# Fibonacci sequence: Difference between revisions

Fibonacci sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The Fibonacci sequence is a sequence   Fn   of natural numbers defined recursively:

      F0 = 0
F1 = 1
Fn = Fn-1 + Fn-2, if n>1


Write a function to generate the   nth   Fibonacci number.

Solutions can be iterative or recursive (though recursive solutions are generally considered too slow and are mostly used as an exercise in recursion).

The sequence is sometimes extended into negative numbers by using a straightforward inverse of the positive definition:

      Fn = Fn+2 - Fn+1, if n<0


support for negative     n     in the solution is optional.

References

## 0815

%<:0D:>~$<:01:~%>=<:a94fad42221f2702:>~> }:_s:{x{={~$x+%{=>~>x~-x<:0D:~>~>~^:_s:?

## 11l

Translation of: Python
F fib_iter(n)
I n < 2
R n
V fib_prev = 1
V fib = 1
L 2 .< n
(fib_prev, fib) = (fib, fib + fib_prev)
R fib

L(i) 1..20
print(fib_iter(i), end' ‘ ’)
print()
Output:
1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765


## 360 Assembly

For maximum compatibility, programs use only the basic instruction set.

### using fullword integers

*        Fibonacci sequence    05/11/2014
*        integer (31 bits) = 10 decimals -> max fibo(46)
FIBONACC CSECT
USING FIBONACC,R12    base register
SAVEAREA B     STM-SAVEAREA(R15) skip savearea
DC    17F'0'          savearea
DC    CL8'FIBONACC'   eyecatcher
STM      STM   R14,R12,12(R13) save previous context
*        ----
LA    R1,0            f(n-2)=0
LA    R2,1            f(n-1)=1
LA    R4,2            n=2
LA    R6,1            step
LH    R7,NN           limit
LOOP     EQU   *               for n=2 to nn
LR    R3,R2             f(n)=f(n-1)
AR    R3,R1             f(n)=f(n-1)+f(n-2)
CVD   R4,PW             n  convert binary to packed (PL8)
UNPK  ZW,PW             packed (PL8) to zoned (ZL16)
MVC   CW,ZW             zoned (ZL16) to  char (CL16)
OI    CW+L'CW-1,X'F0'   zap sign
MVC   WTOBUF+5(2),CW+14 output
CVD   R3,PW             f(n) binary to packed decimal (PL8)
ED    ZN,PW             packed dec (PL8) to char (CL20)
MVC   WTOBUF+9(14),ZN+6 output
WTO   MF=(E,WTOMSG)     write buffer
LR    R1,R2             f(n-2)=f(n-1)
LR    R2,R3             f(n-1)=f(n)
BXLE  R4,R6,LOOP      endfor n
*        ----
LM    R14,R12,12(R13) restore previous savearea pointer
XR    R15,R15         return code set to 0
*        ----  DATA
NN       DC    H'46'           nn max n
PW       DS    PL8             15num
ZW       DS    ZL16
CW       DS    CL16
ZN       DS    CL20
*                  ' b 0 0 0 , 0 0 0 , 0 0 0 , 0 0 0 , 0 0 0'  15num
WTOMSG   DS    0F
DC    H'80',XL2'0000'
*                   fibo(46)=1836311903
WTOBUF   DC    CL80'fibo(12)=1234567890'
REGEQU
END   FIBONACC
Output:
...
fibo(41)=   165,580,141
fibo(42)=   267,914,296
fibo(43)=   433,494,437
fibo(44)=   701,408,733
fibo(45)= 1,134,903,170
fibo(46)= 1,836,311,903


### using packed decimals

*        Fibonacci sequence        31/07/2018
*        packed dec (PL8) = 15 decimals => max fibo(73)
FIBOWTOP CSECT
USING  FIBOWTOP,R13       base register
B      72(R15)            skip savearea
DC     17F'0'             savearea
SAVE   (14,12)            save previous context
*        ----
ZAP    FNM2,=P'0'         f(0)=0
ZAP    FNM1,=P'1'         f(1)=1
LA     R4,2               n=2
LA     R6,1               step
LH     R7,NN              limit
LOOP     EQU    *                  for n=2 to nn
ZAP    FN,FNM1              f(n)=f(n-2)
AP     FN,FNM2              f(n)=f(n-1)+f(n-2)
CVD    R4,PW                n
ED     ZN,PW                packed dec (PL8) to char (CL16)
MVC    WTOBUF+5(2),ZN+L'ZN-2  output
ED     ZN,FN                packed dec (PL8) to char (CL16)
MVC    WTOBUF+9(L'ZN),ZN        output
WTO    MF=(E,WTOMSG)        write buffer
ZAP    FNM2,FNM1            f(n-2)=f(n-1)
ZAP    FNM1,FN              f(n-1)=f(n)
BXLE   R4,R6,LOOP         endfor n
*        ----
L      R13,4(0,R13)       restore previous savearea pointer
RETURN (14,12),RC=0       restore registers from calling sav
*        ----   DATA
NN       DC     H'73'              nn
FNM2     DS     PL8                f(n-2)
FNM1     DS     PL8                f(n-1)
FN       DS     PL8                f(n)
PW       DS     PL8                15num
ZN       DS     CL20
*                   ' b 0 0 0 , 0 0 0 , 0 0 0 , 0 0 0 , 0 0 0'  15num
WTOMSG   DS     0F
DC     H'80',XL2'0000'
*                    fibo(73)=806515533049393
WTOBUF   DC     CL80'fibo(12)=123456789012345 '
REGEQU
END    FIBOWTOP
Output:
...
fibo(68)=  72,723,460,248,141
fibo(69)= 117,669,030,460,994
fibo(70)= 190,392,490,709,135
fibo(71)= 308,061,521,170,129
fibo(72)= 498,454,011,879,264
fibo(73)= 806,515,533,049,393


## 6502 Assembly

This subroutine stores the first n—by default the first ten—Fibonacci numbers in memory, beginning (because, why not?) at address 3867 decimal = F1B hex. Intermediate results are stored in three sequential addresses within the low 256 bytes of memory, which are the most economical to access.

The results are calculated and stored, but are not output to the screen or any other physical device: how to do that would depend on the hardware and the operating system.

       LDA  #0
STA  $F0 ; LOWER NUMBER LDA #1 STA$F1     ; HIGHER NUMBER
LDX  #0
LOOP:  LDA  $F1 STA$0F1B,X
STA  $F2 ; OLD HIGHER NUMBER ADC$F0
STA  $F1 ; NEW HIGHER NUMBER LDA$F2
STA  $F0 ; NEW LOWER NUMBER INX CPX #$0A    ; STOP AT FIB(10)
BMI  LOOP
RTS          ; RETURN FROM SUBROUTINE

## 68000 Assembly

Translation of: ARM Assembly

Input is in D0, and the output is also in D0. There is no protection from 32-bit overflow, so use at your own risk. (I used this C compiler to create this in ARM Assembly and translated it manually into 68000 Assembly. It wasn't that difficult since both CPUs have similar architectures.)

fib:
MOVEM.L D4-D5,-(SP)
MOVE.L D0,D4
MOVEQ #0,D5
CMP.L #2,D0
BCS .bar
MOVEQ #0,D5
.foo:
MOVE.L D4,D0
SUBQ.L #1,D0
JSR fib
SUBQ.L #2,D4
CMP.L #1,D4
BHI .foo
.bar:
MOVE.L D5,D0
MOVEM.L (SP)+,D4-D5
RTS


## 8080 Assembly

This subroutine expects to be called with the value of ${\displaystyle n}$ in register A, and returns ${\displaystyle f(n)}$ also in A. You may want to take steps to save the previous contents of B, C, and D. The routine only works with fairly small values of ${\displaystyle n}$.

FIBNCI: MOV  C,  A  ; C will store the counter
DCR  C      ; decrement, because we know f(1) already
MVI  A,  1
MVI  B,  0
LOOP:   MOV  D,  A
ADD  B      ; A := A + B
MOV  B,  D
DCR  C
JNZ  LOOP   ; jump if not zero
RET         ; return from subroutine

## 8086 Assembly

### Calculating the values at runtime

Is it cheating to write it in C for 64-bit x86 then port it to 16-bit x86?

The max input is 24 before you start having 16-bit overflow.

fib:
; 	WRITTEN IN C WITH X86-64 CLANG 3.3 AND DOWNGRADED TO 16-BIT X86
; 	INPUT: DI = THE NUMBER YOU WISH TO CALC THE FIBONACCI NUMBER FOR.
;	OUTPUTS TO AX

push    BP
push    BX
push    AX
mov     BX, DI			;COPY INPUT TO BX
xor     AX, AX			;MOV AX,0
test    BX, BX			;SET FLAGS ACCORDING TO BX
je      LBB0_4			;IF BX == 0 RETURN 0
cmp     BX, 1			;IF BX == 1 RETURN 1
jne     LBB0_3
mov     AX, 1			;ELSE, SET AX = 1 AND RETURN
jmp     LBB0_4
LBB0_3:
lea     DI, WORD PTR [BX - 1]   ;DI = BX - 1
call    fib			;RETURN FIB(BX-1)
mov     BP, AX			;STORE THIS IN BP
mov     DI, BX
call    fib			;GET FIB(DI - 2)
add     AX, BP		        ;RETURN FIB(DI - 1) + FIB(DI - 2)
LBB0_4:

pop     BX
pop     BP
ret


### Using A Lookup Table

With old computers it was common to use lookup tables to fetch pre-calculated values that would otherwise take some time to compute. The elements of the table are ordered by index, so you can simply create a function that takes an offset as the parameter and returns the element of the array at that offset.

Although lookup tables are very fast, there are some drawbacks to using them. For one, you end up taking up a lot of space. We're wasting a lot of bytes to store very low numbers at the beginning (each takes up 4 bytes regardless of how many digits you see). Unfortunately, when using lookup tables you have very little choice, since trying to conditionally change the scaling of the index would more than likely take more code than encoding all data as the maximum size regardless of the contents, as was done here. This keeps it simple for the CPU, which isn't aware of the intended size of each entry of the table.

For the purpose of this example, assume that both this code and the table are in the .CODE segment.

getfib:
;input: BX = the desired fibonacci number (in other words, the "n" in "F(n)")
;       DS must point to the segment where the fibonacci table is stored
;outputs to DX:AX (DX = high word, AX = low word)
push ds
cmp bx,41  ;bounds check
ja IndexOutOfBounds
shl bx,1
shl bx,1 ;multiply by 4, since this is a table of dwords
mov ax,@code
mov ds,ax
mov si,offset fib
mov ax,[ds:si]    ;fetch the low word into AX
mov dx,2[ds:si]   ;fetch the high word into DX
pop ds
ret

IndexOutOfBounds:
stc             ;set carry to indicate an error
mov ax,0FFFFh   ;return FFFF as the error code
pop ds
ret

;table of the first 41 fibonacci numbers
fib dword 0, 1, 1, 2, 3, 5, 8, 13
dword 21, 34, 55, 89, 144, 233, 377, 610
dword 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657
dword 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269
dword 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986
dword 102334155


## 8th

An iterative solution:

: fibon \ n -- fib(n)
>r 0 1
( tuck n:+ ) \ fib(n-2) fib(n-1) -- fib(n-1) fib(n)
r> n:1- times ;

: fib \ n -- fib(n)
dup 1 n:= if 1 ;; then
fibon nip ;


## ABAP

### Iterative

FORM fibonacci_iter USING index TYPE i
CHANGING number_fib TYPE i.
DATA: lv_old type i,
lv_cur type i.
Do index times.
If sy-index = 1 or sy-index = 2.
lv_cur = 1.
lv_old = 0.
endif.
number_fib = lv_cur + lv_old.
lv_old = lv_cur.
lv_cur = number_fib.
enddo.
ENDFORM.


### Impure Functional

Works with: ABAP version 7.4 SP08 Or above only
cl_demo_output=>display( REDUCE #( INIT fibnm = VALUE stringtab( ( |0| ) ( |1| ) )
n TYPE string
x = 0
y = 1
FOR i = 1 WHILE i <= 100
NEXT n = ( x + y )
fibnm = VALUE #( BASE fibnm ( n ) )
x = y
y = n ) ).


## ACL2

Fast, tail recursive solution:

(defun fast-fib-r (n a b)
(if (or (zp n) (zp (1- n)))
b
(fast-fib-r (1- n) b (+ a b))))

(defun fast-fib (n)
(fast-fib-r n 1 1))

(defun first-fibs-r (n i)
(declare (xargs :measure (nfix (- n i))))
(if (zp (- n i))
nil
(cons (fast-fib i)
(first-fibs-r n (1+ i)))))

(defun first-fibs (n)
(first-fibs-r n 0))

Output:
>(first-fibs 20)
(1 1 2 3 5 8 13 21 34 55 89
144 233 377 610 987 1597 2584 4181 6765)


## Action!

Action! language does not support recursion. Therefore an iterative approach has been proposed.

INT FUNC Fibonacci(INT n)
INT curr,prev,tmp

IF n>=-1 AND n<=1 THEN
RETURN (n)
FI

prev=0
IF n>0 THEN
curr=1
DO
tmp=prev
prev=curr
curr==+tmp
n==-1
UNTIL n=1
OD
ELSE
curr=-1
DO
tmp=prev
prev=curr
curr==+tmp
n==+1
UNTIL n=-1
OD
FI
RETURN (curr)

PROC Main()
BYTE n
INT f

Put(125) ;clear screen

FOR n=0 TO 22
DO
f=Fibonacci(n)
Position(2,n+1)
PrintF("Fib(%I)=%I",n,f)

IF n>0 THEN
f=Fibonacci(-n)
Position(21,n+1)
PrintF("Fib(%I)=%I",-n,f)
FI
OD
RETURN
Output:
Fib(0)=0
Fib(1)=1           Fib(-1)=-1
Fib(2)=1           Fib(-2)=-1
Fib(3)=2           Fib(-3)=-2
Fib(4)=3           Fib(-4)=-3
Fib(5)=5           Fib(-5)=-5
Fib(6)=8           Fib(-6)=-8
Fib(7)=13          Fib(-7)=-13
Fib(8)=21          Fib(-8)=-21
Fib(9)=34          Fib(-9)=-34
Fib(10)=55         Fib(-10)=-55
Fib(11)=89         Fib(-11)=-89
Fib(12)=144        Fib(-12)=-144
Fib(13)=233        Fib(-13)=-233
Fib(14)=377        Fib(-14)=-377
Fib(15)=610        Fib(-15)=-610
Fib(16)=987        Fib(-16)=-987
Fib(17)=1597       Fib(-17)=-1597
Fib(18)=2584       Fib(-18)=-2584
Fib(19)=4181       Fib(-19)=-4181
Fib(20)=6765       Fib(-20)=-6765
Fib(21)=10946      Fib(-21)=-10946
Fib(22)=17711      Fib(-22)=-17711


## ActionScript

public function fib(n:uint):uint
{
if (n < 2)
return n;

return fib(n - 1) + fib(n - 2);
}


### Recursive

with Ada.Text_IO, Ada.Command_Line;

procedure Fib is

function Fib(P: Positive) return Positive is
begin
if P <= 2 then
return 1;
else
return Fib(P-1) + Fib(P-2);
end if;
end Fib;

begin
Ada.Text_IO.Put("Fibonacci(" & Integer'Image(X) & " ) = ");
end Fib;


### Iterative, build-in integers

with Ada.Text_IO;  use Ada.Text_IO;

procedure Test_Fibonacci is
function Fibonacci (N : Natural) return Natural is
This : Natural := 0;
That : Natural := 1;
Sum  : Natural;
begin
for I in 1..N loop
Sum  := This + That;
That := This;
This := Sum;
end loop;
return This;
end Fibonacci;
begin
for N in 0..10 loop
Put_Line (Positive'Image (Fibonacci (N)));
end loop;
end Test_Fibonacci;

Output:
 0
1
1
2
3
5
8
13
21
34
55


### Iterative, long integers

Using the big integer implementation from a cryptographic library [1].

with Ada.Text_IO, Ada.Command_Line, Crypto.Types.Big_Numbers;

procedure Fibonacci is

Bit_Length: Positive := 1 + (696 * X) / 1000;
-- that number of bits is sufficient to store the full result.

package LN is new Crypto.Types.Big_Numbers
(Bit_Length + (32 - Bit_Length mod 32));
-- the actual number of bits has to be a multiple of 32
use LN;

function Fib(P: Positive) return Big_Unsigned is
Previous: Big_Unsigned := Big_Unsigned_Zero;
Result:   Big_Unsigned := Big_Unsigned_One;
Tmp:      Big_Unsigned;
begin
-- Result = 1 = Fibonacci(1)
for I in 1 .. P-1 loop
Tmp := Result;
Result := Previous + Result;
Previous := Tmp;
-- Result = Fibonacci(I+1))
end loop;
return Result;
end Fib;

begin
Ada.Text_IO.Put("Fibonacci(" & Integer'Image(X) & " ) = ");
end Fibonacci;

Output:
> ./fibonacci 777
Fibonacci( 777 ) = 1081213530912648191985419587942084110095342850438593857649766278346130479286685742885693301250359913460718567974798268702550329302771992851392180275594318434818082

### Fast method using fast matrix exponentiation

with ada.text_io;

procedure fast_fibo is
-- We work with biggest natural integers in a 64 bits machine
type Big_Int is mod 2**64;

-- We provide an index type for accessing the fibonacci sequence terms
type Index is new Big_Int;

-- fibo is a generic function that needs a modulus type since it will return
-- the n'th term of the fibonacci sequence modulus this type (use Big_Int to get the
-- expected behaviour in this particular task)
generic
type ring_element is mod <>;
with function "*" (a, b : ring_element) return ring_element is <>;
function fibo (n : Index) return ring_element;
function fibo (n : Index) return ring_element is

type matrix is array (1 .. 2, 1 .. 2) of ring_element;

-- f is the matrix you apply to a column containing (F_n, F_{n+1}) to get
-- the next one containing (F_{n+1},F_{n+2})
-- could be a more general matrix (given as a generic parameter) to deal with
-- other linear sequences of order 2
f : constant matrix := (1 => (0, 1), 2 => (1, 1));

function "*" (a, b : matrix) return matrix is
(1 => (a(1,1)*b(1,1)+a(1,2)*b(2,1), a(1,1)*b(1,2)+a(1,2)*b(2,2)),
2 => (a(2,1)*b(1,1)+a(2,2)*b(2,1), a(2,1)*b(1,2)+a(2,2)*b(2,2)));

function square (m : matrix) return matrix is (m * m);

-- Fast_Pow could be non recursive but it doesn't really matter since
-- the number of calls is bounded up by the size (in bits) of Big_Int (e.g 64)
function fast_pow (m : matrix; n : Index) return matrix is
(if n = 0 then (1 => (1, 0), 2 => (0, 1)) -- = identity matrix
elsif n mod 2 = 0 then square (fast_pow (m, n / 2))
else m * square (fast_pow (m, n / 2)));

begin
return fast_pow (f, n)(2, 1);
end fibo;

function Big_Int_Fibo is new fibo (Big_Int);
begin
-- calculate instantly F_n with n=10^15 (modulus 2^64 )
put_line (Big_Int_Fibo (10**15)'img);
end fast_fibo;


### Recursive

#include "totvs.ch"
User Function fibb(a,b,n)
return(if(--n>0,fibb(b,a+b,n),a))

### Iterative

#include "totvs.ch"
User Function fibb(n)
local fnow:=0, fnext:=1, tempf
while (--n>0)
tempf:=fnow+fnext
fnow:=fnext
fnext:=tempf
end while
return(fnext)

## Aime

integer
fibs(integer n)
{
integer w;

if (n == 0) {
w = 0;
} elif (n == 1) {
w = 1;
} else {
integer a, b, i;

i = 1;
a = 0;
b = 1;
while (i < n) {
w = a + b;
a = b;
b = w;
i += 1;
}
}

return w;
}

## ALGOL 60

Works with: A60
begin
comment Fibonacci sequence;
integer procedure fibonacci(n); value n; integer n;
begin
integer i, fn, fn1, fn2;
fn2 := 1;
fn1 := 0;
fn  := 0;
for i := 1 step 1 until n do begin
fn  := fn1 + fn2;
fn2 := fn1;
fn1 := fn
end;
fibonacci := fn
end fibonacci;

integer i;
for i := 0 step 1 until 20 do outinteger(1,fibonacci(i))
end
Output:
 0  1  1  2  3  5  8  13  21  34  55  89  144  233  377  610  987  1597  2584  4181  6765


## ALGOL 68

### Analytic

Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d
PROC analytic fibonacci = (LONG INT n)LONG INT:(
LONG REAL sqrt 5 = long sqrt(5);
LONG REAL p = (1 + sqrt 5) / 2;
LONG REAL q = 1/p;
ROUND( (p**n + q**n) / sqrt 5 )
);

FOR i FROM 1 TO 30 WHILE
print(whole(analytic fibonacci(i),0));
# WHILE # i /= 30 DO
print(", ")
OD;
print(new line)
Output:
1, 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


### Iterative

Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d
PROC iterative fibonacci = (INT n)INT:
CASE n+1 IN
0, 1, 1, 2, 3, 5
OUT
INT even:=3, odd:=5;
FOR i FROM odd+1 TO n DO
(ODD i|odd|even) := odd + even
OD;
(ODD n|odd|even)
ESAC;

FOR i FROM 0 TO 30 WHILE
print(whole(iterative fibonacci(i),0));
# WHILE # i /= 30 DO
print(", ")
OD;
print(new line)
Output:
0, 1, 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


### Recursive

Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d
PROC recursive fibonacci = (INT n)INT:
( n < 2 | n | fib(n-1) + fib(n-2));

### Generative

Translation of: Python – Note: This specimen retains the original Python coding style.
Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d
MODE YIELDINT = PROC(INT)VOID;

PROC gen fibonacci = (INT n, YIELDINT yield)VOID: (
INT even:=0, odd:=1;
yield(even);
yield(odd);
FOR i FROM odd+1 TO n DO
yield( (ODD i|odd|even) := odd + even )
OD
);

main:(
# FOR INT n IN # gen fibonacci(30, # ) DO ( #
##   (INT n)VOID:(
print((" ",whole(n,0)))
# OD # ));
print(new line)
)
Output:
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


### Array (Table) Lookup

Works with: ALGOL 68 version Revision 1 - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d

This uses a pre-generated list, requiring much less run-time processor usage, but assumes that INT is only 31 bits wide.

[]INT const fibonacci = []INT( -1836311903, 1134903170,
-701408733, 433494437, -267914296, 165580141, -102334155,
63245986, -39088169, 24157817, -14930352, 9227465, -5702887,
3524578, -2178309, 1346269, -832040, 514229, -317811, 196418,
-121393, 75025, -46368, 28657, -17711, 10946, -6765, 4181,
-2584, 1597, -987, 610, -377, 233, -144, 89, -55, 34, -21, 13,
-8, 5, -3, 2, -1, 1, 0, 1, 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
)[@-46];

PROC VOID value error := stop;

PROC lookup fibonacci = (INT i)INT: (
IF LWB const fibonacci <= i AND i<= UPB const fibonacci THEN
const fibonacci[i]
ELSE
value error; SKIP
FI
);

FOR i FROM 0 TO 30 WHILE
print(whole(lookup fibonacci(i),0));
# WHILE # i /= 30 DO
print(", ")
OD;
print(new line)
Output:
0, 1, 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


## ALGOL W

begin
% return the nth Fibonacci number %
integer procedure Fibonacci( integer value n ) ;
begin
integer fn, fn1, fn2;
fn2 := 1;
fn1 := 0;
fn  := 0;
for i := 1 until n do begin
fn  := fn1 + fn2;
fn2 := fn1;
fn1 := fn
end ;
fn
end Fibonacci ;

for i := 0 until 10 do writeon( i_w := 3, s_w := 0, Fibonacci( i ) )

end.
Output:
  0  1  1  2  3  5  8 13 21 34 55


## ALGOL-M

Note that the 21st Fibonacci number (= 10946) is the largest that can be calculated without overflowing ALGOL-M's integer data type.

#### Iterative

INTEGER FUNCTION FIBONACCI( X ); INTEGER X;
BEGIN
INTEGER M, N, A, I;
M := 0;
N := 1;
FOR I := 2 STEP 1 UNTIL X DO
BEGIN
A := N;
N := M + N;
M := A;
END;
FIBONACCI := N;
END;

#### Naively recursive

INTEGER FUNCTION FIBONACCI( X ); INTEGER X;
BEGIN
IF X < 3 THEN
FIBONACCI := 1
ELSE
FIBONACCI := FIBONACCI( X - 2 ) + FIBONACCI( X - 1 );
END;

## Alore

def fib(n as Int) as Int
if n < 2
return 1
end
return fib(n-1) + fib(n-2)
end

## Amazing Hopper

Analitic, Recursive and Iterative mode.

#include <hbasic.h>

#define TERM1    1.61803398874989
#define TERM2    -0.61803398874989

#context get Fibonacci number with analitic mode
GetArgs(n)
get Inv of (M_SQRT5), Mul by( Pow (TERM 1, n), Minus( Pow(TERM 2, n) )  );
then Return\\

#proto fibonacci_recursive(__X__)
#synon _fibonacci_recursive    getFibonaccinumberwithrecursivemodeof

#proto fibonacci_iterative(__X__)
#synon _fibonacci_iterative    getFibonaccinumberwithiterativemodeof

Begin
Option Stack 1024

get Arg Number(2, n), and Take( n );
then, get Fibonacci number with analitic mode, and Print It with a Newl.
secondly, get Fibonacci number with recursive mode of(n), and Print It with a Newl.
finally, get Fibonacci number with iterative mode of (n), and Print It with a Newl.
End

Subrutines

fibonacci_recursive(n)
Iif ( var(n) Is Le? (2), 1 , \
get Fibonacci number with recursive mode of( var(n) Minus (1));\
get Fibonacci number with recursive mode of( var(n) Minus (2)); and Add It )
Return

fibonacci_iterative(n)
A=0
B=1
For Up( I:=2, n, 1 )
C=B
Let ( B: = var(A) Plus (B) )
A=C
Next
Return(B)
Output:
$hopper src/fibo1.bas 25 75025 75025 75025  ## AntLang /Sequence fib:{<0;1> {x,<x[-1]+x[-2]>}/ range[x]} /nth fibn:{fib[x][x]} ## Apex /* author: snugsfbay date: March 3, 2016 description: Create a list of x numbers in the Fibonacci sequence. - user may specify the length of the list - enforces a minimum of 2 numbers in the sequence because any fewer is not a sequence - enforces a maximum of 47 because further values are too large for integer data type - Fibonacci sequence always starts with 0 and 1 by definition */ public class FibNumbers{ final static Integer MIN = 2; //minimum length of sequence final static Integer MAX = 47; //maximum length of sequence /* description: method to create a list of numbers in the Fibonacci sequence param: user specified integer representing length of sequence should be 2-47, inclusive. - Sequence starts with 0 and 1 by definition so the minimum length could be as low as 2. - For 48th number in sequence or greater, code would require a Long data type rather than an Integer. return: list of integers in sequence. */ public static List<Integer> makeSeq(Integer len){ List<Integer> fib = new List<Integer>{0,1}; // initialize list with first two values Integer i; if(len<MIN || len==null || len>MAX) { if (len>MAX){ len=MAX; //set length to maximum if user entered too high a value }else{ len=MIN; //set length to minimum if user entered too low a value or none } } //This could be refactored using teneray operator, but we want code coverage to be reflected for each condition //start with initial list size to find previous two values in the sequence, continue incrementing until list reaches user defined length for(i=fib.size(); i<len; i++){ fib.add(fib[i-1]+fib[i-2]); //create new number based on previous numbers and add that to the list } return fib; } } ## APL #### Naive Recursive Works with: Dyalog APL fib←{⍵≤1:⍵ ⋄ (∇ ⍵-1)+∇ ⍵-2}  Read this as: In the variable "fib", store the function that says, if the argument is less than or equal to 1, return the argument. Else, calculate the value you get when you recursively call the current function with the argument of the current argument minus one and add that to the value you get when you recursively call the current function with the argument of the current function minus two. This naive solution requires Dyalog APL because GNU APL does not support this syntax for conditional guards. #### Array Works with: Dyalog APL Works with: GNU APL Since APL is an array language we'll use the following identity: ${\displaystyle \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}.}$ In APL: ↑+.×/N/⊂2 2⍴1 1 1 0  Plugging in 4 for N gives the following result: ${\displaystyle \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}}$ Here's what happens: We replicate the 2-by-2 matrix N times and then apply inner product-replication. The First removes the shell from the Enclose. At this point we're basically done, but we need to pick out only ${\displaystyle F_n}$ in order to complete the task. Here's one way: ↑0 1↓↑+.×/N/⊂2 2⍴1 1 1 0  #### Analytic Works with: Dyalog APL Works with: GNU APL An alternative approach, using Binet's formula (which was apparently known long before Binet): ⌊.5+(((1+PHI)÷2)*⍳N)÷PHI←5*.5  ## AppleScript ### Imperative set fibs to {} set x to (text returned of (display dialog "What fibbonaci number do you want?" default answer "3")) set x to x as integer repeat with y from 1 to x if (y = 1 or y = 2) then copy 1 to the end of fibs else copy ((item (y - 1) of fibs) + (item (y - 2) of fibs)) to the end of fibs end if end repeat return item x of fibs  ### Functional The simple recursive version is famously slow: on fib(n) if n < 1 then 0 else if n < 3 then 1 else fib(n - 2) + fib(n - 1) end if end fib  but we can combine enumFromTo(m, n) with the accumulator of a higher-order fold/reduce function to memoize the series: Translation of: JavaScript (ES6 memoized fold example) Translation of: Haskell (Memoized fold example) -------------------- FIBONACCI SEQUENCE -------------------- -- fib :: Int -> Int on fib(n) -- lastTwo : (Int, Int) -> (Int, Int) script lastTwo on |λ|([a, b]) [b, a + b] end |λ| end script item 1 of foldl(lastTwo, {0, 1}, enumFromTo(1, n)) end fib --------------------------- TEST --------------------------- on run fib(32) --> 2178309 end run -------------------- GENERIC FUNCTIONS --------------------- -- enumFromTo :: Int -> Int -> [Int] on enumFromTo(m, n) if m ≤ n then set lst to {} repeat with i from m to n set end of lst to i end repeat lst else {} end if end enumFromTo -- foldl :: (a -> b -> a) -> a -> [b] -> a on 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 tell end foldl -- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f) if class of f is script then f else script property |λ| : f end script end if end mReturn  Output: 2178309 ## Arendelle ( fibonacci , 1; 1 ) [ 98 , // 100 numbers of fibonacci ( fibonacci[ @fibonacci? ] , @fibonacci[ @fibonacci - 1 ] + @fibonacci[ @fibonacci - 2 ] ) "Index: | @fibonacci? | => | @fibonacci[ @fibonacci? - 1 ] |" ] ## ARM Assembly Expects to be called with ${\displaystyle n}$ in R0, and will return ${\displaystyle f(n)}$ in the same register. fibonacci: push {r1-r3} mov r1, #0 mov r2, #1 fibloop: mov r3, r2 add r2, r1, r2 mov r1, r3 sub r0, r0, #1 cmp r0, #1 bne fibloop mov r0, r2 pop {r1-r3} mov pc, lr ## ArnoldC IT'S SHOWTIME HEY CHRISTMAS TREE f1 YOU SET US UP @I LIED TALK TO THE HAND f1 HEY CHRISTMAS TREE f2 YOU SET US UP @NO PROBLEMO HEY CHRISTMAS TREE f3 YOU SET US UP @I LIED STICK AROUND @NO PROBLEMO GET TO THE CHOPPER f3 HERE IS MY INVITATION f1 GET UP f2 ENOUGH TALK TALK TO THE HAND f3 GET TO THE CHOPPER f1 HERE IS MY INVITATION f2 ENOUGH TALK GET TO THE CHOPPER f2 HERE IS MY INVITATION f3 ENOUGH TALK CHILL YOU HAVE BEEN TERMINATED ## Arturo ### Recursive fib:$[x][
if? x<2 [1]
else [(fib x-1) + (fib x-2)]
]

loop 1..25 [x][
print ["Fibonacci of" x "=" fib x]
]

## AsciiDots

/--#$--\ | | >-*>{+}/ | \+-/ 1 | # 1 | # | | . . ## ATS ### Recursive fun fib_rec(n: int): int = if n >= 2 then fib_rec(n-1) + fib_rec(n-2) else n ### Iterative (* ** This one is also referred to as being tail-recursive *) fun fib_trec(n: int): int = if n > 0 then (fix loop (i:int, r0:int, r1:int): int => if i > 1 then loop (i-1, r1, r0+r1) else r1)(n, 0, 1) else 0 ### Iterative and Verified (* ** This implementation is verified! *) dataprop FIB (int, int) = | FIB0 (0, 0) | FIB1 (1, 1) | {n:nat} {r0,r1:int} FIB2 (n+2, r0+r1) of (FIB (n, r0), FIB (n+1, r1)) // end of [FIB] // end of [dataprop] fun fibats{n:nat} (n: int (n)) : [r:int] (FIB (n, r) | int r) = let fun loop {i:nat | i <= n}{r0,r1:int} ( pf0: FIB (i, r0), pf1: FIB (i+1, r1) | ni: int (n-i), r0: int r0, r1: int r1 ) : [r:int] (FIB (n, r) | int r) = if (ni > 0) then loop{i+1}(pf1, FIB2 (pf0, pf1) | ni - 1, r1, r0 + r1) else (pf0 | r0) // end of [if] // end of [loop] in loop {0} (FIB0 (), FIB1 () | n, 0, 1) end // end of [fibats] ### Matrix-based (* ****** ****** *) // // How to compile: // patscc -o fib fib.dats // (* ****** ****** *) // #include "share/atspre_staload.hats" // (* ****** ****** *) // abst@ype int3_t0ype = (int, int, int) // typedef int3 = int3_t0ype // (* ****** ****** *) extern fun int3 : (int, int, int) -<> int3 extern fun int3_1 : int3 -<> int extern fun mul_int3_int3: (int3, int3) -<> int3 (* ****** ****** *) local assume int3_t0ype = (int, int, int) in (* in-of-local *) // implement int3 (x, y, z) = @(x, y, z) // implement int3_1 (xyz) = xyz.1 // implement mul_int3_int3 ( @(a,b,c), @(d,e,f) ) = (a*d + b*e, a*e + b*f, b*e + c*f) // end // end of [local] (* ****** ****** *) // implement gnumber_int<int3> (n) = int3(n, 0, n) // implement gmul_val<int3> = mul_int3_int3 // (* ****** ****** *) // fun fib (n: intGte(0)): int = int3_1(gpow_int_val<int3> (n, int3(1, 1, 0))) // (* ****** ****** *) implement main0 () = { // val N = 10 val () = println! ("fib(", N, ") = ", fib(N)) val N = 20 val () = println! ("fib(", N, ") = ", fib(N)) val N = 30 val () = println! ("fib(", N, ") = ", fib(N)) val N = 40 val () = println! ("fib(", N, ") = ", fib(N)) // } (* end of [main0] *) ## AutoHotkey Search autohotkey.com: sequence ### Iterative Translation of: C Loop, 5 MsgBox % fib(A_Index) Return fib(n) { If (n < 2) Return n i := last := this := 1 While (i <= n) { new := last + this last := this this := new i++ } Return this }  ### Recursive and iterative Source: AutoHotkey forum by Laszlo /* Important note: the recursive version would be very slow without a global or static array. The iterative version handles also negative arguments properly. */ FibR(n) { ; n-th Fibonacci number (n>=0, recursive with static array Fibo) Static Return n<2 ? n : Fibo%n% ? Fibo%n% : Fibo%n% := FibR(n-1)+FibR(n-2) } Fib(n) { ; n-th Fibonacci number (n < 0 OK, iterative) a := 0, b := 1 Loop % abs(n)-1 c := b, b += a, a := c Return n=0 ? 0 : n>0 || n&1 ? b : -b }  ## AutoIt ### Iterative #AutoIt Version: 3.2.10.0$n0 = 0
$n1 = 1$n = 10
MsgBox (0,"Iterative Fibonacci ", it_febo($n0,$n1,$n)) Func it_febo($n_0,$n_1,$N)
$first =$n_0
$second =$n_1
$next =$first + $second$febo = 0
For $i = 1 To$N-3
$first =$second
$second =$next
$next =$first + $second Next if$n==0 Then
$febo = 0 ElseIf$n==1 Then
$febo =$n_0
ElseIf $n==2 Then$febo = $n_1 Else$febo = $next EndIf Return$febo
EndFunc


### Recursive

#AutoIt Version: 3.2.10.0
$n0 = 0$n1 = 1
$n = 10 MsgBox (0,"Recursive Fibonacci ", rec_febo($n0,$n1,$n))
Func rec_febo($r_0,$r_1,$R) if$R<3 Then
if $R==2 Then Return$r_1
ElseIf $R==1 Then Return$r_0
ElseIf $R==0 Then Return 0 EndIf Return$R
Else
Return rec_febo($r_0,$r_1,$R-1) + rec_febo($r_0,$r_1,$R-2)
EndIf
EndFunc


## AWK

As in many examples, this one-liner contains the function as well as testing with input from stdin, output to stdout.

$awk 'func fib(n){return(n<2?n:fib(n-1)+fib(n-2))}{print "fib("$1")="fib($1)}' 10 fib(10)=55  ## Axe A recursive solution is not practical in Axe because there is no concept of variable scope in Axe. Iterative solution: Lbl FIB r₁→N 0→I 1→J For(K,1,N) I+J→T J→I T→J End J Return ## Babel In Babel, we can define fib using a stack-based approach that is not recursive: fib { <- 0 1 { dup <- + -> swap } -> times zap } < foo x < puts x in foo. In this case, x is the code list between the curly-braces. This is how you define callable code in Babel. The definition works by initializing the stack with 0, 1. On each iteration of the times loop, the function duplicates the top element. In the first iteration, this gives 0, 1, 1. Then it moves down the stack with the <- operator, giving 0, 1 again. It adds, giving 1. then it moves back up the stack, giving 1, 1. Then it swaps. On the next iteration this gives: 1, 1, 1 (dup) 1, 1, (<-) 2 (+) 2, 1 (->) 1, 2 (swap)  And so on. To test fib: {19 iter - fib !} 20 times collect ! lsnum ! Output: ( 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 ) ## bash ### Iterative $ fib=1;j=1;while((fib<100));do echo $fib;((k=fib+j,fib=j,j=k));done  1 1 2 3 5 8 13 21 34 55 89  ### Recursive fib() { if [$1 -le 0 ]
then
echo 0
return 0
fi
if [ $1 -le 2 ] then echo 1 else a=$(fib $[$1-1])
b=$(fib$[$1-2]) echo$(($a+$b))
fi
}


## BASIC

### Applesoft BASIC

Same code as Commodore BASIC

Entering a value of N > 183, produces an error message:

?OVERFLOW ERROR IN 220

### BASIC256

# Basic-256 ver 1.1.4
# iterative Fibonacci sequence
# Matches sequence A000045 in the OEIS, https://oeis.org/A000045/list

# Return the Nth Fibonacci number

input "N = ",f
limit = 500                        # set upper limit - can be changed, removed
f = int(f)
if f > limit then f = limit
a = 0 : b = 1 : c = 0 : n = 0      # initial values

while n < f
print n + chr(9) + c   # chr(9) = tab
a = b
b = c
c = a + b
n += 1

end while

print " "
print n + chr(9) + c

### Commodore BASIC

100 PRINT CHR$(147); CHR$(18); "****      FIBONACCI GENERATOR       ****"
110 INPUT "MIN, MAX"; N1, N2
120 IF N1 > N2 THEN T = N1: N1 = N2: N2 = T
130 A = 0: B = 1: S = SGN(N1)
140 FOR I = S TO N1 STEP S
150 : IF S > 0 THEN T = A + B: A = B: B = T
160 : IF S < 0 THEN T = B - A: B = A: A = T
170 NEXT I
180 PRINT
190 PRINT STR$(A); : REM STR$() PREVENTS TRAILING SPACE
200 IF N2 = N1 THEN 250
210 FOR I = N1 + 1 TO N2
220 : T = A + B: A = B: B = T
230 : PRINT ","; STR$(A); 240 NEXT I 250 PRINT  Output: **** FIBONACCI GENERATOR **** MIN, MAX? -6,6 -8, 5,-3, 2,-1, 1, 0, 1, 1, 2, 3, 5, 8 READY. ### Integer BASIC Only works with quite small values of ${\displaystyle n}$.  10 INPUT N 20 A=0 30 B=1 40 FOR I=2 TO N 50 C=B 60 B=A+B 70 A=C 80 NEXT I 90 PRINT B 100 END  ### IS-BASIC 100 PROGRAM "Fibonac.bas" 110 FOR I=0 TO 20 120 PRINT "F";I,FIB(I) 130 NEXT 140 DEF FIB(N) 150 NUMERIC I 160 LET A=0:LET B=1 170 FOR I=1 TO N 180 LET T=A+B:LET A=B:LET B=T 190 NEXT 200 LET FIB=A 210 END DEF ### QBasic Works with: QBasic Works with: FreeBASIC #### Iterative FUNCTION itFib (n) n1 = 0 n2 = 1 FOR k = 1 TO ABS(n) sum = n1 + n2 n1 = n2 n2 = sum NEXT k IF n < 0 THEN itFib = n1 * ((-1) ^ ((-n) + 1)) ELSE itFib = n1 END IF END FUNCTION  Next version calculates each value once, as needed, and stores the results in an array for later retreival (due to the use of REDIM PRESERVE, it requires QuickBASIC 4.5 or newer): DECLARE FUNCTION fibonacci& (n AS INTEGER) REDIM SHARED fibNum(1) AS LONG fibNum(1) = 1 '*****sample inputs***** PRINT fibonacci(0) 'no calculation needed PRINT fibonacci(13) 'figure F(2)..F(13) PRINT fibonacci(-42) 'figure F(14)..F(42) PRINT fibonacci(47) 'error: too big '*****sample inputs***** FUNCTION fibonacci& (n AS INTEGER) DIM a AS INTEGER a = ABS(n) SELECT CASE a CASE 0 TO 46 SHARED fibNum() AS LONG DIM u AS INTEGER, L0 AS INTEGER u = UBOUND(fibNum) IF a > u THEN REDIM PRESERVE fibNum(a) AS LONG FOR L0 = u + 1 TO a fibNum(L0) = fibNum(L0 - 1) + fibNum(L0 - 2) NEXT END IF IF n < 0 THEN fibonacci = fibNum(a) * ((-1) ^ (a + 1)) ELSE fibonacci = fibNum(n) END IF CASE ELSE 'limited to signed 32-bit int (LONG) 'F(47)=&hB11924E1 ERROR 6 'overflow END SELECT END FUNCTION  Output: (unhandled error in final input prevents output)  0 233 -267914296  #### Recursive This example can't handle n < 0. FUNCTION recFib (n) IF (n < 2) THEN recFib = n ELSE recFib = recFib(n - 1) + recFib(n - 2) END IF END FUNCTION  #### Array (Table) Lookup This uses a pre-generated list, requiring much less run-time processor usage. (Since the sequence never changes, this is probably the best way to do this in "the real world". The same applies to other sequences like prime numbers, and numbers like pi and e.) DATA -1836311903,1134903170,-701408733,433494437,-267914296,165580141,-102334155 DATA 63245986,-39088169,24157817,-14930352,9227465,-5702887,3524578,-2178309 DATA 1346269,-832040,514229,-317811,196418,-121393,75025,-46368,28657,-17711 DATA 10946,-6765,4181,-2584,1597,-987,610,-377,233,-144,89,-55,34,-21,13,-8,5,-3 DATA 2,-1,1,0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765 DATA 10946,17711,28657,46368,75025,121393,196418,317811,514229,832040,1346269 DATA 2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986 DATA 102334155,165580141,267914296,433494437,701408733,1134903170,1836311903 DIM fibNum(-46 TO 46) AS LONG FOR n = -46 TO 46 READ fibNum(n) NEXT '*****sample inputs***** FOR n = -46 TO 46 PRINT fibNum(n), NEXT PRINT '*****sample inputs*****  #### QB64 Fibonacci from Russia DIM F(80) AS DOUBLE 'FibRus.bas DANILIN F(1) = 0: F(2) = 1 'OPEN "FibRus.txt" FOR OUTPUT AS #1 FOR i = 3 TO 80 F(i) = F(i-1)+F(i-2) NEXT i FOR i = 1 TO 80 f$ = STR$(F(i)): LF = 22 - LEN(f$)
n$= "" FOR j = 1 TO LF: n$ = " " + n$: NEXT f$ = n$+ f$
PRINT i, f$: ' PRINT #1, i, f$
NEXT i

Output:
1                                 0
2                                 1
3                                 1
4                                 2
5                                 3
6                                 5
7                                 8
8                                13
9                                21
...
24                            28657
25                            46368
26                            75025
...
36                          9227465
37                         14930352
38                         24157817
...
48                       2971215073
49                       4807526976
50                       7778742049
...
60                     956722026041
61                    1548008755920
62                    2504730781961
...
76                 2111485077978050
77                 3416454622906707
78                 5527939700884757
79                 8944394323791464
80            1.447233402467622D+16

### Sinclair ZX81 BASIC

#### Analytic

 10 INPUT N
20 PRINT INT (0.5+(((SQR 5+1)/2)**N)/SQR 5)


#### Iterative

 10 INPUT N
20 LET A=0
30 LET B=1
40 FOR I=2 TO N
50 LET C=B
60 LET B=A+B
70 LET A=C
80 NEXT I
90 PRINT B


#### Tail recursive

 10 INPUT N
20 LET A=0
30 LET B=1
40 GOSUB 70
50 PRINT B
60 STOP
70 IF N=1 THEN RETURN
80 LET C=B
90 LET B=A+B
100 LET A=C
110 LET N=N-1
120 GOSUB 70
130 RETURN


## Batch File

Recursive version

::fibo.cmd
@echo off
if "%1" equ "" goto :eof
call :fib %1
echo %errorlevel%
goto :eof

:fib
setlocal enabledelayedexpansion
if %1 geq 2 goto :ge2
exit /b %1

:ge2
set /a r1 = %1 - 1
set /a r2 = %1 - 2
call :fib !r1!
set r1=%errorlevel%
call :fib !r2!
set r2=%errorlevel%
set /a r0 = r1 + r2
exit /b !r0!

Output:
>for /L %i in (1,5,20) do fibo.cmd %i

>fibo.cmd 1
1

>fibo.cmd 6
8

>fibo.cmd 11
89

>fibo.cmd 16
987

## Battlestar

// Fibonacci sequence, recursive version
fun fibb
loop
a = funparam[0]
break (a < 2)

a--

// Save "a" while calling fibb
a -> stack

// Set the parameter and call fibb
funparam[0] = a
call fibb

// Handle the return value and restore "a"
b = funparam[0]
stack -> a

// Save "b" while calling fibb again
b -> stack

a--

// Set the parameter and call fibb
funparam[0] = a
call fibb

// Handle the return value and restore "b"
c = funparam[0]
stack -> b

// Sum the results
b += c
a = b

funparam[0] = a

break
end
end

// vim: set syntax=c ts=4 sw=4 et:


## BBC BASIC

      PRINT FNfibonacci_r(1),  FNfibonacci_i(1)
PRINT FNfibonacci_r(13), FNfibonacci_i(13)
PRINT FNfibonacci_r(26), FNfibonacci_i(26)
END

DEF FNfibonacci_r(N)
IF N < 2 THEN = N
= FNfibonacci_r(N-1) + FNfibonacci_r(N-2)

DEF FNfibonacci_i(N)
LOCAL F, I, P, T
IF N < 2 THEN = N
P = 1
FOR I = 1 TO N
T = F
F += P
P = T
NEXT
= F

Output:
         1         1
233       233
121393    121393

## bc

### iterative

#! /usr/bin/bc -q

define fib(x) {
if (x <= 0) return 0;
if (x == 1) return 1;

a = 0;
b = 1;
for (i = 1; i < x; i++) {
c = a+b; a = b; b = c;
}
return c;
}
fib(1000)
quit


## BCPL

get "libhdr"

let fib(n) = n<=1 -> n, valof
$( let a=0 and b=1 for i=2 to n$(  let c=a
a := b
b := a+c
$) resultis b$)

let start() be
for i=0 to 10 do
writef("F_%N*T= %N*N", i, fib(i))
Output:
F_0     = 0
F_1     = 1
F_2     = 1
F_3     = 2
F_4     = 3
F_5     = 5
F_6     = 8
F_7     = 13
F_8     = 21
F_9     = 34
F_10    = 55

## beeswax

                        #>'#{;
_Enter n: TNFib({)=X~P~K#{;
#>~P~L#MM@>+@'q@{;
b~@M<

Example output:

Notice the UInt64 wrap-around at Fib(94)!

julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i0

Fib(0)=0
Program finished!

julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i10

Fib(10)=55
Program finished!

julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i92

Fib(92)=7540113804746346429
Program finished!

julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i93

Fib(93)=12200160415121876738
Program finished!

julia> beeswax("n-th Fibonacci number.bswx")
Enter n: i94

Fib(94)=1293530146158671551
Program finished!


## Befunge

00:.1:.>:"@"8**++\1+:67+#@_v
^ .:\/*8"@"\%*8"@":\ <


## BlitzMax

local a:int = 0, b:int = 1, c:int = 1, n:int

n = int(input( "Enter n: "))
if n = 0 then
print 0
end
else if n = 1
print 1
end
end if

while n>2
a = b
b = c
c = a + b
n = n - 1
wend
print c


## Blue

: fib ( nth:ecx -- result:edi ) 1 0
: compute ( times:ecx accum:eax scratch:edi -- result:edi ) xadd latest loop ;

: example ( -- ) 11 fib drop ;

## BQN

All given functions return the nth element in the sequence.

### Recursive

A primitive recursive can be done with predicates:

Fib ← {𝕩>1 ? (𝕊 𝕩-1) + 𝕊 𝕩-2; 𝕩}

Or, it can be done with the Choose(◶) modifier:

Fib2 ← {(𝕩-1) (𝕩>1)◶⟨𝕩, +○𝕊⟩ 𝕩-2}

### Iterative

An iterative solution can be made with the Repeat(⍟) modifier:

{⊑(+⌽)⍟𝕩 0‿1}

## Bracmat

### Recursive

fib=.!arg:<2|fib$(!arg+-2)+fib$(!arg+-1)
 fib$30 832040  ### Iterative (fib= last i this new . !arg:<2 | 0:?last:?i & 1:?this & whl ' ( !i+1:<!arg:?i & !last+!this:?new & !this:?last & !new:?this ) & !this )  fib$777
1081213530912648191985419587942084110095342850438593857649766278346130479286685742885693301250359913460718567974798268702550329302771992851392180275594318434818082


## Brainf***

Works with: Brainf*** version implementations with unbounded cell size

The first cell contains n (10), the second cell will contain fib(n) (55), and the third cell will contain fib(n-1) (34).

++++++++++
>>+<<[->[->+>+<<]>[-<+>]>[-<+>]<<<]


The following generates n fibonacci numbers and prints them, though not in ascii. It does have a limit due to the cells usually being 1 byte in size.

+++++ +++++	#0 set to n
>> +		Init #2 to 1
<<
[
-	#Decrement counter in #0
>>.	Notice: This doesn't print it in ascii
To look at results you can pipe into a file and look with a hex editor

Copying sequence to save #2 in #4 using #5 as restore space
>>[-]	Move to #4 and clear
>[-]	Clear #5
<<<	#2
[	Move loop
- >> + > + <<<	Subtract #2 and add #4 and #5
]
>>>
[	Restore loop
- <<< + >>>	Subtract from #5 and add to #2
]

<<<<	Back to #1
Non destructive add sequence using #3 as restore value
- > + > + <<	Subtract #1 and add to value #2 and restore space #3
]
>>
[	Loop to restore #1 from #3
- << + >>	Subtract from restore space #3 and add in #1
]

<< [-]	Clear #1
>>>
[	Loop to move #4 to #1
- <<< + >>>	Subtract from #4 and add to #1
]
<<<<	Back to #0
]


## Brat

### Recursive

fibonacci = { x |
true? x < 2, x, { fibonacci(x - 1) + fibonacci(x - 2) }
}

### Tail Recursive

fib_aux = { x, next, result |
true? x == 0,
result,
{ fib_aux x - 1, next + result, next }
}

fibonacci = { x |
fib_aux x, 1, 0
}

### Memoization

cache = hash.new

fibonacci = { x |
true? cache.key?(x)
{ cache[x] }
{true? x < 2, x, { cache[x] = fibonacci(x - 1) + fibonacci(x - 2) }}
}

## Burlesque

{0 1}{^^++[+[-^^-]\/}30.*\[e!vv
0 1{{.+}c!}{1000.<}w!

## C

### Recursive

long long fibb(long long a, long long b, int n) {
return (--n>0)?(fibb(b, a+b, n)):(a);
}


### Iterative

long long int fibb(int n) {
int fnow = 0, fnext = 1, tempf;
while(--n>0){
tempf = fnow + fnext;
fnow = fnext;
fnext = tempf;
}
return fnext;
}


### Analytic

#include <tgmath.h>
#define PHI ((1 + sqrt(5))/2)

long long unsigned fib(unsigned n) {
return floor( (pow(PHI, n) - pow(1 - PHI, n))/sqrt(5) );
}


### Generative

Translation of: Python
Works with: gcc version version 4.1.2 20080704 (Red Hat 4.1.2-44)
#include <stdio.h>
typedef enum{false=0, true=!0} bool;
typedef void iterator;

#include <setjmp.h>
/* declare label otherwise it is not visible in sub-scope */
#define LABEL(label) jmp_buf label; if(setjmp(label))goto label;
#define GOTO(label) longjmp(label, true)

/* the following line is the only time I have ever required "auto" */
#define FOR(i, iterator) { auto bool lambda(i); yield_init = (void *)&lambda; iterator; bool lambda(i)
#define DO {
#define     YIELD(x) if(!yield(x))return
#define     BREAK    return false
#define     CONTINUE return true
#define OD CONTINUE; } }

static volatile void *yield_init; /* not thread safe */
#define YIELDS(type) bool (*yield)(type) = yield_init

iterator fibonacci(int stop){
YIELDS(int);
int f[] = {0, 1};
int i;
for(i=0; i<stop; i++){
YIELD(f[i%2]);
f[i%2]=f[0]+f[1];
}
}

main(){
printf("fibonacci: ");
FOR(int i, fibonacci(16)) DO
printf("%d, ",i);
OD;
printf("...\n");
}

Output:
fibonacci: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, ...


### Fast method for a single large value

#include <stdlib.h>
#include <stdio.h>
#include <gmp.h>

typedef struct node node;
struct node {
int n;
mpz_t v;
node *next;
};

#define CSIZE 37
node *cache[CSIZE];

// very primitive linked hash table
node * find_cache(int n)
{
int idx = n % CSIZE;
node *p;

for (p = cache[idx]; p && p->n != n; p = p->next);
if (p) return p;

p = malloc(sizeof(node));
p->next = cache[idx];
cache[idx] = p;

if (n < 2) {
p->n = n;
mpz_init_set_ui(p->v, 1);
} else {
p->n = -1; // -1: value not computed yet
mpz_init(p->v);
}
return p;
}

mpz_t tmp1, tmp2;
mpz_t *fib(int n)
{
int x;
node *p = find_cache(n);

if (p->n < 0) {
p->n = n;
x = n / 2;

mpz_mul(tmp1, *fib(x-1), *fib(n - x - 1));
mpz_mul(tmp2, *fib(x), *fib(n - x));
}
return &p->v;
}

int main(int argc, char **argv)
{
int i, n;
if (argc < 2) return 1;

mpz_init(tmp1);
mpz_init(tmp2);

for (i = 1; i < argc; i++) {
n = atoi(argv[i]);
if (n < 0) {
continue;
}

// about 75% of time is spent in printing
gmp_printf("%Zd\n", *fib(n));
}
return 0;
}

Output:
% ./a.out 0 1 2 3 4 5
1
1
2
3
5
8
% ./a.out 10000000 | wc -c    # count length of output, including the newline
1919488


## C#

### Recursive

public static ulong Fib(uint n) {
return (n < 2)? n : Fib(n - 1) + Fib(n - 2);
}


### Tail-Recursive

public static ulong Fib(uint n) {
return Fib(0, 1, n);
}

private static ulong Fib(ulong a, ulong b, uint n) {
return (n < 1)? a :(n == 1)?  b : Fib(b, a + b, n - 1);
}


### Iterative

public static ulong Fib(uint x) {
if (x == 0) return 0;

ulong prev = 0;
ulong next = 1;
for (int i = 1; i < x; i++)
{
ulong sum = prev + next;
prev = next;
next = sum;
}
return next;
}


### Eager-Generative

public static IEnumerable<long> Fibs(uint x) {
IList<ulong> fibs = new List<ulong>();

ulong prev = -1;
ulong next = 1;
for (int i = 0; i < x; i++)
{
long sum = prev + next;
prev = next;
next = sum;
}
return fibs;
}


### Lazy-Generative

public static IEnumerable<ulong> Fibs(uint x) {
ulong prev = -1;
ulong next = 1;
for (uint i = 0; i < x; i++) {
ulong sum = prev + next;
prev = next;
next = sum;
yield return sum;
}
}


### Analytic

This returns digits up to the 93rd Fibonacci number, but the digits become inaccurate past the 71st. There is custom rounding applied to the result that allows the function to be accurate at the 71st number instead of topping out at the 70th.

static double r5 = Math.Sqrt(5.0), Phi = (r5 + 1.0) / 2.0;

static ulong fib(uint n) {
if (n > 71) throw new ArgumentOutOfRangeException("n", n, "Needs to be smaller than 72.");
double r = Math.Pow(Phi, n) / r5;
return (ulong)(n < 64 ? Math.Round(r) : Math.Floor(r)); }

To get to the 93rd Fibonacci number, one must use the decimal type, rather than the double type, like this:
static decimal Sqrt_dec(decimal x, decimal g) { decimal t, lg;
do { t = x / g; lg = g; g = (t + g) / 2M; } while (lg != g);
return g; }

static decimal Pow_dec (decimal bas, uint exp) {
if (exp == 0) return 1M;
decimal tmp = Pow_dec(bas, exp >> 1); tmp *= tmp;
if ((exp & 1) == 1) tmp *= bas; return tmp; }

static decimal r5 = Sqrt_dec(5.0M, (decimal)Math.Sqrt(5.0)),
Phi = (r5 + 1.0M) / 2.0M;

static ulong fib(uint n) {
if (n > 93) throw new ArgumentOutOfRangeException("n", n, "Needs to be smaller than 94.");
decimal r = Pow_dec(Phi, n) / r5;
return (ulong)(n < 64 ? Math.Round(r) : Math.Floor(r)); }

Note that the Math.Pow() function and the Math.Sqrt() function must be replaced with ones returning the decimal type.

If one allows the fib() function to return the decimal type, one can reach the 138th Fibonacci number. However, the accuracy is lost after the 128th.
static decimal Sqrt_dec(decimal x, decimal g) { decimal t, lg;
do { t = x / g; lg = g; g = (t + g) / 2M; } while (lg != g);
return g; }

static decimal Pow_dec (decimal bas, uint exp) {
if (exp == 0) return 1M;
decimal tmp = Pow_dec(bas, exp >> 1); tmp *= tmp;
if ((exp & 1) == 1) tmp *= bas; return tmp; }

static decimal r5 = Sqrt_dec(5.0M, (decimal)Math.Sqrt(5.0)),
Phi = (r5 + 1.0M) / 2.0M;

static decimal fib(uint n) {
if (n > 128) throw new ArgumentOutOfRangeException("n", n, "Needs to be smaller than 129.");
decimal r = Pow_dec(Phi, n) / r5;
return n < 64 ? Math.Round(r) : Math.Floor(r); }


### Matrix

Algorithm is based on

${\displaystyle \begin{pmatrix}1&1\\1&0\end{pmatrix}^n = \begin{pmatrix}F(n+1)&F(n)\\F(n)&F(n-1)\end{pmatrix}}$.

Needs System.Windows.Media.Matrix or similar Matrix class. Calculates in ${\displaystyle O(n)}$.

public static ulong Fib(uint n) {
var M = new Matrix(1,0,0,1);
var N = new Matrix(1,1,1,0);
for (uint i = 1; i < n; i++) M *= N;
return (ulong)M[0][0];
}


Needs System.Windows.Media.Matrix or similar Matrix class. Calculates in ${\displaystyle O(\log{n})}$.

private static Matrix M;
private static readonly Matrix N = new Matrix(1,1,1,0);

public static ulong Fib(uint n) {
M = new Matrix(1,0,0,1);
MatrixPow(n-1);
return (ulong)M[0][0];
}

private static void MatrixPow(double n){
if (n > 1) {
MatrixPow(n/2);
M *= M;
}
if (n % 2 == 0) M *= N;
}


### Array (Table) Lookup

private static int[] fibs = new int[]{ -1836311903, 1134903170,
-701408733, 433494437, -267914296, 165580141, -102334155,
63245986, -39088169, 24157817, -14930352, 9227465, -5702887,
3524578, -2178309, 1346269, -832040, 514229, -317811, 196418,
-121393, 75025, -46368, 28657, -17711, 10946, -6765, 4181,
-2584, 1597, -987, 610, -377, 233, -144, 89, -55, 34, -21, 13,
-8, 5, -3, 2, -1, 1, 0, 1, 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};

public static int Fib(int n) {
if(n < -46 || n > 46) throw new ArgumentOutOfRangeException("n", n, "Has to be between -46 and 47.")
return fibs[n+46];
}


### Arbitrary Precision

This large step recurrence routine can calculate the two millionth Fibonacci number in under 1 / 5 second at tio.run. This routine can generate the fifty millionth Fibonacci number in under 30 seconds at tio.run. The unused conventional iterative method times out at two million on tio.run, you can only go to around 1,290,000 or so to keep the calculation time (plus string conversion time) under the 60 second timeout limit there. When using this large step recurrence method, it takes around 5 seconds to convert the two millionth Fibonacci number (417975 digits) into a string (so that one may count those digits).

using System;
using System.Collections.Generic;
using BI = System.Numerics.BigInteger;

class Program
{
// A sparse array of values calculated along the way
static SortedList<int, BI> sl = new SortedList<int, BI>();

// This routine is semi-recursive, but doesn't need to evaluate every number up to n.
// Algorithm from here: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibFormula.html#section3
static BI Fsl(int n)
{
if (n < 2) return n;
int n2 = n >> 1, pm = n2 + ((n & 1) << 1) - 1; IfNec(n2); IfNec(pm);
return n2 > pm ? (2 * sl[pm] + sl[n2]) * sl[n2] : sqr(sl[n2]) + sqr(sl[pm]);
// Helper routine for Fsl(). It adds an entry to the sorted list when necessary
void IfNec(int x) { if (!sl.ContainsKey(x)) sl.Add(x, Fsl(x)); }
// Helper function to square a BigInteger
BI sqr(BI x) { return x * x; }
}

// Conventional iteration method (not used here)
public static BI Fm(BI n)
{
if (n < 2) return n; BI cur = 0, pre = 1;
for (int i = 0; i <= n - 1; i++) { BI sum = cur + pre; pre = cur; cur = sum; }
return cur;
}

public static void Main()
{
int num = 2_000_000, digs = 35, vlen;
var sw = System.Diagnostics.Stopwatch.StartNew(); var v = Fsl(num); sw.Stop();
Console.Write("{0:n3} ms to calculate the {1:n0}th Fibonacci number, ",
sw.Elapsed.TotalMilliseconds, num);
Console.WriteLine("number of digits is {0}", vlen = (int)Math.Ceiling(BI.Log10(v)));
if (vlen < 10000) {
sw.Restart(); Console.WriteLine(v); sw.Stop();
Console.WriteLine("{0:n3} ms to write it to the console.", sw.Elapsed.TotalMilliseconds);
} else
Console.Write("partial: {0}...{1}", v / BI.Pow(10, vlen - digs), v % BI.Pow(10, digs));
}
}

Output:
137.209 ms to calculate the 2,000,000th Fibonacci number, number of digits is 417975
partial: 85312949175076415430516606545038251...91799493108960825129188777803453125


### Shift PowerMod

Illustrated here is an algorithm to compute a Fibonacci number directly, without needing to calculate any of the Fibonacci numbers less than the desired result. It uses shifting and the power mod function (BigInteger.ModPow() in C#). It calculates more quickly than the large step recurrence routine (illustrated above) for smaller Fibonacci numbers (less than 2800 digits or so, around Fibonacci(13000)), but gets slower for larger ones, as the intermediate BigIntegers created are very large, much larger than the Fibonacci result.

Also included is a routine that returns an array of Fibonacci numbers (fibTab()). It reuses the intermediate large shifted BigIntegers on suceeding iterations, therfore it is a little more efficient than calling the oneshot (oneFib()) routine repeatedly from a loop.

using System;
using BI = System.Numerics.BigInteger;

class Program {

// returns the nth Fibonacci number without calculating 0..n-1
static BI oneFib(int n) {
BI z = (BI)1 << ++n;
return BI.ModPow(z, n, (z << n) - z - 1) % z;
}

// returns an array of Fibonacci numbers from the 0th to the nth
static BI[] fibTab(int n) {
var res = new BI[++n];
BI z = (BI)1 << 1, zz = z << 1;
for (int i = 0; i < n; ) {
res[i] = BI.ModPow(z, ++i, zz - z - 1) % z;
z <<= 1; zz <<= 2;
}
return res;
}

static void Main(string[] args) {
int n = 20;
Console.WriteLine("Fibonacci numbers 0..{0}: {1}", n, string.Join(" ",fibTab(n)));
n = 1000;
Console.WriteLine("Fibonacci({0}): {1}", n, oneFib(n));
}
}

Output:
Fibonacci numbers 0..20: 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765
Fibonacci(1000): 43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875

## C++

Using unsigned int, this version only works up to 48 before fib overflows.

#include <iostream>

int main()
{
unsigned int a = 1, b = 1;
unsigned int target = 48;
for(unsigned int n = 3; n <= target; ++n)
{
unsigned int fib = a + b;
std::cout << "F("<< n << ") = " << fib << std::endl;
a = b;
b = fib;
}

return 0;
}


Library: GMP

This version does not have an upper bound.

#include <iostream>
#include <gmpxx.h>

int main()
{
mpz_class a = mpz_class(1), b = mpz_class(1);
mpz_class target = mpz_class(100);
for(mpz_class n = mpz_class(3); n <= target; ++n)
{
mpz_class fib = b + a;
if ( fib < b )
{
std::cout << "Overflow at " << n << std::endl;
break;
}
std::cout << "F("<< n << ") = " << fib << std::endl;
a = b;
b = fib;
}
return 0;
}


Version using transform:

#include <algorithm>
#include <vector>
#include <functional>
#include <iostream>

unsigned int fibonacci(unsigned int n) {
if (n == 0) return 0;
std::vector<int> v(n+1);
v[1] = 1;
transform(v.begin(), v.end()-2, v.begin()+1, v.begin()+2, std::plus<int>());
// "v" now contains the Fibonacci sequence from 0 up
return v[n];
}


#include <numeric>
#include <vector>
#include <functional>
#include <iostream>

unsigned int fibonacci(unsigned int n) {
if (n == 0) return 0;
std::vector<int> v(n, 1);
// "array" now contains the Fibonacci sequence from 1 up
return v[n-1];
}


Version which computes at compile time with metaprogramming:

#include <iostream>

template <int n> struct fibo
{
enum {value=fibo<n-1>::value+fibo<n-2>::value};
};

template <> struct fibo<0>
{
enum {value=0};
};

template <> struct fibo<1>
{
enum {value=1};
};

int main(int argc, char const *argv[])
{
std::cout<<fibo<12>::value<<std::endl;
std::cout<<fibo<46>::value<<std::endl;
return 0;
}


The following version is based on fast exponentiation:

#include <iostream>

inline void fibmul(int* f, int* g)
{
int tmp = f[0]*g[0] + f[1]*g[1];
f[1] = f[0]*g[1] + f[1]*(g[0] + g[1]);
f[0] = tmp;
}

int fibonacci(int n)
{
int f[] = { 1, 0 };
int g[] = { 0, 1 };
while (n > 0)
{
if (n & 1) // n odd
{
fibmul(f, g);
--n;
}
else
{
fibmul(g, g);
n >>= 1;
}
}
return f[1];
}

int main()
{
for (int i = 0; i < 20; ++i)
std::cout << fibonacci(i) << " ";
std::cout << std::endl;
}

Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181


### Using Zeckendorf Numbers

The nth fibonacci is represented as Zeckendorf 1 followed by n-1 zeroes. Here I define a class N which defines the operations increment ++() and comparison <=(other N) for Zeckendorf Numbers.

// Use Zeckendorf numbers to display Fibonacci sequence.
// Nigel Galloway October 23rd., 2012
int main(void) {
char NG[22] = {'1',0};
int x = -1;
N G;
for (int fibs = 1; fibs <= 20; fibs++) {
for (;G <= N(NG); ++G) x++;
NG[fibs] = '0';
NG[fibs+1] = 0;
std::cout << x << " ";
}
std::cout << std::endl;
return 0;
}

Output:
1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946


### Using Standard Template Library

Possibly less "Far-fetched version".

// Use Standard Template Library to display Fibonacci sequence.
// Nigel Galloway March 30th., 2013
#include <algorithm>
#include <iostream>
#include <iterator>
int main()
{
int x = 1, y = 1;
generate_n(std::ostream_iterator<int>(std::cout, " "), 21, [&]{int n=x; x=y; y+=n; return n;});
return 0;
}

Output:

1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765 10946

## Cat

define fib {
dup 1 <=
[]
[dup 1 - fib swap 2 - fib +]
if
}

## Chapel

iter fib() {
var a = 0, b = 1;

while true {
yield a;
(a, b) = (b, b + a);
}
}


## Chef

Stir-Fried Fibonacci Sequence.

An unobfuscated iterative implementation.
It prints the first N + 1 Fibonacci numbers,
where N is taken from standard input.

Ingredients.
0 g last
1 g this
0 g new
0 g input

Method.
Take input from refrigerator.
Put this into 4th mixing bowl.
Loop the input.
Clean the 3rd mixing bowl.
Put last into 3rd mixing bowl.
Add this into 3rd mixing bowl.
Fold new into 3rd mixing bowl.
Clean the 1st mixing bowl.
Put this into 1st mixing bowl.
Fold last into 1st mixing bowl.
Clean the 2nd mixing bowl.
Put new into 2nd mixing bowl.
Fold this into 2nd mixing bowl.
Put new into 4th mixing bowl.
Endloop input until looped.
Pour contents of the 4th mixing bowl into baking dish.

Serves 1.

## Clio

Clio is pure and functions are lazy and memoized by default

fn fib n:
if n < 2: n
else: (n - 1 -> fib) + (n - 2 -> fib)

[0:100] -> * fib -> * print

## Clojure

### Lazy Sequence

This is implemented idiomatically as an infinitely long, lazy sequence of all Fibonacci numbers:

(defn fibs []
(map first (iterate (fn [[a b]] [b (+ a b)]) [0 1])))


Thus to get the nth one:

(nth (fibs) 5)


So long as one does not hold onto the head of the sequence, this is unconstrained by length.

The one-line implementation may look confusing at first, but on pulling it apart it actually solves the problem more "directly" than a more explicit looping construct.

(defn fibs []
(map first ;; throw away the "metadata" (see below) to view just the fib numbers
(iterate ;; create an infinite sequence of [prev, curr] pairs
(fn [[a b]] ;; to produce the next pair, call this function on the current pair
[b (+ a b)]) ;; new prev is old curr, new curr is sum of both previous numbers
[0 1]))) ;; recursive base case: prev 0, curr 1


A more elegant solution is inspired by the Haskell implementation of an infinite list of Fibonacci numbers:

(def fib (lazy-cat [0 1] (map + fib (rest fib))))


Then, to see the first ten,

user> (take 10 fib)
(0 1 1 2 3 5 8 13 21 34)


### Iterative

Here's a simple interative process (using a recursive function) that carries state along with it (as args) until it reaches a solution:

;; max is which fib number you'd like computed (0th, 1st, 2nd, etc.)
;; n is which fib number you're on for this call (0th, 1st, 2nd, etc.)
;; j is the nth fib number (ex. when n = 5, j = 5)
;; i is the nth - 1 fib number
(defn- fib-iter
[max n i j]
(if (= n max)
j
(recur max
(inc n)
j
(+ i j))))

(defn fib
[max]
(if (< max 2)
max
(fib-iter max 1 0N 1N)))


"defn-" means that the function is private (for use only inside this library). The "N" suffixes on integers tell Clojure to use arbitrary precision ints for those.

### Doubling Algorithm (Fast)

Based upon the doubling algorithm which computes in O(log (n)) time as described here https://www.nayuki.io/page/fast-fibonacci-algorithms Implementation credit: https://stackoverflow.com/questions/27466311/how-to-implement-this-fast-doubling-fibonacci-algorithm-in-clojure/27466408#27466408

(defn fib [n]
(letfn [(fib* [n]
(if (zero? n)
[0 1]
(let [[a b] (fib* (quot n 2))
c (*' a (-' (*' 2 b) a))
d (+' (*' b b) (*' a a))]
(if (even? n)
[c d]
[d (+' c d)]))))]
(first (fib* n))))


### Recursive

A naive slow recursive solution:

(defn fib [n]
(case n
0 0
1 1
(+ (fib (- n 1))
(fib (- n 2)))))


This can be improved to an O(n) solution, like the iterative solution, by memoizing the function so that numbers that have been computed are cached. Like a lazy sequence, this also has the advantage that subsequent calls to the function use previously cached results rather than recalculating.

(def fib
(memoize
(fn [n]
(case n
0 0
1 1
(+ (fib (- n 1))
(fib (- n 2)))))))


### Using core.async

(ns fib.core)
(require '[clojure.core.async
:refer [<! >! >!! <!! timeout chan alt! go]])

(defn fib [c]
(loop [a 0 b 1]
(>!! c a)
(recur b (+ a b))))

(defn -main []
(let [c (chan)]
(go (fib c))
(dorun
(for [i (range 10)]
(println (<!! c))))))


## CLU

% Generate Fibonacci numbers
fib = iter () yields (int)
a: int := 0
b: int := 1

while true do
yield (a)
a, b := b, a+b
end
end fib

% Grab the n'th value from an iterator
nth = proc [T: type] (g: itertype () yields (T), n: int) returns (T)
for v: T in g() do
if n<=0 then return (v) end
n := n-1
end
end nth

% Print a few values
start_up = proc ()
po: stream := stream$primary_output() % print values coming out of the fibonacci iterator % (which are generated one after the other without delay) count: int := 0 for f: int in fib() do stream$putl(po, "F(" || int$unparse(count) || ") = " || int$unparse(f))
count := count + 1
if count = 15 then break end
end

% print a few random fibonacci numbers
% (to do this it has to restart at the beginning for each
% number, making it O(N))
fibs: sequence[int] := sequence[int]$[20,30,50] for n: int in sequence[int]$elements(fibs) do
stream$putl(po, "F(" || int$unparse(n) || ") = "
|| int$unparse(nth[int](fib, n))) end end start_up Output: F(0) = 0 F(1) = 1 F(2) = 1 F(3) = 2 F(4) = 3 F(5) = 5 F(6) = 8 F(7) = 13 F(8) = 21 F(9) = 34 F(10) = 55 F(11) = 89 F(12) = 144 F(13) = 233 F(14) = 377 F(20) = 6765 F(30) = 832040 F(50) = 12586269025 ## CMake Iteration uses a while() loop. Memoization uses global properties. set_property(GLOBAL PROPERTY fibonacci_0 0) set_property(GLOBAL PROPERTY fibonacci_1 1) set_property(GLOBAL PROPERTY fibonacci_next 2) # var = nth number in Fibonacci sequence. function(fibonacci var n) # If the sequence is too short, compute more Fibonacci numbers. get_property(next GLOBAL PROPERTY fibonacci_next) if(NOT next GREATER${n})
# a, b = last 2 Fibonacci numbers
math(EXPR i "${next} - 2") get_property(a GLOBAL PROPERTY fibonacci_${i})
math(EXPR i "${next} - 1") get_property(b GLOBAL PROPERTY fibonacci_${i})

while(NOT next GREATER ${n}) math(EXPR i "${a} + ${b}") # i = next Fibonacci number set_property(GLOBAL PROPERTY fibonacci_${next} ${i}) set(a${b})
set(b ${i}) math(EXPR next "${next} + 1")
endwhile()
set_property(GLOBAL PROPERTY fibonacci_next ${next}) endif() get_property(answer GLOBAL PROPERTY fibonacci_${n})
set(${var}${answer} PARENT_SCOPE)
endfunction(fibonacci)

# Test program: print 0th to 9th and 25th to 30th Fibonacci numbers.
set(s "")
foreach(i RANGE 0 9)
fibonacci(f ${i}) set(s "${s} ${f}") endforeach(i) set(s "${s} ... ")
foreach(i RANGE 25 30)
fibonacci(f ${i}) set(s "${s} ${f}") endforeach(i) message(${s})

 0 1 1 2 3 5 8 13 21 34 ... 75025 121393 196418 317811 514229 832040

## COBOL

### Iterative

Program-ID. Fibonacci-Sequence.
Data Division.
Working-Storage Section.
01  FIBONACCI-PROCESSING.
05  FIBONACCI-NUMBER  PIC 9(36)   VALUE 0.
05  FIB-ONE           PIC 9(36)   VALUE 0.
05  FIB-TWO           PIC 9(36)   VALUE 1.
01  DESIRED-COUNT       PIC 9(4).
01  FORMATTING.
05  INTERM-RESULT     PIC Z(35)9.
05  FORMATTED-RESULT  PIC X(36).
05  FORMATTED-SPACE   PIC x(35).
Procedure Division.
000-START-PROGRAM.
Display "What place of the Fibonacci Sequence would you like (<173)? " with no advancing.
Accept DESIRED-COUNT.
If DESIRED-COUNT is less than 1
Stop run.
If DESIRED-COUNT is less than 2
Move FIBONACCI-NUMBER to INTERM-RESULT
Move INTERM-RESULT to FORMATTED-RESULT
Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT
Display FORMATTED-RESULT
Stop run.
Subtract 1 from DESIRED-COUNT.
Move FIBONACCI-NUMBER to INTERM-RESULT.
Move INTERM-RESULT to FORMATTED-RESULT.
Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT.
Display FORMATTED-RESULT.
Perform 100-COMPUTE-FIBONACCI until DESIRED-COUNT = zero.
Stop run.
100-COMPUTE-FIBONACCI.
Compute FIBONACCI-NUMBER = FIB-ONE + FIB-TWO.
Move FIB-TWO to FIB-ONE.
Move FIBONACCI-NUMBER to FIB-TWO.
Subtract 1 from DESIRED-COUNT.
Move FIBONACCI-NUMBER to INTERM-RESULT.
Move INTERM-RESULT to FORMATTED-RESULT.
Unstring FORMATTED-RESULT delimited by all spaces into FORMATTED-SPACE,FORMATTED-RESULT.
Display FORMATTED-RESULT.


### Recursive

Works with: GNU Cobol version 2.0
       >>SOURCE FREE
IDENTIFICATION DIVISION.
PROGRAM-ID. fibonacci-main.

DATA DIVISION.
WORKING-STORAGE SECTION.
01  num                                 PIC 9(6) COMP.
01  fib-num                             PIC 9(6) COMP.

PROCEDURE DIVISION.
ACCEPT num
CALL "fibonacci" USING CONTENT num RETURNING fib-num
DISPLAY fib-num
.
END PROGRAM fibonacci-main.

IDENTIFICATION DIVISION.
PROGRAM-ID. fibonacci RECURSIVE.

DATA DIVISION.
LOCAL-STORAGE SECTION.
01  1-before                            PIC 9(6) COMP.
01  2-before                            PIC 9(6) COMP.

01  num                                 PIC 9(6) COMP.

01  fib-num                             PIC 9(6) COMP BASED.

PROCEDURE DIVISION USING num RETURNING fib-num.
ALLOCATE fib-num
EVALUATE num
WHEN 0
MOVE 0 TO fib-num
WHEN 1
MOVE 1 TO fib-num
WHEN OTHER
SUBTRACT 1 FROM num
CALL "fibonacci" USING CONTENT num RETURNING 1-before
SUBTRACT 1 FROM num
CALL "fibonacci" USING CONTENT num RETURNING 2-before
ADD 1-before TO 2-before GIVING fib-num
END-EVALUATE
.
END PROGRAM fibonacci.


## CoffeeScript

### Analytic

fib_ana = (n) ->
sqrt = Math.sqrt
phi = ((1 + sqrt(5))/2)
Math.round((Math.pow(phi, n)/sqrt(5)))


### Iterative

fib_iter = (n) ->
return n if n < 2
[prev, curr] = [0, 1]
[prev, curr] = [curr, curr + prev] for i in [1..n]
curr


### Recursive

fib_rec = (n) ->
if n < 2 then n else fib_rec(n-1) + fib_rec(n-2)


## Comefrom0x10

Recursion is is not possible in Comefrom0x10.

### Iterative

stop = 6
a = 1
i = 1  # start
a      # print result

fib
comefrom if i is 1  # start
b = 1
comefrom fib        # start of loop
i = i + 1
next_b = a + b
a = b
b = next_b

comefrom fib if i > stop

## Common Lisp

Note that Common Lisp uses bignums, so this will never overflow.

### Iterative

(defun fibonacci-iterative (n &aux (f0 0) (f1 1))
(case n
(0 f0)
(1 f1)
(t (loop for n from 2 to n
for a = f0 then b and b = f1 then result
for result = (+ a b)
finally (return result)))))


Simpler one:

(defun fibonacci (n)
(let ((a 0) (b 1) (c n))
(loop for i from 2 to n do
(setq c (+ a b)
a b
b c))
c))

Not a function, just printing out the entire (for some definition of "entire") sequence with a for var =  loop:
(loop for x = 0 then y and y = 1 then (+ x y) do (print x))


### Recursive

(defun fibonacci-recursive (n)
(if (< n 2)
n
(+ (fibonacci-recursive (- n 2)) (fibonacci-recursive (- n 1)))))


(defun fibonacci-tail-recursive ( n &optional (a 0) (b 1))
(if (= n 0)
a
(fibonacci-tail-recursive (- n 1) b (+ a b))))


Tail recursive and squaring:

(defun fib (n &optional (a 1) (b 0) (p 0) (q 1))
(if (= n 1) (+ (* b p) (* a q))
(fib (ash n -1)
(if (evenp n) a (+ (* b q) (* a (+ p q))))
(if (evenp n) b (+ (* b p) (* a q)))
(+ (* p p) (* q q))
(+ (* q q) (* 2 p q))))) ;p is Fib(2^n-1), q is Fib(2^n).

(print (fib 100000))


### Alternate solution

I use Allegro CL 10.1

;; Project : Fibonacci sequence

(defun fibonacci (nr)
(cond ((= nr 0) 1)
((= nr 1) 1)
(t (+ (fibonacci (- nr 1))
(fibonacci (- nr 2))))))
(format t "~a" "First 10 Fibonacci numbers")
(dotimes (n 10)
(if (< n 1) (terpri))
(if (< n 9) (format t "~a" " "))
(write(+ n 1)) (format t "~a" ": ")
(write (fibonacci n)) (terpri))


Output:

First 10 Fibonacci numbers
1: 1
2: 1
3: 2
4: 3
5: 5
6: 8
7: 13
8: 21
9: 34
10: 55


### Solution with methods and eql specializers

(defmethod fib (n)
(declare ((integer 0 *) n))
(+ (fib (- n 1))
(fib (- n 2))))

(defmethod fib ((n (eql 0))) 0)

(defmethod fib ((n (eql 1))) 1)


### List-based iterative

This solution uses a list to keep track of the Fibonacci sequence for 0 or a positive integer.

(defun fibo (n)
(cond ((< n 0) nil)
((< n 2) n)
(t (let ((leo '(1 0)))
(loop for i from 2 upto n do
(setf leo (cons (+ (first leo)
(second leo))
leo))
finally (return (first leo)))))))

Output:
> (fibo 0)
0
> (fibo 1)
1
> (fibo 10)
55
> (fibo 100)
354224848179261915075
> (fibo 1000)
43466557686937456435688527675040625802564660517371780402481729089536555417949051890403879840079255169295922593080322634775209689623239873322471161642996440906533187938298969649928516003704476137795166849228875
> (fibo -10)
NIL

### List-based recursive

This solution computes Fibonacci numbers as either:

1. a list starting from the first element;
2. a single number;
3. an interval from i-th to j-th element.

Options #2 and #3 can take negative parameters, but i (lowest index in range) must be greater than j (highest index in range).

Values are represented internally by a reversed list that grows from the head (and that's why we reverse it back when we return it).

(defparameter *fibo-start* '(1 1)) ; elements 1 and 2

;;; Helper functions
(defun grow-fibo (fibo)
(cons (+ (first fibo) (second fibo)) fibo))

(defun generate-fibo (fibo n) ; n must be > 1
(if (equal (list-length fibo) n)
fibo
(generate-fibo (grow-fibo fibo)  n)))

;;; User functions
(defun fibo (n)
(cond ((= n 0) 0)
((= (abs n) 1) 1)
(t (let ((result (first (generate-fibo *fibo-start* (abs n)))))
(if (and (< n -1) (evenp n))
(- result)
result)))))

(defun fibo-list (n)
(cond ((< n 1) nil)
((= n 1) '(1))
(t (reverse (generate-fibo *fibo-start* n)))))

(defun fibo-range (lower upper)
(if (<= upper lower)
nil
(reverse (generate-fibo
(list
(fibo (1+ lower))
(fibo  lower))
(1+ (- upper lower))))))

Output:
> (fibo 100)
354224848179261915075
> (fibo -150)
-9969216677189303386214405760200
> (fibo-list 20)
(1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765)
> (fibo-range -10 15)
(-55 34 -21 13 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610)
> (fibo-range 0 20)
(0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181 6765)

## Computer/zero Assembly

To find the ${\displaystyle n}$th Fibonacci number, set the initial value of count equal to ${\displaystyle n}$–2 and run the program. The machine will halt with the answer stored in the accumulator. Since Computer/zero's word length is only eight bits, the program will not work with values of ${\displaystyle n}$ greater than 13.

loop:   LDA  y      ; higher No.
STA  temp
STA  y
LDA  temp
STA  x

LDA  count
SUB  one
BRZ  done

STA  count
JMP  loop

done:   LDA  y
STP

one:         1
count:       8      ; n = 10
x:           1
y:           1
temp:        0

## Corescript

print Fibonacci Sequence:
var previous = 1
var number = 0
var temp = (blank)

:fib
if number > 50000000000:kill
print (number)
set temp = (add number previous)
set previous = (number)
set number = (temp)
goto fib

:kill
stop

## Cowgol

include "cowgol.coh";

sub fibonacci(n: uint32): (a: uint32) is
a := 0;
var b: uint32 := 1;
while n > 0 loop
var c := a + b;
a := b;
b := c;
n := n - 1;
end loop;
end sub;

# test
var i: uint32 := 0;
while i < 20 loop
print_i32(fibonacci(i));
print_char(' ');
i := i + 1;
end loop;
print_nl();
Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 987 1597 2584 4181

## Crystal

### Recursive

def fib(n)
n < 2 ? n : fib(n - 1) + fib(n - 2)
end


### Iterative

def fibIterative(n, prevFib = 0, fib = 1)
return n if n < 2

n.times do
prevFib, fib = fib, prevFib + fib
end

prevFib
end


### Tail Recursive

def fibTailRecursive(n, prevFib = 0, fib = 1)
n == 0 ? prevFib : fibTailRecursive(n - 1, fib, prevFib + fib)
end


### Analytic

def fibBinet(n)
(((5 ** 0.5 + 1) / 2) ** n / 5 ** 0.5).round.to_i
end


## D

Here are four versions of Fibonacci Number calculating functions. FibD has an argument limit of magnitude 84 due to floating point precision, the others have a limit of 92 due to overflow (long).The traditional recursive version is inefficient. It is optimized by supplying a static storage to store intermediate results. A Fibonacci Number generating function is added. All functions have support for negative arguments.

import std.stdio, std.conv, std.algorithm, std.math;

long sgn(alias unsignedFib)(int n) { // break sign manipulation apart
immutable uint m = (n >= 0) ? n : -n;
if (n < 0 && (n % 2 == 0))
return -unsignedFib(m);
else
return unsignedFib(m);
}

long fibD(uint m) { // Direct Calculation, correct for abs(m) <= 84
enum sqrt5r =  1.0L / sqrt(5.0L);         //  1 / sqrt(5)
enum golden = (1.0L + sqrt(5.0L)) / 2.0L; // (1 + sqrt(5)) / 2
return roundTo!long(pow(golden, m) * sqrt5r);
}

long fibI(in uint m) pure nothrow { // Iterative
long thisFib = 0;
long nextFib = 1;
foreach (i; 0 .. m) {
long tmp = nextFib;
nextFib += thisFib;
thisFib  = tmp;
}
return thisFib;
}

long fibR(uint m) { // Recursive
return (m < 2) ? m : fibR(m - 1) + fibR(m - 2);
}

long fibM(uint m) { // memoized Recursive
static long[] fib = [0, 1];
while (m >= fib.length )
fib ~= fibM(m - 2) + fibM(m - 1);
return fib[m];
}

alias sgn!fibD sfibD;
alias sgn!fibI sfibI;
alias sgn!fibR sfibR;
alias sgn!fibM sfibM;

auto fibG(in int m) { // generator(?)
immutable int sign = (m < 0) ? -1 : 1;
long yield;

return new class {
final int opApply(int delegate(ref int, ref long) dg) {
int idx = -sign; // prepare for pre-increment
foreach (f; this)
if (dg(idx += sign, f))
break;
return 0;
}

final int opApply(int delegate(ref long) dg) {
long f0, f1 = 1;
foreach (p; 0 .. m * sign + 1) {
if (sign == -1 && (p % 2 == 0))
yield = -f0;
else
yield = f0;
if (dg(yield)) break;
auto temp = f1;
f1 = f0 + f1;
f0 = temp;
}
return 0;
}
};
}

void main(in string[] args) {
int k = args.length > 1 ? to!int(args[1]) : 10;
writefln("Fib(%3d) = ", k);
writefln("D : %20d <- %20d + %20d",
sfibD(k), sfibD(k - 1), sfibD(k - 2));
writefln("I : %20d <- %20d + %20d",
sfibI(k), sfibI(k - 1), sfibI(k - 2));
if (abs(k) < 36 || args.length > 2)
// set a limit for recursive version
writefln("R : %20d <- %20d + %20d",
sfibR(k), sfibM(k - 1), sfibM(k - 2));
writefln("O : %20d <- %20d + %20d",
sfibM(k), sfibM(k - 1), sfibM(k - 2));
foreach (i, f; fibG(-9))
writef("%d:%d | ", i, f);
}

Output:
for n = 85
Fib( 85) =
D :   259695496911122586 <-   160500643816367088 +    99194853094755497
I :   259695496911122585 <-   160500643816367088 +    99194853094755497
O :   259695496911122585 <-   160500643816367088 +    99194853094755497
0:0 | -1:1 | -2:-1 | -3:2 | -4:-3 | -5:5 | -6:-8 | -7:13 | -8:-21 | -9:34 | 

### Matrix Exponentiation Version

import std.bigint;

T fibonacciMatrix(T=BigInt)(size_t n) {
int[size_t.sizeof * 8] binDigits;
size_t nBinDigits;
while (n > 0) {
binDigits[nBinDigits] = n % 2;
n /= 2;
nBinDigits++;
}

T x=1, y, z=1;
foreach_reverse (b; binDigits[0 .. nBinDigits]) {
if (b) {
x = (x + z) * y;
y = y ^^ 2 + z ^^ 2;
} else {
auto x_old = x;
x = x ^^ 2 + y ^^ 2;
y = (x_old + z) * y;
}
z = x + y;
}

return y;
}

void main() {
10_000_000.fibonacciMatrix;
}


### Faster Version

For N = 10_000_000 this is about twice faster (run-time about 2.20 seconds) than the matrix exponentiation version.

import std.bigint, std.math;

// Algorithm from: Takahashi, Daisuke,
// "A fast algorithm for computing large Fibonacci numbers".
// Information Processing Letters 75.6 (30 November 2000): 243-246.
// Implementation from:
// pythonista.wordpress.com/2008/07/03/pure-python-fibonacci-numbers
BigInt fibonacci(in ulong n)
in {
assert(n > 0, "fibonacci(n): n must be > 0.");
} body {
if (n <= 2)
return 1.BigInt;
BigInt F = 1;
BigInt L = 1;
int sign = -1;
immutable uint n2 = cast(uint)n.log2.floor;
auto mask = 2.BigInt ^^ (n2 - 1);
foreach (immutable i; 1 .. n2) {
auto temp = F ^^ 2;
F = (F + L) / 2;
F = 2 * F ^^ 2 - 3 * temp - 2 * sign;
L = 5 * temp + 2 * sign;
sign = 1;
temp = F;
F = (F + L) / 2;
L = F + 2 * temp;
sign = -1;
}
}
if ((n & mask) == 0) {
F *= L;
} else {
F = (F + L) / 2;
F = F * L - sign;
}
return F;
}

void main() {
10_000_000.fibonacci;
}


## Dart

int fib(int n) {
if (n==0 || n==1) {
return n;
}
var prev=1;
var current=1;
for (var i=2; i<n; i++) {
var next = prev + current;
prev = current;
current = next;
}
return current;
}

int fibRec(int n) => n==0 || n==1 ? n : fibRec(n-1) + fibRec(n-2);

main() {
print(fib(11));
print(fibRec(11));
}


## Datalog

Simple recurive implementation for Souffle.

.decl Fib(i:number, x:number)
Fib(0, 0).
Fib(1, 1).
Fib(i+2,x+y) :- Fib(i+1, x), Fib(i, y), i+2<=40, i+2>=2.
Fib(i-2,y-x) :- Fib(i-1, x), Fib(i, y), i-2>=-40, i-2<0.

## DBL

;
;       Fibonacci sequence for DBL version 4 by Dario B.
;
RECORD

FIB1,  D10
FIB2,  D10
FIBN,  D10

J,     D5
A2,    A2
A5,    A5
PROC
;----------------------------------------------------------------
XCALL FLAGS (0007000000,1)         ;Suppress STOP message

OPEN (1,O,'TT:')
DISPLAY (1,'First 10 Fibonacci Numbers:',10)

FIB2=1

FOR J=1 UNTIL 10
DO BEGIN
FIBN=FIB1+FIB2

A2=J,'ZX'
A5=FIBN,'ZZZZX'
DISPLAY (1,A2,' : ',A5,10)

FIB1=FIB2
FIB2=FIBN
END

CLOSE 1
END


## Dc

This needs a modern Dc with r (swap) and # (comment). It easily can be adapted to an older Dc, but it will impact readability a lot.

[               # todo: n(<2) -- 1 and break 2 levels
d -           # 0
1 +           # 1
q
] s1

[               # todo: n(>-1) -- F(n)
d 0=1         # n(!=0)
d 1=1         # n(!in {0,1})
2 - d 1 +     # (n-2) (n-1)
lF x          # (n-2) F(n-1)
r             # F(n-1) (n-2)
lF x          # F(n-1)+F(n-2)
+
] sF

33 lF x f
Output:
5702887


## Delphi

### Iterative

function FibonacciI(N: Word): UInt64;
var
Last, New: UInt64;
I: Word;
begin
if N < 2 then
Result := N
else begin
Last := 0;
Result := 1;
for I := 2 to N do
begin
New := Last + Result;
Last := Result;
Result := New;
end;
end;
end;


### Recursive

function Fibonacci(N: Word): UInt64;
begin
if N < 2 then
Result := N
else
Result := Fibonacci(N - 1) + Fibonacci(N - 2);
end;


### Matrix

Algorithm is based on

${\displaystyle \begin{pmatrix}1&1\\1&0\end{pmatrix}^n = \begin{pmatrix}F(n+1)&F(n)\\F(n)&F(n-1)\end{pmatrix}}$.
function fib(n: Int64): Int64;

type TFibMat = array[0..1] of array[0..1] of Int64;

function FibMatMul(a,b: TFibMat): TFibMat;
var i,j,k: integer;
tmp: TFibMat;
begin
for i := 0 to 1 do
for j := 0 to 1 do
begin
tmp[i,j] := 0;
for k := 0 to 1 do tmp[i,j] := tmp[i,j] + a[i,k] * b[k,j];
end;
FibMatMul := tmp;
end;

function FibMatExp(a: TFibMat; n: Int64): TFibmat;
begin
if n <= 1 then fibmatexp := a
else if (n mod 2 = 0) then FibMatExp := FibMatExp(FibMatMul(a,a), n div 2)
else if (n mod 2 = 1) then FibMatExp := FibMatMul(a, FibMatExp(FibMatMul(a,a), n div 2));
end;

var
matrix: TFibMat;

begin
matrix[0,0] := 1;
matrix[0,1] := 1;
matrix[1,0] := 1;
matrix[1,1] := 0;
if n > 1 then
matrix := fibmatexp(matrix,n-1);
fib := matrix[0,0];
end;


## DIBOL-11

      START    ;First 10 Fibonacci NUmbers

RECORD
FIB1,    D10,   0
FIB2,    D10,   1
FIBNEW,  D10
LOOPCNT, D2,    1

,            A32, "First 10 Fibonacci Numbers."

RECORD OUTPUT
LOOPOUT, A2
,         A3, " : "
FIBOUT, A10

PROC

OPEN(8,O,'TT:')

LOOP,
FIBNEW = FIB1 + FIB2
LOOPOUT = LOOPCNT, 'ZX'
FIBOUT = FIBNEW, 'ZZZZZZZZZX'

WRITES(8,OUTPUT)

FIB1 = FIB2
FIB2 = FIBNEW

LOOPCNT = LOOPCNT + 1
IF LOOPCNT .LE. 10 GOTO LOOP

CLOSE 8
END

## DWScript

function fib(N : Integer) : Integer;
begin
if N < 2 then Result := 1
else Result := fib(N-2) + fib(N-1);
End;


## Dyalect

func fib(n) {
if n < 2 {
return n
} else {
return fib(n - 1) + fib(n - 2)
}
}

print(fib(30))

## E

def fib(n) {
var s := [0, 1]
for _ in 0..!n {
def [a, b] := s
s := [b, a+b]
}
return s[0]
}

(This version defines fib(0) = 0 because OEIS A000045 does.)

## EasyLang

func fib n . res .
if n < 2
res = n
.
prev = 0
val = 1
for _ range n - 1
res = prev + val
prev = val
val = res
.
.
call fib 36 r
print r


Recursive (inefficient):

func fib n . res .
if n < 2
res = n
else
call fib n - 1 a
call fib n - 2 b
res = a + b
.
.
call fib 36 r
print r


## EchoLisp

Use memoization with the recursive version.

(define (fib n)
(if (< n 2) n
(+ (fib (- n 2)) (fib (1- n)))))

(remember 'fib #(0 1))

(for ((i 12)) (write (fib i)))
0 1 1 2 3 5 8 13 21 34 55 89


## ECL

### Analytic

//Calculates Fibonacci sequence up to n steps using Binet's closed form solution

FibFunction(UNSIGNED2 n) := FUNCTION
REAL Sqrt5 := Sqrt(5);
REAL Phi := (1+Sqrt(5))/2;
REAL Phi_Inv := 1/Phi;
UNSIGNED FibValue := ROUND( ( POWER(Phi,n)-POWER(Phi_Inv,n) ) /Sqrt5);
RETURN FibValue;
END;

FibSeries(UNSIGNED2 n) := FUNCTION

Fib_Layout := RECORD
UNSIGNED5 FibNum;
UNSIGNED5 FibValue;
END;

FibSeq := DATASET(n+1,
TRANSFORM
( Fib_Layout
, SELF.FibNum := COUNTER-1
, SELF.FibValue := IF(SELF.FibNum<2,SELF.FibNum, FibFunction(SELF.FibNum) )
)
);

RETURN FibSeq;

END; }


## EDSAC order code

This program calculates the nth—by default the tenth—number in the Fibonacci sequence and displays it (in binary) in the first word of storage tank 3.

[ Fibonacci sequence
==================

A program for the EDSAC

Calculates the nth Fibonacci
number and displays it at the
top of storage tank 3

The default value of n is 10

To calculate other Fibonacci
numbers, set the starting value
of the count to n-2

Works with Initial Orders 2 ]

T56K  [ set load point  ]
GK    [ set theta       ]

[ Orders ]

[  0 ]  T20@  [ a = 0           ]
A17@  [ a += y          ]
U18@  [ temp = a        ]
A16@  [ a += x          ]
T17@  [ y = a; a = 0    ]
A18@  [ a += temp       ]
T16@  [ x = a; a = 0    ]

A19@  [ a = count       ]
S15@  [ a -= 1          ]
U19@  [ count = a       ]
E@    [ if a>=0 go to θ ]

T20@  [ a = 0           ]
A17@  [ a += y          ]
T96F  [ C(96) = a; a = 0]

ZF    [ halt ]

[ Data ]

[ 15 ]  P0D   [ const: 1        ]
[ 16 ]  P0F   [ var: x = 0      ]
[ 17 ]  P0D   [ var: y = 1      ]
[ 18 ]  P0F   [ var: temp = 0   ]
[ 19 ]  P4F   [ var: count = 8  ]
[ 20 ]  P0F   [ used to clear a ]

EZPF  [ begin execution ]
Output:
00000000000110111

## Eiffel

class
APPLICATION

create
make

feature

fibonacci (n: INTEGER): INTEGER
require
non_negative: n >= 0
local
i, n2, n1, tmp: INTEGER
do
n2 := 0
n1 := 1
from
i := 1
until
i >= n
loop
tmp := n1
n1 := n2 + n1
n2 := tmp
i := i + 1
end
Result := n1
if n = 0 then
Result := 0
end
end

feature {NONE} -- Initialization

make
-- Run application.
do
print (fibonacci (0))
print (" ")
print (fibonacci (1))
print (" ")
print (fibonacci (2))
print (" ")
print (fibonacci (3))
print (" ")
print (fibonacci (4))
print ("%N")
end

end


## Ela

Tail-recursive function:

fib = fib' 0 1
where fib' a b 0 = a
fib' a b n = fib' b (a + b) (n - 1)

Infinite (lazy) list:

fib = fib' 1 1
where fib' x y = & x :: fib' y (x + y)

## Elena

Translation of: Smalltalk

ELENA 5.0 :

import extensions;

fibu(n)
{
int[] ac := new int[]{ 0,1 };
if (n < 2)
{
^ ac[n]
}
else
{
for(int i := 2, i <= n, i+=1)
{
int t := ac[1];
ac[1] := ac[0] + ac[1];
ac[0] := t
};

^ ac[1]
}
}

public program()
{
for(int i := 0, i <= 10, i+=1)
{
console.printLine(fibu(i))
}
}
Output:
0
1
1
2
3
5
8
13
21
34
55


### Alternative version using yieldable method

import extensions;

public FibonacciGenerator
{
yieldable next()
{
long n_2 := 1l;
long n_1 := 1l;

yield:n_2;
yield:n_1;

while(true)
{
long n := n_2 + n_1;

yield:n;

n_2 := n_1;
n_1 := n
}
}
}

public program()
{
auto e := new FibonacciGenerator();

for(int i := 0, i < 10, i += 1) {
console.printLine(e.next())
};

}

## Elixir

defmodule Fibonacci do
def fib(0), do: 0
def fib(1), do: 1
def fib(n), do: fib(0, 1, n-2)

def fib(_, prv, -1), do: prv
def fib(prvprv, prv, n) do
next = prv + prvprv
fib(prv, next, n-1)
end
end

IO.inspect Enum.map(0..10, fn i-> Fibonacci.fib(i) end)


Using Stream:

Stream.unfold({0,1}, fn {a,b} -> {a,{b,a+b}} end) |> Enum.take(10)

Output:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55]


## Elm

Naïve recursive implementation.

fibonacci : Int -> Int
fibonacci n = if n < 2 then
n
else
fibonacci(n - 2) + fibonacci(n - 1)


## Emacs Lisp

### version 1

(defun fib (n a b c)
(cond
((< c n) (fib n b (+ a b) (+ 1 c)))
((= c n) b)
(t a)))

(defun fibonacci (n)
(if (< n 2)
n
(fib n 0 1 1)))


### version 2

(defun fibonacci (n)
(let (vec i j k)
(if (< n 2)
n
(setq vec (make-vector (+ n 1) 0)
i 0
j 1
k 2)
(setf (aref vec 1) 1)
(while (<= k n)
(setf (aref vec k) (+ (elt vec i) (elt vec j)))
(setq i (1+ i)
j (1+ j)
k (1+ k)))
(elt vec n))))


Eval:

(insert
(mapconcat (lambda (n) (format "%d" (fibonacci n)))
(number-sequence 0 15) " "))

Output:
0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610


## Erlang

### Recursive

-module(fib).
-export([fib/1]).

fib(0) -> 0;
fib(1) -> 1;
fib(N) -> fib(N-1) + fib(N-2).


### Iterative

-module(fiblin).
-export([fib/1])

fib(0) -> 0;
fib(1) -> 1;
fib(2) -> 1;
fib(3) -> 2;
fib(4) -> 3;
fib(5) -> 5;

fib(N) when is_integer(N) -> fib(N - 6, 5, 8).
fib(N, A, B) -> if N < 1 -> B; true -> fib(N-1, B, A+B) end.


Evaluate:

io:write([fiblin:fib(X) || X <- lists:seq(1,10) ]).


Output:


[1,1,2,3,5,8,13,21,34,55]ok


### Iterative 2

fib(N) -> fib(N, 0, 1).

fib(0, Result, _Next) -> Result;
fib(Iter, Result, Next) -> fib(Iter-1, Next, Result+Next).


## ERRE

!-------------------------------------------
! derived from my book "PROGRAMMARE IN ERRE"
! iterative solution
!-------------------------------------------

PROGRAM FIBONACCI

!$DOUBLE !VAR F1#,F2#,TEMP#,COUNT%,N% BEGIN !main INPUT("Number",N%) F1=0 F2=1 REPEAT TEMP=F2 F2=F1+F2 F1=TEMP COUNT%=COUNT%+1 UNTIL COUNT%=N% PRINT("FIB(";N%;")=";F2) ! Obviously a FOR loop or a WHILE loop can ! be used to solve this problem END PROGRAM Output: Number? 20 FIB( 20 )= 6765  ## Euphoria ### 'Recursive' version Works with: Euphoria version any version function fibor(integer n) if n<2 then return n end if return fibor(n-1)+fibor(n-2) end function ### 'Iterative' version Works with: Euphoria version any version function fiboi(integer n) integer f0=0, f1=1, f if n<2 then return n end if for i=2 to n do f=f0+f1 f0=f1 f1=f end for return f end function ### 'Tail recursive' version Works with: Euphoria version 4.0.0 function fibot(integer n, integer u = 1, integer s = 0) if n < 1 then return s else return fibot(n-1,u+s,u) end if end function -- example: ? fibot(10) -- says 55 ### 'Paper tape' version Works with: Euphoria version 4.0.0 include std/mathcons.e -- for PINF constant enum ADD, MOVE, GOTO, OUT, TEST, TRUETO global sequence tape = { 0, 1, { ADD, 2, 1 }, { TEST, 1, PINF }, { TRUETO, 0 }, { OUT, 1, "%.0f\n" }, { MOVE, 2, 1 }, { MOVE, 0, 2 }, { GOTO, 3 } } global integer ip global integer test global atom accum procedure eval( sequence cmd ) atom i = 1 while i <= length( cmd ) do switch cmd[ i ] do case ADD then accum = tape[ cmd[ i + 1 ] ] + tape[ cmd[ i + 2 ] ] i += 2 case OUT then printf( 1, cmd[ i + 2], tape[ cmd[ i + 1 ] ] ) i += 2 case MOVE then if cmd[ i + 1 ] = 0 then tape[ cmd[ i + 2 ] ] = accum else tape[ cmd[ i + 2 ] ] = tape[ cmd[ i + 1 ] ] end if i += 2 case GOTO then ip = cmd[ i + 1 ] - 1 -- due to ip += 1 in main loop i += 1 case TEST then if tape[ cmd[ i + 1 ] ] = cmd[ i + 2 ] then test = 1 else test = 0 end if i += 2 case TRUETO then if test then if cmd[ i + 1 ] = 0 then abort(0) else ip = cmd[ i + 1 ] - 1 end if end if end switch i += 1 end while end procedure test = 0 accum = 0 ip = 1 while 1 do -- embedded sequences (assumed to be code) are evaluated -- atoms (assumed to be data) are ignored if sequence( tape[ ip ] ) then eval( tape[ ip ] ) end if ip += 1 end while ## Excel ### LAMBDA Binding the name FIBONACCI to the following lambda in the Excel worksheet Name Manager: FIBONACCI =LAMBDA(n, APPLYN(n - 2)( LAMBDA(xs, APPENDROWS(xs)( SUM( LASTNROWS(2)(xs) ) ) ) )({1;1}) )  And assuming that the following names are also bound to reusable generic lambdas in the Name manager: APPENDROWS =LAMBDA(xs, LAMBDA(ys, LET( nx, ROWS(xs), rowIndexes, SEQUENCE(nx + ROWS(ys)), colIndexes, SEQUENCE( 1, MAX(COLUMNS(xs), COLUMNS(ys)) ), IFERROR( IF(rowIndexes <= nx, INDEX(xs, rowIndexes, colIndexes), INDEX(ys, rowIndexes - nx, colIndexes) ), NA() ) ) ) ) APPLYN =LAMBDA(n, LAMBDA(f, LAMBDA(x, IF(0 < n, APPLYN(n - 1)(f)( f(x) ), x ) ) ) ) LASTNROWS =LAMBDA(n, LAMBDA(xs, LET( nRows, COUNTA(xs), x, MIN(nRows, n), IF(0 < n, INDEX( xs, SEQUENCE( x, 1, 1 + nRows - x, 1 ) ), NA() ) ) ) )  Output: The FIBONACCI(n) lambda defines a column of integers. Here we obtain a row, by composing FIBONACCI with the built-in TRANSPOSE function:  =TRANSPOSE(FIBONACCI(15)) fx A B C D E F G H I J K L M N O P 1 2 15 Fibonacci terms: 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 Or as a fold, obtaining just the Nth term of the Fibonacci series: FIBONACCI2 =LAMBDA(n, INDEX( FOLDL( LAMBDA(ab, LAMBDA(_, APPEND(INDEX(ab, 2))(SUM(ab)) ) ) )({0;1})( ENUMFROMTO(1)(n) ), 1 ) )  Assuming the following generic bindings in the Excel worksheet Name manager: APPEND =LAMBDA(xs, LAMBDA(ys, LET( nx, ROWS(xs), rowIndexes, SEQUENCE(nx + ROWS(ys)), colIndexes, SEQUENCE( 1, MAX(COLUMNS(xs), COLUMNS(ys)) ), IF(rowIndexes <= nx, INDEX(xs, rowIndexes, colIndexes), INDEX(ys, rowIndexes - nx, colIndexes) ) ) ) ) ENUMFROMTO =LAMBDA(a, LAMBDA(z, SEQUENCE(1 + z - a, 1, a, 1) ) ) FOLDL =LAMBDA(op, LAMBDA(a, LAMBDA(xs, IF( 2 > ROWS(xs), op(a)(xs), FOLDL(op)( op(a)( HEAD(xs) ) )( TAIL(xs) ) ) ) ) ) HEAD =LAMBDA(xs, INDEX(xs, 1, SEQUENCE(1, COLUMNS(xs))) ) TAIL =LAMBDA(xs, INDEX( xs, SEQUENCE(ROWS(xs) - 1, 1, 2, 1), SEQUENCE(1, COLUMNS(xs)) ) )  Output:  =FIBONACCI2(A2) fx A B 1 N Fibonacci 2 32 2178309 3 64 10610209857723 ## F# This is a fast [tail-recursive] approach using the F# big integer support: let fibonacci n : bigint = let rec f a b n = match n with | 0 -> a | 1 -> b | n -> (f b (a + b) (n - 1)) f (bigint 0) (bigint 1) n > fibonacci 100;; val it : bigint = 354224848179261915075I  Lazy evaluated using sequence workflow: let rec fib = seq { yield! [0;1]; for (a,b) in Seq.zip fib (Seq.skip 1 fib) -> a+b}  The above is extremely slow due to the nested recursions on sequences, which aren't very efficient at the best of times. The above takes seconds just to compute the 30th Fibonacci number! Lazy evaluation using the sequence unfold anamorphism is much much better as to efficiency: let fibonacci = Seq.unfold (fun (x, y) -> Some(x, (y, x + y))) (0I,1I) fibonacci |> Seq.nth 10000  Approach similar to the Matrix algorithm in C#, with some shortcuts involved. Since it uses exponentiation by squaring, calculations of fib(n) where n is a power of 2 are particularly quick. Eg. fib(2^20) was calculated in a little over 4 seconds on this poster's laptop. open System open System.Diagnostics open System.Numerics /// Finds the highest power of two which is less than or equal to a given input. let inline prevPowTwo (x : int) = let mutable n = x n <- n - 1 n <- n ||| (n >>> 1) n <- n ||| (n >>> 2) n <- n ||| (n >>> 4) n <- n ||| (n >>> 8) n <- n ||| (n >>> 16) n <- n + 1 match x with | x when x = n -> x | _ -> n/2 /// Evaluates the nth Fibonacci number using matrix arithmetic and /// exponentiation by squaring. let crazyFib (n : int) = let powTwo = prevPowTwo n /// Applies 2n rule repeatedly until another application of the rule would /// go over the target value (or the target value has been reached). let rec iter1 i q r s = match i with | i when i < powTwo -> iter1 (i*2) (q*q + r*r) (r * (q+s)) (r*r + s*s) | _ -> i, q, r, s /// Applies n+1 rule until the target value is reached. let rec iter2 (i, q, r, s) = match i with | i when i < n -> iter2 ((i+1), (q+r), q, r) | _ -> q match n with | 0 -> 1I | _ -> iter1 1 1I 1I 0I |> iter2  ## Factor ### Iterative : fib ( n -- m ) dup 2 < [ [ 0 1 ] dip [ swap [ + ] keep ] times drop ] unless ;  ### Recursive : fib ( n -- m ) dup 2 < [ [ 1 - fib ] [ 2 - fib ] bi + ] unless ;  ### Tail-Recursive : fib2 ( x y n -- a ) dup 1 < [ 2drop ] [ [ swap [ + ] keep ] dip 1 - fib2 ] if ; : fib ( n -- m ) [ 0 1 ] dip fib2 ;  ### Matrix Translation of: Ruby USE: math.matrices : fib ( n -- m ) dup 2 < [ [ { { 0 1 } { 1 1 } } ] dip 1 - m^n second second ] unless ;  ## Falcon ### Iterative function fib_i(n) if n < 2: return n fibPrev = 1 fib = 1 for i in [2:n] tmp = fib fib += fibPrev fibPrev = tmp end return fib end ### Recursive function fib_r(n) if n < 2 : return n return fib_r(n-1) + fib_r(n-2) end ### Tail Recursive function fib_tr(n) return fib_aux(n,0,1) end function fib_aux(n,a,b) switch n case 0 : return a default: return fib_aux(n-1,a+b,a) end end ## FALSE [[$0=~][1-@@\$@@+\$44,.@]#]f:
20n: {First 20 numbers}
0 1 n;f;!%%44,. {Output: "0,1,1,2,3,5..."}

## Fancy

class Fixnum {
def fib {
match self -> {
case 0 -> 0
case 1 -> 1
case _ -> self - 1 fib + (self - 2 fib)
}
}
}

15 times: |x| {
x fib println
}


## Fantom

Ints have a limit of 64-bits, so overflow errors occur after computing Fib(92) = 7540113804746346429.

class Main
{
static Int fib (Int n)
{
if (n < 2) return n
fibNums := [1, 0]
while (fibNums.size <= n)
{
fibNums.insert (0, fibNums[0] + fibNums[1])
}
return fibNums.first
}

public static Void main ()
{
20.times |n|
{
echo ("Fib($n) is${fib(n)}")
}
}
}

## Fermat

Func Fibonacci(n) = if n<0 then -(-1)^n*Fibonacci(-n) else if n<2 then n else
Array fib[n+1];
fib[1] := 0;
fib[2] := 1;
for i = 2, n do
fib[i+1]:=fib[i]+fib[i-1]
od;
Return(fib[n+1]);
fi;
fi;
.

## Fexl

# (fib n) = the nth Fibonacci number
\fib=
(
\loop==
(\x\y\n
le n 0 x;
\z=(+ x y)
\n=(- n 1)
loop y z n
)
loop 0 1
)

# Now test it:
for 0 20 (\n say (fib n))
Output:
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
6765


## Fish

Outputs Fibonacci numbers until stopped.

10::n' 'o&+&$10.  ## FOCAL 01.10 TYPE "FIBONACCI NUMBERS" ! 01.20 ASK "N =", N 01.30 SET A=0 01.40 SET B=1 01.50 FOR I=2,N; DO 2.0 01.60 TYPE "F(N) ", %8, B, ! 01.70 QUIT 02.10 SET T=B 02.20 SET B=A+B 02.30 SET A=T Output: FIBONACCI NUMBERS N =:20 F(N) = 6765 ## Forth : fib ( n -- fib ) 0 1 rot 0 ?do over + swap loop drop ;  Or, for negative-index support: : fib ( n -- Fn ) 0 1 begin rot dup 0 = if drop drop exit then dup 0 > if 1 - rot rot dup rot + else 1 + rot rot over - swap then again ;  Since there are only a fixed and small amount of Fibonacci numbers that fit in a machine word, this FORTH version creates a table of Fibonacci numbers at compile time. It stops compiling numbers when there is arithmetic overflow (the number turns negative, indicating overflow.) : F-start, here 1 0 dup , ; : F-next, over + swap dup 0> IF dup , true ELSE false THEN ; : computed-table ( compile: 'start 'next / run: i -- x ) create >r execute BEGIN r@ execute not UNTIL rdrop does> swap cells + @ ; ' F-start, ' F-next, computed-table fibonacci 2drop here swap - cell/ Constant #F/64 \ # of fibonacci numbers generated 16 fibonacci . 987 ok #F/64 . 93 ok 92 fibonacci . 7540113804746346429 ok \ largest number generated.  ## Fortran ### FORTRAN IV C FIBONACCI SEQUENCE - FORTRAN IV NN=46 DO 1 I=0,NN 1 WRITE(*,300) I,IFIBO(I) 300 FORMAT(1X,I2,1X,I10) END C FUNCTION IFIBO(N) IF(N) 9,1,2 1 IFN=0 GOTO 9 2 IF(N-1) 9,3,4 3 IFN=1 GOTO 9 4 IFNM1=0 IFN=1 DO 5 I=2,N IFNM2=IFNM1 IFNM1=IFN 5 IFN=IFNM1+IFNM2 9 IFIBO=IFN END  Output:  0 0 1 1 2 1 3 2 4 3 5 5 6 8 7 13 8 21 9 34 10 55 ... 45 1134903170 46 1836311903  ### FORTRAN 77  FUNCTION IFIB(N) IF (N.EQ.0) THEN ITEMP0=0 ELSE IF (N.EQ.1) THEN ITEMP0=1 ELSE IF (N.GT.1) THEN ITEMP1=0 ITEMP0=1 DO 1 I=2,N ITEMP2=ITEMP1 ITEMP1=ITEMP0 ITEMP0=ITEMP1+ITEMP2 1 CONTINUE ELSE ITEMP1=1 ITEMP0=0 DO 2 I=-1,N,-1 ITEMP2=ITEMP1 ITEMP1=ITEMP0 ITEMP0=ITEMP2-ITEMP1 2 CONTINUE END IF IFIB=ITEMP0 END  Test program  EXTERNAL IFIB CHARACTER*10 LINE PARAMETER ( LINE = '----------' ) WRITE(*,900) 'N', 'F[N]', 'F[-N]' WRITE(*,900) LINE, LINE, LINE DO 1 N = 0, 10 WRITE(*,901) N, IFIB(N), IFIB(-N) 1 CONTINUE 900 FORMAT(3(X,A10)) 901 FORMAT(3(X,I10)) END  Output:  N F[N] F[-N] ---------- ---------- ---------- 0 0 0 1 1 1 2 1 -1 3 2 2 4 3 -3 5 5 5 6 8 -8 7 13 13 8 21 -21 9 34 34 10 55 -55  ### Recursive In ISO Fortran 90 or later, use a RECURSIVE function: module fibonacci contains recursive function fibR(n) result(fib) integer, intent(in) :: n integer :: fib select case (n) case (:0); fib = 0 case (1); fib = 1 case default; fib = fibR(n-1) + fibR(n-2) end select end function fibR  ### Iterative In ISO Fortran 90 or later:  function fibI(n) integer, intent(in) :: n integer, parameter :: fib0 = 0, fib1 = 1 integer :: fibI, back1, back2, i select case (n) case (:0); fibI = fib0 case (1); fibI = fib1 case default fibI = fib1 back1 = fib0 do i = 2, n back2 = back1 back1 = fibI fibI = back1 + back2 end do end select end function fibI end module fibonacci  Test program program fibTest use fibonacci do i = 0, 10 print *, fibr(i), fibi(i) end do end program fibTest  Output: 0 0 1 1 1 1 2 2 3 3 5 5 8 8 13 13 21 21 34 34 55 55  ## Free Pascal See also: Pascal type /// domain for Fibonacci function /// where result is within nativeUInt // You can not name it fibonacciDomain, // since the Fibonacci function itself // is defined for all whole numbers // but the result beyond F(n) exceeds high(nativeUInt). fibonacciLeftInverseRange = {$ifdef CPU64} 0..93 {$else} 0..47 {$endif};

{**
implements Fibonacci sequence iteratively

\param n the index of the Fibonacci number to calculate
\returns the Fibonacci value at n
}
function fibonacci(const n: fibonacciLeftInverseRange): nativeUInt;
type
/// more meaningful identifiers than simple integers
relativePosition = (previous, current, next);
var
/// temporary iterator variable
i: longword;
/// holds preceding fibonacci values
f: array[relativePosition] of nativeUInt;
begin
f[previous] := 0;
f[current] := 1;

// note, in Pascal for-loop-limits are inclusive
for i := 1 to n do
begin
f[next] := f[previous] + f[current];
f[previous] := f[current];
f[current] := f[next];
end;

// assign to previous, bc f[current] = f[next] for next iteration
fibonacci := f[previous];
end;


## FreeBASIC

Extended sequence coded big integer.

'Fibonacci extended
'Freebasic version 24  Windows
Dim Shared ADDQmod(0 To 19) As Ubyte
Dim Shared ADDbool(0 To 19) As Ubyte

For z As Integer=0 To 19
Next z

Function plusINT(NUM1 As String,NUM2 As String) As String
Dim As Byte flag
#macro finish()
three=Ltrim(three,"0")
If three="" Then Return "0"
If flag=1 Then Swap NUM2,NUM1
Return three
Exit Function
#endmacro
var lenf=Len(NUM1)
var lens=Len(NUM2)
If lens>lenf Then
Swap NUM2,NUM1
Swap lens,lenf
flag=1
End If

var diff=lenf-lens-Sgn(lenf-lens)
var three="0"+NUM1
var two=String(lenf-lens,"0")+NUM2
Dim As Integer n2

For n2=lenf-1 To diff Step -1
Next n2
finish()
End If
If n2=-1 Then
finish()
End If

For n2=n2 To 0 Step -1
Next n2
finish()
End Function

Function  fibonacci(n As Integer) As String
Dim As String sl,l,term
sl="0": l="1"
If n=1 Then Return "0"
If n=2 Then Return "1"
n=n-2
For x As Integer= 1 To n
term=plusINT(l,sl)
sl=l
l=term
Next x
Function =term
End Function

'==============  EXAMPLE ===============
print "THE SEQUENCE TO 10:"
print
For n As Integer=1 To 10
Print "term";n;": "; fibonacci(n)
Next n
print
print "Selected Fibonacci number"
print "Fibonacci 500"
print
print fibonacci(500)
Sleep

Output:
THE SEQUENCE TO 10:

term 1: 0
term 2: 1
term 3: 1
term 4: 2
term 5: 3
term 6: 5
term 7: 8
term 8: 13
term 9: 21
term 10: 34

Selected Fibonacci number
Fibonacci 500

86168291600238450732788312165664788095941068326060883324529903470149056115823592
713458328176574447204501

## Frink

All of Frink's integers can be arbitrarily large.

fibonacciN[n] :=
{
a = 0
b = 1
count = 0
while count < n
{
[a,b] = [b, a + b]
count = count + 1
}
return a
}

## FRISC Assembly

To find the nth Fibonacci number, call this subroutine with n in register R0: the answer will be returned in R0 too. Contents of other registers are preserved.

FIBONACCI   PUSH   R1
PUSH   R2
PUSH   R3

MOVE   0,  R1
MOVE   1,  R2

FIB_LOOP    SUB    R0,  1, R0
JP_Z   FIB_DONE

MOVE   R2, R3
MOVE   R3, R1

JP     FIB_LOOP

FIB_DONE    MOVE   R2, R0

POP    R3
POP    R2
POP    R1

RET

## FunL

### Recursive

def
fib( 0 ) = 0
fib( 1 ) = 1
fib( n ) = fib( n - 1 ) + fib( n - 2 )

### Tail Recursive

def fib( n ) =
def
_fib( 0, prev, _ )    = prev
_fib( 1, _,    next ) = next
_fib( n, prev, next ) = _fib( n - 1, next, next + prev )

_fib( n, 0, 1 )

### Lazy List

val fib =
def _fib( a, b ) = a # _fib( b, a + b )

_fib( 0, 1 )

println( fib(10000) )
Output:
33644764876431783266621612005107543310302148460680063906564769974680081442166662368155595513633734025582065332680836159373734790483865268263040892463056431887354544369559827491606602099884183933864652731300088830269235673613135117579297437854413752130520504347701602264758318906527890855154366159582987279682987510631200575428783453215515103870818298969791613127856265033195487140214287532698187962046936097879900350962302291026368131493195275630227837628441540360584402572114334961180023091208287046088923962328835461505776583271252546093591128203925285393434620904245248929403901706233888991085841065183173360437470737908552631764325733993712871937587746897479926305837065742830161637408969178426378624212835258112820516370298089332099905707920064367426202389783111470054074998459250360633560933883831923386783056136435351892133279732908133732642652633989763922723407882928177953580570993691049175470808931841056146322338217465637321248226383092103297701648054726243842374862411453093812206564914032751086643394517512161526545361333111314042436854805106765843493523836959653428071768775328348234345557366719731392746273629108210679280784718035329131176778924659089938635459327894523777674406192240337638674004021330343297496902028328145933418826817683893072003634795623117103101291953169794607632737589253530772552375943788434504067715555779056450443016640119462580972216729758615026968443146952034614932291105970676243268515992834709891284706740862008587135016260312071903172086094081298321581077282076353186624611278245537208532365305775956430072517744315051539600905168603220349163222640885248852433158051534849622434848299380905070483482449327453732624567755879089187190803662058009594743150052402532709746995318770724376825907419939632265984147498193609285223945039707165443156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875


### Iterative

def fib( n ) =
a, b = 0, 1

for i <- 1..n
a, b = b, a+b

a

### Binet's Formula

import math.sqrt

def fib( n ) =
phi = (1 + sqrt( 5 ))/2
int( (phi^n - (-phi)^-n)/sqrt(5) + .5 )

### Matrix Exponentiation

def mul( a, b ) =
res = array( a.length(), b(0).length() )

for i <- 0:a.length(), j <- 0:b(0).length()
res( i, j ) = sum( a(i, k)*b(k, j) | k <- 0:b.length() )

vector( res )

def
pow( _, 0 ) = ((1, 0), (0, 1))
pow( x, 1 ) = x
pow( x, n )
| 2|n = pow( mul(x, x), n\2 )
| otherwise = mul(x, pow( mul(x, x), (n - 1)\2 ) )

def fib( n ) = pow( ((0, 1), (1, 1)), n )(0, 1)

for i <- 0..10
println( fib(i) )
Output:
0
1
1
2
3
5
8
13
21
34
55


## Futhark

### Iterative

fun main(n: int): int =
loop((a,b) = (0,1)) = for _i < n do
(b, a + b)
in a


## FutureBasic

### Iterative

window 1, @"Fibonacci Sequence", (0,0,480,620)

local fn Fibonacci( n as long ) as long
static long s1
static long s2
long        temp

if ( n < 2 )
s1 = n
exit fn
else
temp = s1 + s2
s2 = s1
s1 = temp
exit fn
end if
end fn = s1

long i
CFTimeInterval t

t = fn CACurrentMediaTime

for i = 0 to 40
print i;@".\t";fn Fibonacci(i)
next i

print : printf @"Compute time: %.3f ms",(fn CACurrentMediaTime-t)*1000

HandleEvents

Output:

0.	0
1.	1
2.	1
3.	2
4.	3
5.	5
6.	8
7.	13
8.	21
9.	34
10.	55
11.	89
12.	144
13.	233
14.	377
15.	610
16.	987
17.	1597
18.	2584
19.	4181
20.	6765
21.	10946
22.	17711
23.	28657
24.	46368
25.	75025
26.	121393
27.	196418
28.	317811
29.	514229
30.	832040
31.	1346269
32.	2178309
33.	3524578
34.	5702887
35.	9227465
36.	14930352
37.	24157817
38.	39088169
39.	63245986
40.	102334155

Compute time: 2.143 ms

### Recursive

Cost is a time penalty

local fn Fibonacci( n as NSInteger ) as NSInteger
NSInteger result
if n < 2 then result = n : exit fn
result = fn Fibonacci( n-1 ) + fn Fibonacci( n-2 )
end fn = result

window 1

NSInteger i
CFTimeInterval t

t = fn CACurrentMediaTime
for i = 0 to 40
print i;@".\t";fn Fibonacci(i)
next
print : printf @"Compute time: %.3f ms",(fn CACurrentMediaTime-t)*1000

HandleEvents
Output:
0.	0
1.	1
2.	1
3.	2
4.	3
5.	5
6.	8
7.	13
8.	21
9.	34
10.	55
11.	89
12.	144
13.	233
14.	377
15.	610
16.	987
17.	1597
18.	2584
19.	4181
20.	6765
21.	10946
22.	17711
23.	28657
24.	46368
25.	75025
26.	121393
27.	196418
28.	317811
29.	514229
30.	832040
31.	1346269
32.	2178309
33.	3524578
34.	5702887
35.	9227465
36.	14930352
37.	24157817
38.	39088169
39.	63245986
40.	102334155

Compute time: 2844.217 ms


## Fōrmulæ

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.

## GAP

fib := function(n)
local a;
a := [[0, 1], [1, 1]]^n;
return a[1][2];
end;


GAP has also a buit-in function for that.

Fibonacci(n);


## Gecho

0 1 dup wover + dup wover + dup wover + dup wover +

Prints the first several fibonacci numbers...

## GFA Basic

'
' Compute nth Fibonacci number
'
' open a window for display
OPENW 1
CLEARW 1
' Display some fibonacci numbers
' Fib(46) is the largest number GFA Basic can reach
' (long integers are 4 bytes)
FOR i%=0 TO 46
PRINT "fib(";i%;")=";@fib(i%)
NEXT i%
' wait for a key press and tidy up
~INP(2)
CLOSEW 1
'
' Function to compute nth fibonacci number
' n must be in range 0 to 46, inclusive
'
FUNCTION fib(n%)
LOCAL n0%,n1%,nn%,i%
n0%=0
n1%=1
SELECT n%
CASE 0
RETURN n0%
CASE 1
RETURN n1%
DEFAULT
FOR i%=2 TO n%
nn%=n0%+n1%
n0%=n1%
n1%=nn%
NEXT i%
RETURN nn%
ENDSELECT
ENDFUNC


## GML

///fibonacci(n)
//Returns the nth fibonacci number

var n, numb;
n = argument0;

if (n == 0)
{
numb = 0;
}
else
{
var fm2, fm1;
fm2 = 0;
fm1 = 1;
numb = 1;
repeat(n-1)
{
numb = fm2+fm1;
fm2 = fm1;
fm1 = numb;
}
}

return numb;

## Go

### Recursive

func fib(a int) int {
if a < 2 {
return a
}
return fib(a - 1) + fib(a - 2)
}


### Iterative

import (
"math/big"
)

func fib(n uint64) *big.Int {
if n < 2 {
return big.NewInt(int64(n))
}
a, b := big.NewInt(0), big.NewInt(1)
for n--; n > 0; n-- {
a, b = b, a
}
return b
}


### Iterative using a closure

func fibNumber() func() int {
fib1, fib2 := 0, 1
return func() int {
fib1, fib2 = fib2, fib1 + fib2
return fib1
}
}

func fibSequence(n int) int {
f := fibNumber()
fib := 0
for i := 0; i < n; i++ {
fib = f()
}
return fib
}


### Using a goroutine and channel

func fib(c chan int) {
a, b := 0, 1
for {
c <- a
a, b = b, a+b
}
}

func main() {
c := make(chan int)
go fib(c)
for i := 0; i < 10; i++ {
fmt.Println(<-c)
}
}


## Groovy

Full "extra credit" solutions.

### Recursive

A recursive closure must be pre-declared.

def rFib
rFib = {
it == 0   ? 0
: it == 1 ? 1
: it > 1  ? rFib(it-1) + rFib(it-2)
/*it < 0*/: rFib(it+2) - rFib(it+1)

}


### Iterative

def iFib = {
it == 0   ? 0
: it == 1 ? 1
: it > 1  ? (2..it).inject([0,1]){i, j -> [i[1], i[0]+i[1]]}[1]
/*it < 0*/: (-1..it).inject([0,1]){i, j -> [i[1]-i[0], i[0]]}[0]
}


### Analytic

final φ = (1 + 5**(1/2))/2
def aFib = { (φ**it - (-φ)**(-it))/(5**(1/2)) as BigInteger }


Test program:

def time = { Closure c ->
def start = System.currentTimeMillis()
def result = c()
def elapsedMS = (System.currentTimeMillis() - start)/1000
printf '(%6.4fs elapsed)', elapsedMS
result
}

print "  F(n)      elapsed time   "; (-10..10).each { printf ' %3d', it }; println()
print "--------- -----------------"; (-10..10).each { print ' ---' }; println()
[recursive:rFib, iterative:iFib, analytic:aFib].each { name, fib ->
printf "%9s ", name
def fibList = time { (-10..10).collect {fib(it)} }
fibList.each { printf ' %3d', it }
println()
}

Output:
  F(n)      elapsed time    -10  -9  -8  -7  -6  -5  -4  -3  -2  -1   0   1   2   3   4   5   6   7   8   9  10
--------- ----------------- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
recursive (0.0080s elapsed) -55  34 -21  13  -8   5  -3   2  -1   1   0   1   1   2   3   5   8  13  21  34  55
iterative (0.0040s elapsed) -55  34 -21  13  -8   5  -3   2  -1   1   0   1   1   2   3   5   8  13  21  34  55
analytic (0.0030s elapsed) -55  34 -21  13  -8   5  -3   2  -1   1   0   1   1   2   3   5   8  13  21  34  55

## Harbour

### Recursive

#include "harbour.ch"
Function fibb(a,b,n)
return(if(--n>0,fibb(b,a+b,n),a))

### Iterative

#include "harbour.ch"
Function fibb(n)
local fnow:=0, fnext:=1, tempf
while (--n>0)
tempf:=fnow+fnext
fnow:=fnext
fnext:=tempf
end while
return(fnext)

### Analytic

Works with: exact-real version 0.12.5.1

Using Binet's formula and exact real arithmetic library we can calculate arbitrary Fibonacci number exactly.

import Data.CReal

phi = (1 + sqrt 5) / 2

fib :: (Integral b) => b -> CReal 0
fib n = (phi^^n - (-phi)^^(-n))/sqrt 5


Let's try it for large numbers:

λ> fib 10 :: CReal 0
55
(0.01 secs, 137,576 bytes)
λ> fib 100 :: CReal 0
354224848179261915075
(0.01 secs, 253,152 bytes)
λ> fib 10000 :: CReal 0
33644764876431783266621612005107543310302148460680063906564769974680081442166662368155595513633734025582065332680836159373734790483865268263040892463056431887354544369559827491606602099884183933864652731300088830269235673613135117579297437854413752130520504347701602264758318906527890855154366159582987279682987510631200575428783453215515103870818298969791613127856265033195487140214287532698187962046936097879900350962302291026368131493195275630227837628441540360584402572114334961180023091208287046088923962328835461505776583271252546093591128203925285393434620904245248929403901706233888991085841065183173360437470737908552631764325733993712871937587746897479926305837065742830161637408969178426378624212835258112820516370298089332099905707920064367426202389783111470054074998459250360633560933883831923386783056136435351892133279732908133732642652633989763922723407882928177953580570993691049175470808931841056146322338217465637321248226383092103297701648054726243842374862411453093812206564914032751086643394517512161526545361333111314042436854805106765843493523836959653428071768775328348234345557366719731392746273629108210679280784718035329131176778924659089938635459327894523777674406192240337638674004021330343297496902028328145933418826817683893072003634795623117103101291953169794607632737589253530772552375943788434504067715555779056450443016640119462580972216729758615026968443146952034614932291105970676243268515992834709891284706740862008587135016260312071903172086094081298321581077282076353186624611278245537208532365305775956430072517744315051539600905168603220349163222640885248852433158051534849622434848299380905070483482449327453732624567755879089187190803662058009594743150052402532709746995318770724376825907419939632265984147498193609285223945039707165443156421328157688908058783183404917434556270520223564846495196112460268313970975069382648706613264507665074611512677522748621598642530711298441182622661057163515069260029861704945425047491378115154139941550671256271197133252763631939606902895650288268608362241082050562430701794976171121233066073310059947366875
(0.02 secs, 4,847,128 bytes)
λ> fib (-10) :: CReal 0
-55
(0.01 secs, 138,408 bytes)


### Recursive

Simple definition, very inefficient.

fib x =
if x < 1
then 0
else if x < 2
then 1
else fib (x - 1) + fib (x - 2)


### Recursive with Memoization

Very fast.

fib x =
if x < 1
then 0
else if x == 1
then 1
else fibs !! (x - 1) + fibs !! (x - 2)
where
fibs = map fib [0 ..]


### Recursive with Memoization using memoized library

Even faster and simpler is to use a defined memoizer (e.g. from MemoTrie package):

import Data.MemoTrie
fib :: Integer -> Integer
fib = memo f where
f 0 = 0
f 1 = 1
f n = fib (n-1) + fib (n-2)


You can rewrite this without introducing f explicitly

import Data.MemoTrie
fib :: Integer -> Integer
fib = memo $\x -> case x of 0 -> 0 1 -> 1 n -> fib (n-1) + fib (n-2)  Or using LambdaCase extension you can write it even shorter: {-# Language LambdaCase #-} import Data.MemoTrie fib :: Integer -> Integer fib = memo$ \case
0 -> 0
1 -> 1
n -> fib (n-1) + fib (n-2)


The version that supports negative numbers:

{-# Language LambdaCase #-}
import Data.MemoTrie
fib :: Integer -> Integer
foldl' --'
($$a, b) _ -> (b, a + b)) (0, 1) [1 .. n]  ### With matrix exponentiation Adapting the (rather slow) code from Matrix exponentiation operator, we can simply write: import Data.List (transpose) fib :: (Integral b, Num a) => b -> a fib 0 = 0 -- this line is necessary because "something ^ 0" returns "fromInteger 1", which unfortunately -- in our case is not our multiplicative identity (the identity matrix) but just a 1x1 matrix of 1 fib n = (last . head . unMat) (Mat [[1, 1], [1, 0]] ^ n) -- Code adapted from Matrix exponentiation operator task --------------------- (<+>) :: Num c => [c] -> [c] -> [c] (<+>) = zipWith (+) (<*>) :: Num a => [a] -> [a] -> a (<*>) = (sum .) . zipWith (*) newtype Mat a = Mat { unMat :: [[a]] } deriving (Eq) instance Show a => Show (Mat a) where show xm = "Mat " ++ show (unMat xm) instance Num a => Num (Mat a) where negate xm = Mat  map (map negate)  unMat xm xm + ym = Mat  zipWith (<+>) (unMat xm) (unMat ym) xm * ym = Mat [ [ xs Main.<*> ys -- to distinguish from standard applicative operator | ys <- transpose  unMat ym ] | xs <- unMat xm ] fromInteger n = Mat [[fromInteger n]] abs = undefined signum = undefined -- TEST ---------------------------------------------------------------------- main :: IO () main = (print . take 10 . show . fib) (10 ^ 5)  So, for example, the hundred-thousandth Fibonacci number starts with the digits: Output: "2597406934" ### With recurrence relations Using Fib[m=3n+r] recurrence identities: import Control.Arrow ((&&&)) fibstep :: (Integer, Integer) -> (Integer, Integer) fibstep (a, b) = (b, a + b) fibnums :: [Integer] fibnums = map fst  iterate fibstep (0, 1) fibN2 :: Integer -> (Integer, Integer) fibN2 m | m < 10 = iterate fibstep (0, 1) !! fromIntegral m fibN2 m = fibN2_next (n, r) (fibN2 n) where (n, r) = quotRem m 3 fibN2_next (n, r) (f, g) | r == 0 = (a, b) -- 3n ,3n+1 | r == 1 = (b, c) -- 3n+1,3n+2 | r == 2 = (c, d) -- 3n+2,3n+3 (*) where a = 5 * f ^ 3 + if even n then 3 * f else (-3 * f) -- 3n b = g ^ 3 + 3 * g * f ^ 2 - f ^ 3 -- 3n+1 c = g ^ 3 + 3 * g ^ 2 * f + f ^ 3 -- 3n+2 d = 5 * g ^ 3 + if even n then (-3 * g) else 3 * g -- 3(n+1) (*) main :: IO () main = print  (length &&& take 20) . show . fst  fibN2 (10 ^ 2)  Output: (21,"35422484817926191507") (fibN2 n) directly calculates a pair (f,g) of two consecutive Fibonacci numbers, (Fib[n], Fib[n+1]), from recursively calculated such pair at about n/3:  *Main> (length &&& take 20) . show . fst  fibN2 (10^6) (208988,"19532821287077577316")  The above should take less than 0.1s to calculate on a modern box. Other identities that could also be used are here. In particular, for (n-1,n) ---> (2n-1,2n) transition which is equivalent to the matrix exponentiation scheme, we have f (n,(a,b)) = (2*n,(a*a+b*b,2*a*b+b*b)) -- iterate f (1,(0,1)) ; b is nth  and for (n,n+1) ---> (2n,2n+1) (derived from d'Ocagne's identity, for example), g (n,(a,b)) = (2*n,(2*a*b-a*a,a*a+b*b)) -- iterate g (1,(1,1)) ; a is nth  ## Haxe ### Iterative static function fib(steps:Int, handler:Int->Void) { var current = 0; var next = 1; for (i in 1...steps) { handler(current); var temp = current + next; current = next; next = temp; } handler(current); }  ### As Iterator class FibIter { private var current = 0; private var nextItem = 1; private var limit:Int; public function new(limit) this.limit = limit; public function hasNext() return limit > 0; public function next() { limit--; var ret = current; var temp = current + nextItem; current = nextItem; nextItem = temp; return ret; } }  Used like: for (i in new FibIter(10)) Sys.println(i);  ## HicEst REAL :: Fibonacci(10) Fibonacci = (==2) + Fibonacci(-1) + Fibonacci(-2) WRITE(ClipBoard) Fibonacci ! 0 1 1 2 3 5 8 13 21 34 ## Hoon |= n=@ud =/ a=@ud 0 =/ b=@ud 1 |- ?: =(n 0) a (a b, b (add a b), n (dec n)) ## Hope ### Recursive dec f : num -> num; --- f 0 <= 0; --- f 1 <= 1; --- f(n+2) <= f n + f(n+1); ### Tail-recursive dec fib : num -> num; --- fib n <= l (1, 0, n) whererec l == \(a,b,succ c) => if c<1 then a else l((a+b),a,c) |(a,b,0) => 0; ### With lazy lists This language, being one of Haskell's ancestors, also has lazy lists. Here's the (infinite) list of all Fibonacci numbers: dec fibs : list num; --- fibs <= fs whererec fs == 0::1::map (+) (tail fs||fs); The nth Fibonacci number is then just fibs @ n. ## Hy Recursive implementation. (defn fib [n] (if (< n 2) n (+ (fib (- n 2)) (fib (- n 1)))))  ## Icon and Unicon Icon has built-in support for big numbers. First, a simple recursive solution augmented by caching for non-negative input. This examples computes fib(1000) if there is no integer argument. procedure main(args) write(fib(integer(!args) | 1000) end procedure fib(n) static fCache initial { fCache := table() fCache[0] := 0 fCache[1] := 1 } /fCache[n] := fib(n-1) + fib(n-2) return fCache[n] end  The above solution is similar to the one provided fib in memrfncs Now, an O(logN) solution. For large N, it takes far longer to convert the result to a string for output than to do the actual computation. This example computes fib(1000000) if there is no integer argument. procedure main(args) write(fib(integer(!args) | 1000000)) end procedure fib(n) return fibMat(n)[1] end procedure fibMat(n) if n <= 0 then return [0,0] if n = 1 then return [1,0] fp := fibMat(n/2) c := fp[1]*fp[1] + fp[2]*fp[2] d := fp[1]*(fp[1]+2*fp[2]) if n%2 = 1 then return [c+d, d] else return [d, c] end  ## IDL ### Recursive function fib,n if n lt 3 then return,1L else return, fib(n-1)+fib(n-2) end  Execution time O(2^n) until memory is exhausted and your machine starts swapping. Around fib(35) on a 2GB Core2Duo. ### Iterative function fib,n psum = (csum = 1uL) if n lt 3 then return,csum for i = 3,n do begin nsum = psum + csum psum = csum csum = nsum endfor return,nsum end  Execution time O(n). Limited by size of uLong to fib(49) ### Analytic function fib,n q=1/( p=(1+sqrt(5))/2 ) return,round((p^n+q^n)/sqrt(5)) end  Execution time O(1), only limited by the range of LongInts to fib(48). ## Idris ### Analytic fibAnalytic : Nat -> Double fibAnalytic n = floor  ((pow goldenRatio n) - (pow (-1.0/goldenRatio) n)) / sqrt(5) where goldenRatio : Double goldenRatio = (1.0 + sqrt(5)) / 2.0  ### Recursive fibRecursive : Nat -> Nat fibRecursive Z = Z fibRecursive (S Z) = (S Z) fibRecursive (S (S n)) = fibRecursive (S n) + fibRecursive n  ### Iterative fibIterative : Nat -> Nat fibIterative n = fibIterative' n Z (S Z) where fibIterative' : Nat -> Nat -> Nat -> Nat fibIterative' Z a _ = a fibIterative' (S n) a b = fibIterative' n b (a + b)  ### Lazy fibLazy : Lazy (List Nat) fibLazy = 0 :: 1 :: zipWith (+) fibLazy ( case fibLazy of (x::xs) => xs [] => [])  ## J The Fibonacci Sequence essay on the J Wiki presents a number of different ways of obtaining the nth Fibonacci number. Here is one:  fibN=: (-&2 +&: -&1)^:(1&<) M."0  Examples:  fibN 12 144 fibN i.31 0 1 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  (This implementation is doubly recursive except that results are cached across function calls.) ## Java ### Iterative public static long itFibN(int n) { if (n < 2) return n; long ans = 0; long n1 = 0; long n2 = 1; for(n--; n > 0; n--) { ans = n1 + n2; n1 = n2; n2 = ans; } return ans; }  /** * O(log(n)) */ public static long fib(long n) { if (n <= 0) return 0; long i = (int) (n - 1); long a = 1, b = 0, c = 0, d = 1, tmp1,tmp2; while (i > 0) { if (i % 2 != 0) { tmp1 = d * b + c * a; tmp2 = d * (b + a) + c * b; a = tmp1; b = tmp2; } tmp1 = (long) (Math.pow(c, 2) + Math.pow(d, 2)); tmp2 = d * (2 * c + d); c = tmp1; d = tmp2; i = i / 2; } return a + b; }  ### Recursive public static long recFibN(final int n) { return (n < 2) ? n : recFibN(n - 1) + recFibN(n - 2); }  ### Caching-recursive A variant on recursive, that caches previous results, reducing complexity from O(n2) to simply O(n). Leveraging Java’s Map.computeIfAbsent makes this thread-safe, and the implementation pretty trivial. public class Fibonacci { static final Map<Integer, Long> cache = new HashMap<>(); static { cache.put(1, 1L); cache.put(2, 1L); } public static long get(int n) { return (n < 2) ? n : impl(n); } private static long impl(int n) { return cache.computeIfAbsent(n, k -> impl(k-1) + impl(k-2)); } }  ### Analytic This method works up to the 92nd Fibonacci number. After that, it goes out of range. public static long anFibN(final long n) { double p = (1 + Math.sqrt(5)) / 2; double q = 1 / p; return (long) ((Math.pow(p, n) + Math.pow(q, n)) / Math.sqrt(5)); }  ### Tail-recursive public static long fibTailRec(final int n) { return fibInner(0, 1, n); } private static long fibInner(final long a, final long b, final int n) { return n < 1 ? a : n == 1 ? b : fibInner(b, a + b, n - 1); }  ### Streams import java.util.function.LongUnaryOperator; import java.util.stream.LongStream; public class FibUtil { public static LongStream fibStream() { return LongStream.iterate( 1l, new LongUnaryOperator() { private long lastFib = 0; @Override public long applyAsLong( long operand ) { long ret = operand + lastFib; lastFib = operand; return ret; } }); } public static long fib(long n) { return fibStream().limit( n ).reduce((prev, last) -> last).getAsLong(); } }  ## JavaScript ### ES5 #### Recursive Basic recursive function: function fib(n) { return n<2?n:fib(n-1)+fib(n-2); }  Can be rewritten as: function fib(n) { if (n<2) { return n; } else { return fib(n-1)+fib(n-2); } }  One possibility familiar to Scheme programmers is to define an internal function for iteration through anonymous tail recursion: function fib(n) { return function(n,a,b) { return n>0 ? arguments.callee(n-1,b,a+b) : a; }(n,0,1); }  ### Iterative function fib(n) { var a = 0, b = 1, t; while (n-- > 0) { t = a; a = b; b += t; console.log(a); } return a; }  #### Memoization With the keys of a dictionary, var fib = (function(cache){ return cache = cache || {}, function(n){ if (cache[n]) return cache[n]; else return cache[n] = n == 0 ? 0 : n < 0 ? -fib(-n) : n <= 2 ? 1 : fib(n-2) + fib(n-1); }; })();  with the indices of an array, (function () { 'use strict'; function fib(n) { return Array.apply(null, Array(n + 1)) .map(function (_, i, lst) { return lst[i] = ( i ? i < 2 ? 1 : lst[i - 2] + lst[i - 1] : 0 ); })[n]; } return fib(32); })();  Output: 2178309 #### Y-Combinator function Y(dn) { return (function(fn) { return fn(fn); }(function(fn) { return dn(function() { return fn(fn).apply(null, arguments); }); })); } var fib = Y(function(fn) { return function(n) { if (n === 0 || n === 1) { return n; } return fn(n - 1) + fn(n - 2); }; });  #### Generators function* fibonacciGenerator() { var prev = 0; var curr = 1; while (true) { yield curr; curr = curr + prev; prev = curr - prev; } } var fib = fibonacciGenerator();  ### ES6 #### Memoized If we want access to the whole preceding series, as well as a memoized route to a particular member, we can use an accumulating fold. (() => { 'use strict'; // Nth member of fibonacci series // fib :: Int -> Int function fib(n) { return mapAccumL(([a, b]) => [ [b, a + b], b ], [0, 1], range(1, n))[0][0]; }; // GENERIC FUNCTIONS // mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y]) let mapAccumL = (f, acc, xs) => { return xs.reduce((a, x) => { let pair = f(a[0], x); return [pair[0], a[1].concat(pair[1])]; }, [acc, []]); } // range :: Int -> Int -> Maybe Int -> [Int] let range = (m, n) => Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i); // TEST return fib(32); // --> 2178309 })();  Otherwise, a simple fold will suffice. Translation of: Haskell (Memoized fold example) (() => { 'use strict'; // fib :: Int -> Int let fib = n => range(1, n) .reduce(([a, b]) => [b, a + b], [0, 1])[0]; // GENERIC [m..n] // range :: Int -> Int -> [Int] let range = (m, n) => Array.from({ length: Math.floor(n - m) + 1 }, (_, i) => m + i); // TEST return fib(32); // --> 2178309 })();  Output: 2178309 ## Joy ### Recursive DEFINE fib == [small] [] [pred dup pred] [+] binrec. ### Iterative DEFINE fib == [1 0] dip [swap [+] unary] times popd. ## jq Works with: jq Works with gojq, the Go implementation of jq The C implementation of jq does not (yet) have infinite-precision integer arithmetic, and so using it, the following algorithms only give exact answers up to fib(78). By contrast, using the Go implementation of jq and the definition of nth_fib given below: nth_fib(pow(2;20)) | tostring | [length, .[:10], .[-10:]] yields [219140,"1186800606","0691163707"]  in about 20 seconds on a 3GHz machine. Using either the C or Go implementations, at a certain point, integers are converted to floats, but floating point precision for fib(n) fails after n = 1476: in jq, fib(1476) evaluates to 1.3069892237633987e+308 ### Recursive def nth_fib_naive(n): if (n < 2) then n else nth_fib_naive(n - 1) + nth_fib_naive(n - 2) end; ### Tail Recursive Recent versions of jq (after July 1, 2014) include basic optimizations for tail recursion, and nth_fib is defined here to take advantage of TCO. For example, nth_fib(10000000) completes with only 380KB (that's K) of memory. However nth_fib can also be used with earlier versions of jq. def nth_fib(n): # input: [f(i-2), f(i-1), countdown] def fib: (.[0] + .[1]) as sum | .[2] as n | if (n <= 0) then sum else [ .[1], sum, n - 1 ] | fib end; [-1, 1, n] | fib; Example: (range(0;5), 50) | [., nth_fib(.)] yields: [0,0] [1,1] [2,1] [3,2] [4,3] [50,12586269025] ### Binet's Formula def fib_binet(n): (5|sqrt) as rt | ((1 + rt)/2) as phi | ((phi | log) * n | exp) as phin | (if 0 == (n % 2) then 1 else -1 end) as sign | ( (phin - (sign / phin) ) / rt ) + .5 | floor; ### Generator The following is a jq generator which produces the first n terms of the Fibonacci sequence efficiently, one by one. Notice that it is simply a variant of the above tail-recursive function. The function is in effect turned into a generator by changing "( _ | fib )" to "sum, (_ | fib)". # Generator def fibonacci(n): # input: [f(i-2), f(i-1), countdown] def fib: (.[0] + .[1]) as sum | if .[2] == 0 then sum else sum, ([ .[1], sum, .[2] - 1 ] | fib) end; [-1, 1, n] | fib; ## Julia ### Recursive fib(n) = n < 2 ? n : fib(n-1) + fib(n-2)  ### Iterative function fib(n) x,y = (0,1) for i = 1:n x,y = (y, x+y) end x end  ### Matrix form fib(n) = ([1 1 ; 1 0]^n)[1,2]  ## K ### Recursive {:[x<3;1;_f[x-1]+_f[x-2]]} ### Recursive with memoization Using a (global) dictionary c. {c::.();{v:c[a:x];:[x<3;1;:[_n~v;c[a]:_f[x-1]+_f[x-2];v]]}x} ### Analytic phi:(1+_sqrt(5))%2 {_((phi^x)-((1-phi)^x))%_sqrt[5]} ### Sequence to n {(x(|+$$\1 1)[;1]}
{x{x,+/-2#x}/!2}

## Kabap

### Sequence to n

// Calculate the $n'th Fibonacci number // Set this to how many in the sequence to generate$n = 10;

// These are what hold the current calculation
$a = 0;$b = 1;

// This holds the complete sequence that is generated
$sequence = ""; // Prepare a loop$i = 0;
:calcnextnumber;
$i =$i++;

// Do the calculation for this loop iteration
$b =$a + $b;$a = $b -$a;

// Add the result to the sequence
$sequence =$sequence << $a; // Make the loop run a fixed number of times if$i < $n; {$sequence = $sequence << ", "; goto calcnextnumber; } // Use the loop counter as the placeholder$i--;

// Return the sequence
return = "Fibonacci number " << $i << " is " <<$a << " (" << $sequence << ")"; ## Klingphix :Fibonacci dup 0 less ( ["Invalid argument"] [1 1 rot 2 sub [drop over over add] for] ) if ; 30 Fibonacci pstack print nl msec print nl "bertlham " input  ## Kotlin enum class Fibonacci { ITERATIVE { override fun get(n: Int): Long = if (n < 2) { n.toLong() } else { var n1 = 0L var n2 = 1L repeat(n) { val sum = n1 + n2 n1 = n2 n2 = sum } n1 } }, RECURSIVE { override fun get(n: Int): Long = if (n < 2) n.toLong() else this[n - 1] + this[n - 2] }, CACHING { val cache: MutableMap<Int, Long> = mutableMapOf(0 to 0L, 1 to 1L) override fun get(n: Int): Long = if (n < 2) n.toLong() else impl(n) private fun impl(n: Int): Long = cache.computeIfAbsent(n) { impl(it-1) + impl(it-2) } }, ; abstract operator fun get(n: Int): Long } fun main() { val r = 0..30 for (fib in Fibonacci.values()) { print("${fib.name.padEnd(10)}:")
for (i in r) { print(" " + fib[i]) }
println()
}
}

Output:
ITERATIVE:  0 1 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
RECURSIVE:  0 1 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
CACHING   : 0 1 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


## L++

(defn int fib (int n) (return (? (< n 2) n (+ (fib (- n 1)) (fib (- n 2))))))
(main (prn (fib 30)))


## LabVIEW

This image is a VI Snippet, an executable image of LabVIEW code. The LabVIEW version is shown on the top-right hand corner. You can download it, then drag-and-drop it onto the LabVIEW block diagram from a file browser, and it will appear as runnable, editable code.

## lambdatalk

1) basic version
{def fib1
{lambda {:n}
{if {< :n 3}
then 1
else {+ {fib1 {- :n 1}} {fib1 {- :n 2}}} }}}

{fib1 16} -> 987   (CPU ~ 16ms)
{fib1 30} = 832040 (CPU > 12000ms)

2) tail-recursive version
{def fib2
{def fib2.r
{lambda {:a :b :i}
{if {< :i 1}
then :a
else {fib2.r :b {+ :a :b} {- :i 1}} }}}
{lambda {:n} {fib2.r 0 1 :n}}}

{fib2 16} -> 987    (CPU ~ 1ms)
{fib2 30} -> 832040 (CPU ~2ms)
{fib2 1000} -> 4.346655768693743e+208 (CPU ~ 22ms)

3) Dijkstra Algorithm
{def fib3
{def fib3.r
{lambda {:a :b :p :q :count}
{if {= :count 0}
then :b
else {if {= {% :count 2} 0}
then {fib3.r :a :b
{+ {* :p :p} {* :q :q}}
{+ {* :q :q} {* 2 :p :q}}
{/ :count 2}}
else {fib3.r {+ {* :b :q} {* :a :q} {* :a :p}}
{+ {* :b :p} {* :a :q}}
:p :q
{- :count 1}} }}}}
{lambda {:n}
{fib3.r 1 0 0 1 :n} }}

{fib3 16} -> 987    (CPU ~ 2ms)
{fib3 30} -> 832040 (CPU ~ 2ms)
{fib3 1000} -> 4.346655768693743e+208 (CPU ~ 3ms)

4) memoization
{def fib4
{def fib4.m {array.new}}    // init an empty array
{def fib4.r {lambda {:n}
{if {< :n 2}
then {array.get {array.set! {fib4.m} :n 1} :n}      // init with 1,1
else {if {equal? {array.get {fib4.m} :n} undefined} // if not exists
then {array.get {array.set! {fib4.m} :n
{+ {fib4.r {- :n 1}}
{fib4.r {- :n 2}}}} :n}   // compute it
else {array.get {fib4.m} :n} }}}}                   // else get it
{lambda {:n}
{fib4.r :n}
{fib4.m} }} // display the number AND all its predecessors
-> fib4
{fib4 90}
-> 4660046610375530000
[1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121393,196418,
317811,514229,832040,1346269,2178309,3524578,5702887,9227465,14930352,24157817,39088169,63245986,102334155,
165580141,267914296,433494437,701408733,1134903170,1836311903,2971215073,4807526976,7778742049,12586269025,
20365011074,32951280099,53316291173,86267571272,139583862445,225851433717,365435296162,591286729879,956722026041,
1548008755920,2504730781961,4052739537881,6557470319842,10610209857723,17167680177565,27777890035288,44945570212853,
72723460248141,117669030460994,190392490709135,308061521170129,498454011879264,806515533049393,1304969544928657,
2111485077978050,3416454622906707,5527939700884757,8944394323791464,14472334024676220,23416728348467684,
37889062373143900,61305790721611580,99194853094755490,160500643816367070,259695496911122560,420196140727489660,
679891637638612200,1100087778366101900,1779979416004714000,2880067194370816000,4660046610375530000]

5) Binet's formula (non recursive)
{def fib5
{lambda {:n}
{let { {:n :n} {:sqrt5 {sqrt 5}} }
{round {/ {- {pow {/ {+ 1 :sqrt5} 2} :n}
{pow {/ {- 1 :sqrt5} 2} :n}} :sqrt5}}} }}

{fib5 16} -> 987    (CPU ~ 1ms)
{fib5 30} -> 832040 (CPU ~ 1ms)
{fib5 1000} -> 4.346655768693743e+208 (CPU ~ 1ms)


## Lang5

[] '__A set : dip swap __A swap 2 compress collapse '__A set execute
__A -1 extract nip ;  : nip swap drop ;  : tuck swap over ;
: -rot rot rot ; : 0= 0 == ; : 1+ 1 + ; : 1- 1 - ; : sum '+ reduce ;
: bi 'keep dip execute ;  : keep over 'execute dip ;

: fib dup 1 > if dup 1- fib swap 2 - fib + then ;
: fib  dup 1 > if "1- fib" "2 - fib" bi + then ;

## langur

val .fibonacci = f if(.x < 2: .x ; self(.x - 1) + self(.x - 2))

writeln map .fibonacci, series 2..20
Output:
[1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765]

## Lasso

define fibonacci(n::integer) => {

#n < 1 ? return false

local(
swap	= 0,
n1		= 0,
n2		= 1
)

loop(#n) => {
#swap = #n1 + #n2;
#n2 = #n1;
#n1 = #swap;
}
return #n1

}

fibonacci(0) //->output false
fibonacci(1) //->output 1
fibonacci(2) //->output 1
fibonacci(3) //->output 2


## Latitude

### Recursive

fibo := {
takes '[n].
if { n <= 1. } then {
n.
} else {
fibo (n - 1) + fibo (n - 2).
}.
}.

### Memoization

fibo := {
takes '[n].
cache := here cache.
{ cache slot? (n ordinal). } ifFalse {
cache slot (n ordinal) =
if { n <= 1. } then {
n.
} else {
fibo (n - 1) + fibo (n - 2).
}.
}.
cache slot (n ordinal).
} tap {
;; Attach the cache to the method object itself.
#'self cache := Object clone.
}.

## Lean

It runs on Lean 3.4.2:

-- Our first implementation is the usual recursive definition:
def fib1 : ℕ → ℕ
| 0       := 0
| 1       := 1
| (n + 2) := fib1 n + fib1 (n + 1)

-- We can give a second more efficient implementation using an auxiliary function:
def fib_aux : ℕ → ℕ → ℕ → ℕ
| 0 a b       := b
| (n + 1) a b := fib_aux n (a + b) a

def fib2 : ℕ → ℕ
| n := fib_aux n 1 0

-- Use #eval to check computations:
#eval fib1 20
#eval fib2 20


It runs on Lean 4:

-- Naive version
def fib1 (n : Nat) : Nat :=
match n with
| 0 => 0
| 1 => 1
| (k + 2) => fib1 k + fib1 (k + 1)

-- More efficient version
def fib_aux (n : Nat) (a : Nat) (b : Nat) : Nat :=
match n with
| 0 => b
| (k + 1) => fib_aux k (a + b) a

def fib2 (n : Nat) : Nat :=
fib_aux n 1 0

-- Examples
#eval fib1 20
#eval fib2 20


## LFE

### Recursive

(defun fib
((0) 0)
((1) 1)
((n)
(+ (fib (- n 1))
(fib (- n 2)))))


### Iterative

(defun fib
((n) (when (>= n 0))
(fib n 0 1)))

(defun fib
((0 result _)
result)
((n result next)
(fib (- n 1) next (+ result next))))


## Liberty BASIC

### Iterative/Recursive

for i = 0 to 15
print fiboR(i),fiboI(i)
next i

function fiboR(n)
if n <= 1 then
fiboR = n
else
fiboR = fiboR(n-1) + fiboR(n-2)
end if
end function

function fiboI(n)
a = 0
b = 1
for i = 1 to n
temp = a + b
a = b
b = temp
next i
fiboI = a
end function
Output:
0             0
1             1
1             1
2             2
3             3
5             5
8             8
13            13
21            21
34            34
55            55
89            89
144           144
233           233
377           377
610           610


### Iterative/Negative

print "Rosetta Code - Fibonacci sequence": print
print "  n             Fn"
for x=-12 to 12 '68 max
print using("### ", x); using("##############", FibonacciTerm(x))
next x
print
[start]
input "Enter a term#: "; n$n$=lower$(trim$(n$)) if n$="" then print "Program complete.": end
print FibonacciTerm(val(n$)) goto [start] function FibonacciTerm(n) n=int(n) FTa=0: FTb=1: FTc=-1 select case case n=0 : FibonacciTerm=0 : exit function case n=1 : FibonacciTerm=1 : exit function case n=-1 : FibonacciTerm=-1 : exit function case n>1 for x=2 to n FibonacciTerm=FTa+FTb FTa=FTb: FTb=FibonacciTerm next x exit function case n<-1 for x=-2 to n step -1 FibonacciTerm=FTa+FTc FTa=FTc: FTc=FibonacciTerm next x exit function end select end function Output: Rosetta Code - Fibonacci sequence n Fn -12 -144 -11 -89 -10 -55 -9 -34 -8 -21 -7 -13 -6 -8 -5 -5 -4 -3 -3 -2 -2 -1 -1 -1 0 0 1 1 2 1 3 2 4 3 5 5 6 8 7 13 8 21 9 34 10 55 11 89 12 144 Enter a term#: 12 144 Enter a term#: Program complete.  ## Lingo ### Recursive on fib (n) if n<2 then return n return fib(n-1)+fib(n-2) end ### Iterative on fib (n) if n<2 then return n fibPrev = 0 fib = 1 repeat with i = 2 to n tmp = fib fib = fib + fibPrev fibPrev = tmp end repeat return fib end ### Analytic on fib (n) sqrt5 = sqrt(5.0) p = (1+sqrt5)/2 q = 1 - p return integer((power(p,n)-power(q,n))/sqrt5) end ## Lisaac - fib(n : UINTEGER_32) : UINTEGER_64 <- ( + result : UINTEGER_64; (n < 2).if { result := n; } else { result := fib(n - 1) + fib(n - 2); }; result ); ## LiveCode -- Iterative, translation of the basic version. function fibi n put 0 into aa put 1 into b repeat with i = 1 to n put aa + b into temp put b into aa put temp into b end repeat return aa end fibi -- Recursive function fibr n if n <= 1 then return n else return fibr(n-1) + fibr(n-2) end if end fibr ## LLVM ; This is not strictly LLVM, as it uses the C library function "printf". ; LLVM does not provide a way to print values, so the alternative would be ; to just load the string into memory, and that would be boring. ; Additional comments have been inserted, as well as changes made from the output produced by clang such as putting more meaningful labels for the jumps$"PRINT_LONG" = comdat any
@"PRINT_LONG" = linkonce_odr unnamed_addr constant [5 x i8] c"%ld\0A\00", comdat, align 1

;--- The declaration for the external C printf function.
declare i32 @printf(i8*, ...)

;--------------------------------------------------------------------
;-- Function for calculating the nth fibonacci numbers
;--------------------------------------------------------------------
define i32 @fibonacci(i32) {
%2 = alloca i32, align 4            ;-- allocate local copy of n
%3 = alloca i32, align 4            ;-- allocate a
%4 = alloca i32, align 4            ;-- allocate b
store i32 %0, i32* %2, align 4      ;-- store copy of n
store i32 0, i32* %3, align 4       ;-- a := 0
store i32 1, i32* %4, align 4       ;-- b := 1
br label %loop

loop:
%6 = icmp sgt i32 %5, 0             ;-- n > 0
br i1 %6, label %loop_body, label %exit

loop_body:
%9 = add nsw i32 %7, %8             ;-- t = a + b
store i32 %8, i32* %3, align 4      ;-- store a = b
store i32 %9, i32* %4, align 4      ;-- store b = t
%11 = add nsw i32 %10, -1           ;-- decrement n
store i32 %11, i32* %2, align 4     ;-- store n
br label %loop

exit:
ret i32 %12                         ;-- return a
}

;--------------------------------------------------------------------
;-- Main function for printing successive fibonacci numbers
;--------------------------------------------------------------------
define i32 @main() {
%1 = alloca i32, align 4            ;-- allocate index
store i32 0, i32* %1, align 4       ;-- index := 0
br label %loop

loop:
%3 = icmp sle i32 %2, 12            ;-- index <= 12
br i1 %3, label %loop_body, label %exit

loop_body:
%5 = call i32 @fibonacci(i32 %4)
%6 = call i32 (i8*, ...) @printf(i8* getelementptr inbounds ([5 x i8], [5 x i8]* @"PRINT_LONG", i32 0, i32 0), i32 %5)
%8 = add nsw i32 %7, 1              ;-- increment index
store i32 %8, i32* %1, align 4      ;-- store index
br label %loop

exit:
ret i32 0                           ;-- return EXIT_SUCCESS
}

Output:
0
1
1
2
3
5
8
13
21
34
55
89
144

## Logo

to fib :n [:a 0] [:b 1]
if :n < 1 [output :a]
output (fib :n-1 :b :a+:b)
end

## LOLCODE

HAI 1.2
HOW DUZ I fibonacci YR N
EITHER OF BOTH SAEM N AN 1 AN BOTH SAEM N AN 0
O RLY?
YA RLY, FOUND YR 1
NO WAI
I HAS A N1
I HAS A N2
N1 R DIFF OF N AN 1
N2 R DIFF OF N AN 2
N1 R fibonacci N1
N2 R fibonacci N2
FOUND YR SUM OF N1 AN N2
OIC
IF U SAY SO
KTHXBYE

## LSL

Rez a box on the ground, and add the following as a New Script.

integer Fibonacci(integer n) {
if(n<2) {
return n;
} else {
return Fibonacci(n-1)+Fibonacci(n-2);
}
}
default {
state_entry() {
integer x = 0;
for(x=0 ; x<35 ; x++) {
llOwnerSay("Fibonacci("+(string)x+")="+(string)Fibonacci(x));
}
}
}


Output:

Fibonacci(0)=0
Fibonacci(1)=1
Fibonacci(2)=1
Fibonacci(3)=2
Fibonacci(4)=3
Fibonacci(5)=5
Fibonacci(6)=8
Fibonacci(7)=13
Fibonacci(8)=21
Fibonacci(9)=34
Fibonacci(10)=55
Fibonacci(11)=89
Fibonacci(12)=144
Fibonacci(13)=233
Fibonacci(14)=377
Fibonacci(15)=610
Fibonacci(16)=987
Fibonacci(17)=1597
Fibonacci(18)=2584
Fibonacci(19)=4181
Fibonacci(20)=6765
Fibonacci(21)=10946
Fibonacci(22)=17711
Fibonacci(23)=28657
Fibonacci(24)=46368
Fibonacci(25)=75025
Fibonacci(26)=121393
Fibonacci(27)=196418
Fibonacci(28)=317811
Fibonacci(29)=514229
Fibonacci(30)=832040
Fibonacci(31)=1346269
Fibonacci(32)=2178309
Fibonacci(33)=3524578
Fibonacci(34)=5702887


## Lua

### Recursive

--calculates the nth fibonacci number. Breaks for negative or non-integer n.
function fibs(n)
return n < 2 and n or fibs(n - 1) + fibs(n - 2)
end


### Pedantic Recursive

--more pedantic version, returns 0 for non-integer n
function pfibs(n)
if n ~= math.floor(n) then return 0
elseif n < 0 then return pfibs(n + 2) - pfibs(n + 1)
elseif n < 2 then return n
else return pfibs(n - 1) + pfibs(n - 2)
end
end


### Tail Recursive

function a(n,u,s) if n<2 then return u+s end return a(n-1,u+s,u) end
function trfib(i) return a(i-1,1,0) end


### Table Recursive

fib_n = setmetatable({1, 1}, {__index = function(z,n) return n<=0 and 0 or z[n-1] + z[n-2] end})


### Table Recursive 2

-- table recursive done properly (values are actually saved into table;
-- also the first element of Fibonacci sequence is 0, so the initial table should be {0, 1}).
fib_n = setmetatable({0, 1}, {
__index = function(t,n)
if n <= 0 then return 0 end
t[n] = t[n-1] + t[n-2]
return t[n]
end
})


### Iterative

function ifibs(n)
local p0,p1=0,1
for _=1,n do p0,p1 = p1,p0+p1 end
return p0
end


## Luck

function fib(x: int): int = (
let cache = {} in
let fibc x = if x<=1 then x else (
if x not in cache then
cache[x] = fibc(x-1) + fibc(x-2);
cache[x]
) in fibc(x)
);;
for x in range(10) do print(fib(x))

## Lush

(de fib-rec (n)
(if (< n 2)
n
(+ (fib-rec (- n 2)) (fib-rec (- n 1)))))

## M2000 Interpreter

Return decimal type and use an Inventory (as closure) to store known return values. All closures are in scope in every recursive call (we use here lambda(), but we can use fib(), If we make Fib1=fib then we have to use lambda() for recursion.

Inventory K=0:=0,1:=1
fib=Lambda K (x as decimal)-> {
If Exist(K, x) Then =Eval(K) :Exit
Def Ret as Decimal
Ret=If(x>1->Lambda(x-1)+Lambda(x-2), x)
Append K, x:=Ret
=Ret
}
\\ maximum 139
For i=1 to 139 {
Print Fib(i)
}

Here an example where we use a BigNum class to make a Group which hold a stack of values, and take 14 digits per item in stack. We can use inventory to hold groups, so we use the fast fib() function from code above, where we remove the type definition of Ret variable, and set two first items in inventory as groups.

Class BigNum {
a=stack
Function Digits {
=len(.a)*14-(14-len(str$(stackitem(.a,len(.a)) ,""))) } Operator "+" (n) { \\ we get a copy, but .a is pointer \\ we make a copy, and get a new pointer .a<=stack(.a) acc=0 carry=0 const d=100000000000000@ k=min.data(Len(.a), len(n.a)) i=each(.a, 1,k ) j=each(n.a, 1,k) while i, j { acc=stackitem(i)+stackitem(j)+carry carry= acc div d return .a, i^+1:=acc mod d } if len(.a)<len(n.a) Then { i=each(n.a, k+1, -1) while i { acc=stackitem(i)+carry carry= acc div d stack .a {data acc mod d} } } ELse.if len(.a)>len(n.a) Then { i=each(.a, k+1, -1) while i { acc=stackitem(i)+carry carry= acc div d Return .a, i^+1:=acc mod d if carry else exit } } if carry then stack .a { data carry} } Function tostring$ {
if len(.a)=0 then ="0" : Exit
if len(.a)=1 then =str$(Stackitem(.a),"") : Exit document buf$=str$(Stackitem(.a, len(.a)),"") for i=len(.a)-1 to 1 { Stack .a { buf$=str$(StackItem(i), "00000000000000") } } =buf$
}
class:
Module BigNum (s$) { s$=filter$(s$,"+-.,")
if s$<>"" Then { repeat { If len(s$)<14 then Stack .a { Data  val(s$) }: Exit Stack .a { Data val(Right$(s$, 14)) } S$=Left$(S$, len(S$)-14) } Until S$=""
}
}
}

Inventory K=0:=BigNum("0"),1:=BigNum("1")
fib=Lambda K (x as decimal)-> {
If Exist(K, x) Then =Eval(K) :Exit
Ret=If(x>1->Lambda(x-1)+Lambda(x-2), bignum(str$(x,""))) Append K, x:=Ret =Ret } \\ Using this to handle form refresh by code Set Fast! For i=1 to 4000 { N=Fib(i) Print i Print N.tostring$()
Refresh
}

## M4

define(fibo',ifelse(0,$1,0,ifelse(1,$1,1,
eval(fibo(decr($1)) + fibo(decr(decr($1))))')')')dnl
define(loop',ifelse($1,$2,,$3($1) loop(incr($1),$2,$3')')')dnl loop(0,15,fibo') ## MAD  NORMAL MODE IS INTEGER INTERNAL FUNCTION(N) ENTRY TO FIB. A = 0 B = 1 THROUGH LOOP, FOR N=N, -1, N.E.0 C = A + B A = B LOOP B = C FUNCTION RETURN A END OF FUNCTION THROUGH SHOW, FOR I=0, 1, I.GE.20 SHOW PRINT FORMAT FNUM, I, FIB.(I) VECTOR VALUES FNUM =$4HFIB(,I2,4H) = ,I4*\$
END OF PROGRAM
Output:
FIB( 0) =    0
FIB( 1) =    1
FIB( 2) =    1
FIB( 3) =    2
FIB( 4) =    3
FIB( 5) =    5
FIB( 6) =    8
FIB( 7) =   13
FIB( 8) =   21
FIB( 9) =   34
FIB(10) =   55
FIB(11) =   89
FIB(12) =  144
FIB(13) =  233
FIB(14) =  377
FIB(15) =  610
FIB(16) =  987
FIB(17) = 1597
FIB(18) = 2584
FIB(19) = 4181

## Maple

> f := n -> ifelse(n<3,1,f(n-1)+f(n-2));
> f(2);
1
> f(3);
2

## Mathematica / Wolfram Language

The Wolfram Language already has a built-in function Fibonacci, but a simple recursive implementation would be

fib[0] = 0
fib[1] = 1
fib[n_Integer] := fib[n - 1] + fib[n - 2]


An optimization is to cache the values already calculated:

fib[0] = 0
fib[1] = 1
fib[n_Integer] := fib[n] = fib[n - 1] + fib[n - 2]


The above implementations may be too simplistic, as the first is incredibly slow for any reasonable range due to nested recursions and while the second is faster it uses an increasing amount of memory. The following uses recursion much more effectively while not using memory:

fibi[prvprv_Integer, prv_Integer, rm_Integer] :=
If[rm < 1, prvprv, fibi[prv, prvprv + prv, rm - 1]]
fib[n_Integer] := fibi[0, 1, n]


However, the recursive approaches in Mathematica are limited by the limit set for recursion depth (default 1024 or 4096 for the above cases), limiting the range for 'n' to about 1000 or 2000. The following using an iterative approach has an extremely high limit (greater than a million):

fib[n_Integer] := Block[{tmp, prvprv = 0, prv = 1},
For[i = 0, i < n, i++, tmp = prv; prv += prvprv; prvprv = tmp];
Return[prvprv]]


If one wanted a list of Fibonacci numbers, the following is quite efficient:

fibi[{prvprv_Integer, prv_Integer}] := {prv, prvprv + prv}
fibList[n_Integer] := Map[Take[#, 1] &, NestList[fibi, {0, 1}, n]] // Flatten


Output from the last with "fibList[100]":

{0, 1, 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