (equivalent to adding a '1' to the end of a binary number)$ZeckendorfNumber = $ZeckendorfNumber * 2 + 1 # Reduce N by Fibonacci number, skip next Fibonacci number$N -= $Fib[$i--]            }        #  If were aren't finished yet, double Z number        #  (equivalent to adding a '0' to the end of a binary number)        If ( $i ) {$ZeckendorfNumber *= 2 }        }    return $ZeckendorfNumber }   # Get Zeckendorf numbers through 20, convert to binary for display0..20 | ForEach { [convert]::ToString( ( Get-ZeckendorfNumber$_ ), 2 ) }
Output:
0
1
10
100
101
1000
1001
1010
10000
10001
10010
10100
10101
100000
100001
100010
100100
100101
101000
101001
101010


## PureBasic

Procedure.s zeck(n.i)  Dim f.i(1) : Define i.i=1, o$f(0)=1 : f(1)=1 While f(i)<n i+1 : ReDim f(ArraySize(f())+1) : f(i)=f(i-1)+f(i-2) Wend For i=i To 1 Step -1 If n>=f(i) : o$+"1" : n-f(i) : Else : o$+"0" : EndIf Next If Len(o$)>1 : o$=LTrim(o$,"0") : EndIf  ProcedureReturn o$EndProcedure Define n.i, t$OpenConsole("Zeckendorf number representation")PrintN(~"\tNr.\tZeckendorf")For n=0 To 20  t$=zeck(n) If FindString(t$,"11")    PrintN("Error: n= "+Str(n)+~"\tZeckendorf= "+t$) Break Else PrintN(~"\t"+RSet(Str(n),3," ")+~"\t"+RSet(t$,7," "))      EndIfNextInput()
Output:
        Nr.     Zeckendorf
0           0
1           1
2          10
3         100
4         101
5        1000
6        1001
7        1010
8       10000
9       10001
10       10010
11       10100
12       10101
13      100000
14      100001
15      100010
16      100100
17      100101
18      101000
19      101001
20      101010

## Python

def fib():    memo = [1, 2]    while True:        memo.append(sum(memo))        yield memo.pop(0) def sequence_down_from_n(n, seq_generator):    seq = []    for s in seq_generator():        seq.append(s)        if s >= n: break    return seq[::-1] def zeckendorf(n):    if n == 0: return [0]    seq = sequence_down_from_n(n, fib)    digits, nleft = [], n    for s in seq:        if s <= nleft:            digits.append(1)            nleft -= s        else:            digits.append(0)    assert nleft == 0, 'Check all of n is accounted for'    assert sum(x*y for x,y in zip(digits, seq)) == n, 'Assert digits are correct'    while digits[0] == 0:        # Remove any zeroes padding L.H.S.        digits.pop(0)    return digits n = 20print('Fibonacci digit multipliers: %r' % sequence_down_from_n(n, fib))for i in range(n + 1):    print('%3i: %8s' % (i, ''.join(str(d) for d in zeckendorf(i))))
Output:
Fibonacci digit multipliers: [21, 13, 8, 5, 3, 2, 1]
0:        0
1:        1
2:       10
3:      100
4:      101
5:     1000
6:     1001
7:     1010
8:    10000
9:    10001
10:    10010
11:    10100
12:    10101
13:   100000
14:   100001
15:   100010
16:   100100
17:   100101
18:   101000
19:   101001
20:   101010

### Shorter version

n = 20def z(n):    if n == 0 : return [0]    fib = [2,1]    while fib[0] < n: fib[0:0] = [sum(fib[:2])]    dig = []    for f in fib:        if f <= n:            dig, n = dig + [1], n - f        else:            dig += [0]    return dig if dig[0] else dig[1:] for i in range(n + 1):    print('%3i: %8s' % (i, ''.join(str(d) for d in z(i))))
Output:
  0:        0
1:        1
2:       10
3:      100
4:      101
5:     1000
6:     1001
7:     1010
8:    10000
9:    10001
10:    10010
11:    10100
12:    10101
13:   100000
14:   100001
15:   100010
16:   100100
17:   100101
18:   101000
19:   101001
20:   101010

## R

zeckendorf <- function(number) {   # Get an upper limit on Fibonacci numbers needed to cover number  indexOfFibonacciNumber <- function(n) {    if (n < 1) {      2    } else {      Phi <- (1 + sqrt(5)) / 2      invertClosedFormula <- log(n * sqrt(5)) / log(Phi)      ceiling(invertClosedFormula)    }  }   upperLimit <- indexOfFibonacciNumber(number)   # Return the sequence as digits, sorted descending  fibonacciSequenceDigits <- function(n) {    fibGenerator <- function(f, ...) { c(f[2], sum(f)) }    fibSeq <- Reduce(fibGenerator, 1:n, c(0,1), accumulate=TRUE)     fibNums <- unlist(lapply(fibSeq, head, n=1))     # drop last F0 and F1 and reverse sequence    rev(fibNums[-2:-1])  }   digits <- fibonacciSequenceDigits(upperLimit)   isInNumber <- function(digit) {    if (number >= digit) {      number <<- number - digit      1    } else {      0    }  }   zeckSeq <- Map(isInNumber, digits)   # drop leading 0 and convert to String  gsub("^0+1", "1", paste(zeckSeq, collapse=""))} print(unlist(lapply(0:20, zeckendorf)))

This is definitely not the shortest way to implement the Zeckendorf numbers but focus was on the functional aspect of R, so no loops and (almost) no assignments.

Output:
 [1] "0"      "1"      "10"     "100"    "101"    "1000"   "1001"   "1010"
[9] "10000"  "10001"  "10010"  "10100"  "10101"  "100000" "100001" "100010"
[17] "100100" "100101" "101000" "101001" "101010"

## Racket

 #lang racket (require math) (define (fibs n)  (reverse   (for/list ([i (in-naturals 2)] #:break (> (fibonacci i) n))     (fibonacci i)))) (define (zechendorf n)  (match/values   (for/fold ([n n] [xs '()]) ([f (fibs n)])     (if (> f n)         (values n       (cons 0 xs))         (values (- n f) (cons 1 xs))))   [(_ xs) (reverse xs)])) (for/list ([n 21])  (list n (zechendorf n)))

Output:

 '((0 ())  (1 (1))  (2 (1 0))  (3 (1 0 0))  (4 (1 0 1))  (5 (1 0 0 0))  (6 (1 0 0 1))  (7 (1 0 1 0))  (8 (1 0 0 0 0))  (9 (1 0 0 0 1))  (10 (1 0 0 1 0))  (11 (1 0 1 0 0))  (12 (1 0 1 0 1))  (13 (1 0 0 0 0 0))  (14 (1 0 0 0 0 1))  (15 (1 0 0 0 1 0))  (16 (1 0 0 1 0 0))  (17 (1 0 0 1 0 1))  (18 (1 0 1 0 0 0))  (19 (1 0 1 0 0 1))  (20 (1 0 1 0 1 0)))

## REXX

### specific to 20

 /* REXX **************************************************************** 11.10.2012 Walter Pachl                                              **********************************************************************/fib='13 8 5 3 2 1'                                                   Do i=6 To 1 By -1                   /* Prepare Fibonacci Numbers     */  Parse Var fib f.i fib             /* f.1 ... f.7                   */  End                                                                  Do n=0 To 20                        /* for all numbers in the task   */  m=n                               /* copy of number                */  r=''                              /* result for n                  */  Do i=6 To 1 By -1                 /* loop through numbers          */    If m>=f.i Then Do               /* f.i must be used              */      r=r||1                        /* 1 into result                 */      m=m-f.i                       /* subtract                      */      End                                                                  Else                            /* f.i is larger than the rest   */      r=r||0                        /* 0 into result                 */    End                                                                  r=strip(r,'L','0')                /* strip leading zeros           */  If r='' Then r='0'                /* take care of 0                */  Say right(n,2)':  'right(r,6)     /* show result                   */  End

Output:


0:       0
1:       1
2:      10
3:     100
4:     101
5:    1000
6:    1001
7:    1010
8:   10000
9:   10001
10:   10010
11:   10100
12:   10101
13:  100000
14:  100001
15:  100010
16:  100100
17:  100101
18:  101000
19:  101001
20:  101010

### generalized

This generalized REXX version will work for any Zeckendorf number (up to 100,000 decimal digits).

A list of Fibonacci numbers (in ascending order) is generated large enough to handle the   Nth   Zeckendorf number.

/*REXX program  calculates and displays the  first   N   Zeckendorf numbers.            */numeric digits 100000                            /*just in case user gets real ka─razy. */parse arg N .                                    /*let the user specify the upper limit.*/if N=='' | N==","  then n=20;   w= length(N)     /*Not specified?  Then use the default.*/@.1= 1                                           /*start the array with  1   and   2.   */@.2= 2;   do  #=3  until #>=N;  p= #-1;  pp= #-2 /*build a list of Fibonacci numbers.   */          @.#= @.p + @.pp                        /*sum the last two Fibonacci numbers.  */          end   /*#*/                            /* [↑]   #:  contains a Fibonacci list.*/   do j=0  to N;             parse var j x z      /*task:  process zero  ──►  N  numbers.*/     do k=#  by -1  for #;  _= @.k               /*process all the Fibonacci numbers.   */     if x>=_  then do;      z= z'1'              /*is X>the next Fibonacci #?  Append 1.*/                            x= x - _             /*subtract this Fibonacci # from index.*/                   end              else z= z'0'                       /*append zero (0)  to the Fibonacci #. */     end   /*k*/  say '    Zeckendorf'     right(j, w)    "="     right(z+0, 30)     /*display a number.*/  end     /*j*/                                  /*stick a fork in it,  we're all done. */
output   when using the default input:
    Zeckendorf  0 =                              0
Zeckendorf  1 =                              1
Zeckendorf  2 =                             10
Zeckendorf  3 =                            100
Zeckendorf  4 =                            101
Zeckendorf  5 =                           1000
Zeckendorf  6 =                           1001
Zeckendorf  7 =                           1010
Zeckendorf  8 =                          10000
Zeckendorf  9 =                          10001
Zeckendorf 10 =                          10010
Zeckendorf 11 =                          10100
Zeckendorf 12 =                          10101
Zeckendorf 13 =                         100000
Zeckendorf 14 =                         100001
Zeckendorf 15 =                         100010
Zeckendorf 16 =                         100100
Zeckendorf 17 =                         100101
Zeckendorf 18 =                         101000
Zeckendorf 19 =                         101001
Zeckendorf 20 =                         101010


### generic

This generic REXX version will generate up to the   Nth   Zeckendorf numbers (up to 100,000 decimal digits) by
using binary numbers that   don't   have two consecutive   11s   within their binary version.

There isn't any need to generate a Fibonacci series with this method.   This method is extremely fast.

/*REXX program  calculates and displays the  first   N   Zeckendorf numbers.            */numeric digits 100000                            /*just in case user gets real ka─razy. */parse arg N .                                    /*let the user specify the upper limit.*/if N=='' | N==","  then n=20;    w= length(N)    /*Not specified?  Then use the default.*/z=0                                              /*the index of a  Zeckendorf number.   */    do j=0  until z>N;          _=x2b( d2x(j) )  /*task:   process zero  ──►   N.       */    if pos(11, _) \== 0  then iterate            /*are there two consecutive ones (1s) ?*/    say '    Zeckendorf'   right(z, w)    "="     right(_+0, 30)     /*display a number.*/    z= z + 1                                     /*bump the  Zeckendorf  number counter.*/    end   /*j*/                                  /*stick a fork in it,  we're all done. */

{{out|output|text=  is identical to the previous (generalized) version.

## Ring

 # Project : Zeckendorf number representation see "0 0" + nlfor n = 1 to 20     see "" + n + " " + zeckendorf(n) + nlnext func zeckendorf(n)       fib = list(45)       fib[1] = 1       fib[2] = 1       i = 2       o = ""       while fib[i] <= n               i = i + 1               fib[i] = fib[i-1] + fib[i-2]       end       while i != 2               i = i - 1               if n >= fib[i]                   o = o + "1"                   n = n - fib[i]               else                   o = o + "0"               ok        end        return o

Output:

0 0
1 1
2 10
3 100
4 101
5 1000
6 1001
7 1010
8 10000
9 10001
10 10010
11 10100
12 10101
13 100000
14 100001
15 100010
16 100100
17 100101
18 101000
19 101001
20 101010


## Ruby

Featuring a method doubling as an enumerator.

def zeckendorf  return to_enum(__method__) unless block_given?  x = 0  loop do    bin = x.to_s(2)    yield bin unless bin.include?("11")     x += 1  endend zeckendorf.take(21).each_with_index{|x,i| puts "%3d: %8s"% [i, x]}
Output:
  0:        0
1:        1
2:       10
3:      100
4:      101
5:     1000
6:     1001
7:     1010
8:    10000
9:    10001
10:    10010
11:    10100
12:    10101
13:   100000
14:   100001
15:   100010
16:   100100
17:   100101
18:   101000
19:   101001
20:   101010

Translation of: Python
def zeckendorf(n)  return 0 if n.zero?  fib = [1,2]  fib << fib[-2] + fib[-1] while fib[-1] < n  dig = ""  fib.reverse_each do |f|    if f <= n      dig, n = dig + "1", n - f    else      dig += "0"    end  end  dig.to_iend for i in 0..20  puts '%3d: %8d' % [i, zeckendorf(i)]end

(Same output.)

## Scala

def zNum( n:BigInt ) : String = {   if( n == 0 ) return "0"	// Short-circuit this and return zero if we were given zero    val v = n.abs   val fibs : Stream[BigInt] = { def series(i:BigInt,j:BigInt):Stream[BigInt] = i #:: series(j, i+j); series(1,0).tail.tail.tail }    def z( v:BigInt ) : List[BigInt] = if(v == 0) List() else {val m = fibs(fibs.indexWhere(_>v) - 1); m :: z(v-m)}   val zv = z(v)   // Walk the list of fibonacci numbers from the number that matches the most significant down to 1,  // if the zeckendorf matchs then yield '1' otherwise '0'  val s = (for( i <- (fibs.indexWhere(_==zv(0)) to 0 by -1) ) yield {     if( zv.contains(fibs(i))) "1" else "0"   }).mkString   if( n < 0 ) "-" + s		// Using a negative-sign instead of twos-complement   else s}  // A little test...(0 to 20) foreach( i => print( zNum(i) + "\n" ) )
Output:
0
1
10
100
101
1000
1001
1010
10000
10001
10010
10100
10101
100000
100001
100010
100100
100101
101000
101001
101010


## Scheme

Translation of: Java
(import (rnrs)) (define (getFibList maxNum n1 n2 fibs)  (if (> n2 maxNum)      fibs      (getFibList maxNum n2 (+ n1 n2) (cons n2 fibs)))) (define (getZeckendorf num)  (if (<= num 0)      "0"      (let ((fibs (getFibList num 1 2 (list 1))))        (getZeckString "" num fibs)))) (define (getZeckString zeck num fibs)  (let* ((curFib (car fibs))         (placeZeck (>= num curFib))         (outString (string-append zeck (if placeZeck "1" "0")))         (outNum (if placeZeck (- num curFib) num)))    (if (null? (cdr fibs))        outString        (getZeckString outString outNum (cdr fibs))))) (let loop ((i 0))  (when (<= i 20)    (for-each      (lambda (item)        (display item))      (list "Z(" i "):\t" (getZeckendorf i)))    (newline)    (loop (+ i 1))))
Output:
Z(0):   0
Z(1):   1
Z(2):   10
Z(3):   100
Z(4):   101
Z(5):   1000
Z(6):   1001
Z(7):   1010
Z(8):   10000
Z(9):   10001
Z(10):  10010
Z(11):  10100
Z(12):  10101
Z(13):  100000
Z(14):  100001
Z(15):  100010
Z(16):  100100
Z(17):  100101
Z(18):  101000
Z(19):  101001
Z(20):  101010

## Sidef

Translation of: Perl
func fib(n) is cached {    n < 2 ? 1          : (fib(n-1) + fib(n-2))} func zeckendorf(n) {    n == 0 && return '0'    var i = 1    ++i while (fib(i) <= n)    gather {        while (--i > 0) {            var f = fib(i)            f > n ? (take '0')                  : (take '1'; n -= f)        }    }.join} for n (0..20) {    printf("%4d: %8s\n", n, zeckendorf(n))}
Output:
   0:        0
1:        1
2:       10
3:      100
4:      101
5:     1000
6:     1001
7:     1010
8:    10000
9:    10001
10:    10010
11:    10100
12:    10101
13:   100000
14:   100001
15:   100010
16:   100100
17:   100101
18:   101000
19:   101001
20:   101010


## Simula

Translation of: Sinclair ZX81 BASIC
BEGIN   INTEGER N, F0, F1, F2, D;   N := 20;   COMMENT CALCULATE D FROM ANY GIVEN N ;   F1 := 1; F2 := 2; F0 := F1 + F2; D := 2;   WHILE F0 < N DO BEGIN      F1 := F2; F2 := F0; F0 := F1 + F2; D := D + 1;   END;   BEGIN      COMMENT Sinclair ZX81 BASIC Solution ;      TEXT Z1, S1;      INTEGER I, J, Z;      INTEGER ARRAY F(1:D);                  !  10 dim f(6) ;      F(1) := 1;                             !  20 let f(1)=1 ;      F(2) := 2;                             !  30 let f(2)=2 ;      FOR I := 3 STEP 1 UNTIL D DO BEGIN     !  40 for i=3 to 6 ;         F(I) := F(I-2) + F(I-1);            !  50 let f(i)=f(i-2)+f(i-1) ;      END;                                   !  60 next i ;      FOR I := 0 STEP 1 UNTIL N DO BEGIN     !  70 for i=0 to 20 ;         Z1 :- "";                           !  80 let z$="" ; S1 :- " "; ! 90 let s$=" " ;         Z := I;                             ! 100 let z=i ;         FOR J := D STEP -1 UNTIL 1 DO BEGIN ! 110 for j=6 to 1 step -1 ;            IF J=1 THEN S1 :- "0";           ! 120 if j=1 then let s$="0" ; IF NOT (Z<F(J)) THEN BEGIN ! 130 if z<f(j) then goto 180 ; Z1 :- Z1 & "1"; ! 140 let z$=z$+"1" ; Z := Z-F(J); ! 150 let z=z-f(j) ; S1 :- "0"; ! 160 let s$="0" ;            END ELSE                         ! 170 goto 190 ;               Z1 :- Z1 & S1;                ! 180 let z$=z$+s$; END; ! 190 next j ; OUTINT(I, 0); OUTCHAR(' '); ! 200 print i ; !" "; !; IF I<10 THEN OUTCHAR(' '); ! 210 if i<10 then print " "; !; OUTTEXT(Z1); OUTIMAGE; ! 220 print z$ ;      END;                                   ! 230 next i ;   END;END
Output:
0       0
1       1
2      10
3     100
4     101
5    1000
6    1001
7    1010
8   10000
9   10001
10  10010
11  10100
12  10101
13 100000
14 100001
15 100010
16 100100
17 100101
18 101000
19 101001
20 101010

## Sinclair ZX81 BASIC

Works on the 1k RAM model, albeit without much room for manoeuvre. (You'd like the Zeckendorf numbers further over towards the right-hand side of the screen? Sorry, can't spare the video RAM.) If you have 2k or more, you can replace the constant 6 with some higher value wherever it occurs in the program and enable yourself to represent bigger numbers in Zeckendorf form.

 10 DIM F(6) 20 LET F(1)=1 30 LET F(2)=2 40 FOR I=3 TO 6 50 LET F(I)=F(I-2)+F(I-1) 60 NEXT I 70 FOR I=0 TO 20 80 LET Z$="" 90 LET S$=" "100 LET Z=I110 FOR J=6 TO 1 STEP -1120 IF J=1 THEN LET S$="0"130 IF Z<F(J) THEN GOTO 180140 LET Z$=Z$+"1"150 LET Z=Z-F(J)160 LET S$="0"170 GOTO 190180 LET Z$=Z$+S$190 NEXT J200 PRINT I;" ";210 IF I<10 THEN PRINT " ";220 PRINT Z$230 NEXT I
Output:
0       0
1       1
2      10
3     100
4     101
5    1000
6    1001
7    1010
8   10000
9   10001
10  10010
11  10100
12  10101
13 100000
14 100001
15 100010
16 100100
17 100101
18 101000
19 101001
20 101010

## Tcl

package require Tcl 8.5 # Generates the Fibonacci sequence (starting at 1) up to the largest item that# is no larger than the target value. Could use tricks to precompute, but this# is actually a pretty cheap linear operation.proc fibseq target {    set seq {}; set prev 1; set fib 1    for {set n 1;set i 1} {$fib <=$target} {incr n} {	for {} {$i <$n} {incr i} {	    lassign [list $fib [incr fib$prev]] prev fib	}	if {$fib <=$target} {	    lappend seq $fib } } return$seq} # Produce the given Zeckendorf number.proc zeckendorf n {    # Special case: only value that begins with 0    if {$n == 0} {return 0} set zs {} foreach f [lreverse [fibseq$n]] {	lappend zs [set z [expr {$f <=$n}]]	if {$z} {incr n [expr {-$f}]}    }    return [join $zs ""]} Demonstration for {set i 0} {$i <= 20} {incr i} {    puts [format "%2d:%9s" $i [zeckendorf$i]]}
Output:
 0:        0
1:        1
2:       10
3:      100
4:      101
5:     1000
6:     1001
7:     1010
8:    10000
9:    10001
10:    10010
11:    10100
12:    10101
13:   100000
14:   100001
15:   100010
16:   100100
17:   100101
18:   101000
19:   101001
20:   101010

## uBasic/4tH

For x = 0 to 20                        ' Print Zeckendorf numbers 0 - 20  Print x,  Push x : Gosub _Zeckendorf           ' get Zeckendorf number repres.  Print                                ' terminate lineNext End _Fibonacci  Push Tos()                           ' duplicate TOS()  @(0) = 0                             ' This function returns the  @(1) = 1                             ' Fibonacci number which is smaller                                       ' or equal to TOS()  Do While @(1) < Tos() + 1     Push (@(1))     @(1) = @(0) + @(1)                ' get next Fibonacci number     @(0) = Pop()  Loop                                 ' loop if not exceeded TOS()   Gosub _Drop                          ' clear TOS()  Push @(0)                            ' return Fibonacci numberReturn _Zeckendorf  GoSub _Fibonacci                     ' This function breaks TOS() up  Print Tos();                         ' into its Zeckendorf components  Push -(Pop() - Pop())                ' first digit is always there                                       ' the remainder to resolve  Do While Tos()                       ' now go for the next digits    GoSub _Fibonacci    Print " + ";Tos();                 ' print the next digit    Push -(Pop() - Pop())  Loop   Gosub _Drop                          ' clear TOS()Return                                 ' and return _Drop  If Pop()%1 = 0 Then Return           ' This function clears TOS()

Output:

0       0
1       1
2       2
3       3
4       3 + 1
5       5
6       5 + 1
7       5 + 2
8       8
9       8 + 1
10      8 + 2
11      8 + 3
12      8 + 3 + 1
13      13
14      13 + 1
15      13 + 2
16      13 + 3
17      13 + 3 + 1
18      13 + 5
19      13 + 5 + 1
20      13 + 5 + 2

0 OK, 0:901

## VBScript

 Function Zeckendorf(n)	num = n	Set fibonacci = CreateObject("System.Collections.Arraylist")	fibonacci.Add 1 : fibonacci.Add 2	i = 1	Do While fibonacci(i) < num		fibonacci.Add fibonacci(i) + fibonacci(i-1)		i = i + 1	Loop	tmp = ""	For j = fibonacci.Count-1 To 0 Step -1		If fibonacci(j) <= num And (tmp = "" Or Left(tmp,1) <> "1") Then			tmp = tmp & "1"			num = num - fibonacci(j)		Else			tmp = tmp & "0"		End If	Next	Zeckendorf = CLng(tmp)End Function 'testing the functionFor k = 0 To 20	WScript.StdOut.WriteLine k & ": " & Zeckendorf(k)Next
Output:
0: 0
1: 1
2: 10
3: 100
4: 101
5: 1000
6: 1001
7: 1010
8: 10000
9: 10001
10: 10010
11: 10100
12: 10101
13: 100000
14: 100001
15: 100010
16: 100100
17: 100101
18: 101000
19: 101001
20: 101010


## XPL0

include c:\cxpl\codes;          \intrinsic 'code' declarations proc Zeckendorf(N);             \Display Zeckendorf number (N <= 20)int N;int Fib, LZ, I;[Fib:= [1, 2, 3, 5, 8, 13];     \Fibonacci sequenceLZ:= true;                      \suppress leading zerosfor I:= 5 downto 1 do        [if N >= Fib(I) then [N:= N-Fib(I);  ChOut(0, ^1);  LZ:= false]        else ChOut(0, if LZ then ^  else ^0);        ];ChOut(0, N+^0);                 \output final digit, which can be 0]; int N;[for N:= 0 to 20 do        [if N<10 then ChOut(0,^ );  IntOut(0, N);  Text(0, ": ");        Zeckendorf(N);  CrLf(0);        ];]

Output:

 0:      0
1:      1
2:     10
3:    100
4:    101
5:   1000
6:   1001
7:   1010
8:  10000
9:  10001
10:  10010
11:  10100
12:  10101
13: 100000
14: 100001
15: 100010
16: 100100
17: 100101
18: 101000
19: 101001
20: 101010


## Yabasic

sub Zeckendorf(n)	local i, n$, c do n$ = bin$(i) if not instr(n$,"11") then			print c,":\t",n\$			if c = n break			c = c + 1					end if		i = i + 1	loopend sub Zeckendorf(20)

## zkl

    // return powers (0|1) of fib sequence (1,2,3,5,8...) that sum to nfcn zeckendorf(n){ //-->String of 1s & 0s, no consecutive 1's   if(n<=0) return("0");   fibs:=fcn(ab){ ab.append(ab.sum()).pop(0) }.fp(L(1,2));   (0).pump(*,List,fibs,'wrap(fib){ if(fib>n)Void.Stop else fib })   .reverse()   .pump(String,fcn(fib,rn){      if(fib>rn.value)"0" else { rn.set(rn.value-fib); "1" } }.fp1(Ref(n)))}
[0..20].pump(Console.println,fcn(n){ "%2d: %8s".fmt(n,zeckendorf(n)) });
Output:
 0:        0
1:        1
2:       10
3:      100
4:      101
5:     1000
6:     1001
7:     1010
8:    10000
9:    10001
10:    10010
11:    10100
12:    10101
13:   100000
14:   100001
15:   100010
16:   100100
17:   100101
18:   101000
19:   101001
20:   101010