Golf a Custom Fibonacci Sequence

Question

The Fibonacci sequence is a fairly well known thing around here. Heck, it even has its own tag. However, for all that, we sure like to stick to our roots of 1, 1, ... (or is it 0, 1, ...? We may never know...). In this challenge, the rules are the same, but instead of getting the nth item in the Fibonacci sequence, you will get the nth item in the Fibonacci-esque sequence starting with x, y, ....

Input

Three integers, in whatever order you want. n is the index (0 or 1 indexed) of term in the sequence for your output. x and y are the first two items in your current program run's Fibonacci sequence.

Output

The nth term in the Fibonacci sequence starting with x, y.

Test Cases

(0-indexed)

n   x     y     out
5   0     0     0
6   0     1     8
6   1     1     13
2   5     5     10
10  2     2     178
3   3     10    23
13  2308  4261  1325165
0   0     1     0
1   0     1     1

(1-indexed)

n   x     y     out
6   0     0     0
7   0     1     8
7   1     1     13
3   5     5     10
11  2     2     178
4   3     10    23
14  2308  4261  1325165
1   0     1     0
2   0     1     1

Caveats

Assume 0 <= x <= y.

Please note your input order (must be constant).

Answer 1

Jelly, 3 bytes

+¡ạ

Takes x, y, and n (0-indexed) as separate command-line arguments, in that order.

Try it online!

How it works

+¡ạ  Main link. Left argument: x. Right argument: y. Third argument: n

  ạ  Yield abs(x - y) = y - x, the (-1)-th value of the Lucas sequence.
+¡   Add the quicklink's left and right argument (initially x and y-x), replacing
     the right argument with the left one and the left argument with the result.
     Do this n times and return the final value of the left argument.

Answer 2

CJam, 14 9 bytes

l~{_@+}*;

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Input format is "x y n". I'm still a noob at this, so I'm 100% sure there are better ways to do this, but please instead of telling me "do this" try to only give me hints so that I can find the answer myself and get better. Thanks!

Answer 3

JavaScript (ES6), 27 26 bytes

Nothing fancy here, just a standard JS Fibonacci function with the initial values of 0 & 1 removed.

n=>g=(x,y)=>n--?g(y,x+y):x

---

Try it

f=
n=>g=(x,y)=>n--?g(y,x+y):x
o.value=f(i.value=13)(j.value=2308,k.value=4261)
oninput=_=>o.value=f(+i.value)(+j.value,+k.value)

*{font-family:sans-serif;}
input{margin:0 5px 0 0;width:50px;}
#o{width:75px;}

<label for=i>n: </label><input id=i type=number><label for=j>x: </label><input id=j type=number><label for=k>y: </label><input id=k type=number><label for=o>= </label><input id=o>

Answer 4

Python 2, 37 bytes

f=lambda x,y,n:n and f(y,x+y,n-1)or x

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0-indexed, you may need to adjust the recursion limit for n≥999

Answer 5

Python 2, 40 bytes

0-indexed
Try it online

n,a,b=input()
exec'a,b=b,a+b;'*n
print a

Answer 6

Haskell, 30 bytes

x#y=(f!!)where f=x:scanl(+)y f

Try it online! 0-indexed. Use as (x#y)n, e.g. (0#1)5 for the fifth element of the original sequence.

The most likely shortest way to get the Fibonacci sequence in Haskell is f=0:scanl(+)1f, which defines an infinite list f=[0,1,1,2,3,5,8,...] containing the sequence. Replacing 0 and 1 with arguments x and y yields the custom sequence. (f!!) is then a function returning the nth element of f.

Answer 7

Mathematica, 36 bytes

LinearRecurrence[{1,1},{##2},{#+1}]&

input

[n,x,y]

Answer 8

PowerShell, 40 bytes

$n,$x,$y=$args;2..$n|%{$y=$x+($x=$y)};$y

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Answer 9

Brain-Flak, 38 bytes

{({}[()]<(({}<>)<>{}<(<>{}<>)>)>)}{}{}

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{({}[()]<                      >)}     # For n .. 0
         (({}<>)<>            )        # Copy TOS to the other stack and add it to...
                  {}                   # The second value
                    <(<>{}<>)>         # Copy what was TOS back
                                  {}{} # Pop the counter and the n+1th result

Answer 10

Ruby, 27 bytes

->a,b,n{n.times{b=a+a=b};a}

Answer 11

TAESGL, 4 bytes

ēB)Ė

1-indexed

Interpreter

Explanation

Input taken as n,[x,y]

 ēB)Ė
AēB)     get implicit input "A" Fibonacci numbers where "B" is [x,y]
    Ė    pop the last item in the array

Answer 12

Jelly, 6 bytes

;SḊµ¡I

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Explanation

   µ¡  - repeat n times (computes the n+1th and n+2th element):
 S     -  take the sum of the elements of the previous iteration (starting at (x,y))
;      -  append to the end of the previous iteration
  Ḋ    -  remove the first element
     I - Take the difference of the n+1th and n+2th to get the n-th.

Answer 13

Prolog (SWI), 77 bytes

f(N,Y,Z):-M is N-1,f(M,X,Y),Z is X+Y.
l(N,A,B,X):-asserta(f(0,A,B)),f(N,X,_).

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Started off golfing Leaky Nun's answer and arrived at something completely different.

This one has a rule for (Nᵗʰ, (N+1)ᵗʰ) in terms of ((N-1)ᵗʰ, Nᵗʰ) and uses database management to assert 0ᵗʰ and 1ˢᵗ elements at runtime.

f(N,X,Y) means Nᵗʰ element is X and (N+1)ᵗʰ element is Y.

Answer 14

Octave, 24 bytes

@(n,x)(x*[0,1;1,1]^n)(1)

Input format: n,[x,y].

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Answer 15

Braingolf, 15 bytes

VR<2-M[R!+v]R_;

_; is no longer needed on the latest version of Braingolf, however that's as of ~5 minutes ago, so would be non-competing.

Answer 16

Python 2, 112 bytes

1-indexed.

import itertools
def f(x,y):
 while 1:yield x;x,y=y,x+y
def g(x,y,n):return next(itertools.islice(f(x,y),n-1,n))

Try it online!

Answer 17

Prolog (SWI), 85 bytes

l(0,X,Y,X).
l(1,X,Y,Y).
l(N,X,Y,C):-M is N-1,P is N-2,l(M,X,Y,A),l(P,X,Y,B),C is A+B.

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0-indexed.

Answer 18

PHP>=7.1, 55 Bytes

for([,$n,$x,$y]=$argv;$n--;$x=$y,$y=$t)$t=$x+$y;echo$x;

Online Version

PHP>=7.1, 73 Bytes

for([,$n,$x,$y]=$argv,$r=[$x,$y];$i<$n;)$r[]=$r[+$i]+$r[++$i];echo$r[$n];

Online Version

Answer 19

MATL, 7 bytes

:"wy+]x

Output is 0-based.

Try it at MATL Online!

Explanation

Let the inputs be denoted n (index), a, b (initial terms).

:"     % Implicitly input n. Do this n times
       %   At this point in each iteration, the stack contains the two most
       %   recently computed terms of the sequence, say s, t. In the first
       %   iteration the stack is empty, but a, b will be implicitly input
       %   by the next statement
  w    %   Swap. The stack contains t, s
  y    %   Duplicate from below. The stack contains t, s, t
  +    %   Add. The stack contains t, s+t. These are now the new two most
       %   recently comnputed terms
]      % End
x      % Delete (we have computed one term too many). Implicitly display

Answer 20

R, 39 bytes

f=function(x,y,n)'if'(n,f(y,x+y,n-1),x)

A simple recursive function. Funnily enough this is shorter than anything I can come up with for the regular Fibonacci sequence (without built-ins), because this doesn't have to assign 1 to both x and y =P

Calculates n+1 numbers of the sequence, including the initial values. Each recursion is calculates with n-1 and stopped when n==0. The lowest of the two numbers is then returned, giving back the n-th value.

Answer 21

dc, 36 bytes

?sdsbsa[lddlb+sdsbla1-dsa1<c]dscxldp

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0-indexed. Input must be in the format n x y.

Answer 22

Common Lisp, 49 Bytes, 0-indexed

(defun fib(n x y)(if(= 0 n)x(fib(1- n)y(+ x y))))

I'm a Lisp noob so any tips would be appreciated ;)

Explanation:

(defun fib(n x y)                                  | Define a function taking 3 arguments
                 (if(= 0 n)x                       | If n = 0, return x
                            (fib(1- n)y(+ x y))))  | Otherwise, call fib with n-1, y, and x+y

Answer 23

br**nfuck, 39 29 bytes

Thanks to @JoKing for -10!

,<,<,[>[>+>+<<-]<[>+<-]>-]>>.

TIO won't work particularly well for this (or for any BF solution to a problem involving numbers). I strongly suggest @Timwi's EsotericIDE (or implementing BF yourself).

Takes x, then y, then n. 0-indexed. Assumes an unbounded or wrapping tape.

Explanation

,<,<,            Take inputs. Tape: [n, y, x]
[                While n:
  > [->+>+<<]      Add y to x, copying it to the next cell along as well. Tape: [n, 0, x+y, y]
  < [>+<-]         Move n over. Tape: [0, n, x+y, y]
  >-               Decrement n.
] >>.            End loop. Print cell 2 to the right (x for n == 0).

Answer 24

C (gcc), 29 bytes

f(n,x,y){n=n?f(n-1,y,x+y):x;}

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This implementation is 0-based.

Answer 25

05AB1E, 9 bytes

`©GDŠ+}®@

Try it online!

Explanation

`           # split inputs as separate to stack
 ©          # store n in register
  G         # n-1 times do
   D        # duplicate top of stack
    Š       # move down 2 places on stack
     +      # add top 2 values of stack
      }     # end loop
       ®@   # get the value nth value from the bottom of stack

Answer 26

Lua, 44 bytes

0-Indexed

n,x,y=...for i=1,n do
x,y=y,x+y
end
print(x)

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Answer 27

Axiom, 88 57 bytes

f(k,x,y)==(repeat(k<=0=>break;c:=y;y:=x+y;x:=c;k:=k-1);x)

this would pass the test proposed (0 indexed)

(14) -> f(5,0,0)
   (14)  0
                                                 Type: NonNegativeInteger
(15) -> f(6,0,1)
   (15)  8
                                                    Type: PositiveInteger
(16) -> f(2,5,5)
   (16)  10
                                                    Type: PositiveInteger
(17) -> f(10,2,2)
   (17)  178
                                                    Type: PositiveInteger
(18) -> f(3,3,10)
   (18)  23
                                                    Type: PositiveInteger
(19) -> f(13,2308,4261)
   (19)  1325165
                                                    Type: PositiveInteger

Answer 28

Klein, 18 + 3 bytes

This uses the 000 topology

:?\(:(+)$)1-+
((/@

Pass input in the form x y n.

Answer 29

Retina, 37 bytes

\d+
$*1
+`(1*) (1*) 1
$2 $1$2 
 .*

1

Try it online!

0-based, takes x y n separated by space. Calculates in unary.

Answer 30

TI-Basic, 32 bytes

Prompt N,X,Y
While N
X+Y➡Z
Y➡X
Z➡Y
DS<(N,0
End
X

Answer 31

k, 15 bytes

{*x(|+\)/y,z-y}

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Answer 32

Pari/GP, 24 bytes

n->x->(x*[0,1;1,1]^n)[1]

Input format: (n)([x,y]).

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Answer 33

AWK, 39 bytes

{for(n=3;n<$1+2;)$++n=$n+$(n-1);$0=$n}1

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Takes inputs as n x y 0-indexed

Answer 34

Japt, 11 bytes

?ß´UWV+W :V

Try it online

---

Explanation

(Inputs U, V & W line up with n, x & y in the challenge.)
     :Implicit input of integer U.
?    :Ternary to check if U is currently >0.
ß    :If not, call the function again, with the arguments...
´U   :U-1,
W    :W,
V+W  :and V+W.
:    :Ternary "else", i.e. if U=0
V    :Return V

Answer 35

Java, 58 bytes

int f(int n,int x,int y){return n<3?n<2?x:y:f(n-1,y,x+y);}

This function is 1-indexed.

Alternatively, one could write the following lambda with recursion, for 55 bytes:

interface B{F f=(n,x,y)->n<3?n<2?x:y:B.f.f(n-1,y,x+y);}

This requires the following description, which isn't usually counted in Java Lambdas:

interface F{int f(int n,int x,int y);}

Test

public class Main {

  static
  int f(int n,int x,int y){return n<3?n<2?x:y:f(n-1,y,x+y);}

  public static void main(String[] args) {
    int[][] tests = {
      {6, 0, 0, 0},
      {7, 0, 1, 8},
      {7, 1, 1, 13},
      {3, 5, 5, 10},
      {11, 2, 2, 178},
      {4, 3, 10, 23},
      {14, 2308, 4261, 1325165},
      {1, 0, 1, 0},
      {2, 0, 1, 1}
    };

    for (int[] test: tests) {
      System.out.printf("%2d %4d %4d %7d%n", test[0], test[1], test[2], f(test[0],test[1],test[2]));
    }
  }
}

Answer 36

Mathematica, 25 bytes

Although this is quite old, and there has already been a Mathematica answer using LinearRecurrence built-in, mine solution is shorter.

{##2}.Fibonacci/@{#-1,#}&

Answer 37

Forth (gforth), 30 bytes

: f 1- 0 ?do tuck + loop nip ;

Try it online!

Input Order is x y n

Answer 38

Perl 6, 24 bytes

{($^x,$^y,*+*...*)[$^n]}

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Answer 39

SmileBASIC, 44 39 bytes

^{crossed out 44 is still regular 44 ;(}

DEF F N,X,Y
IF!N THEN?X
F N-1,Y,X+Y
END

Recursive function; prints the result once N reaches 0, ends with a stack overflow.

Answer 40

Pyth, 23 bytes

J.)QWgJlQ=aQ+@Q_2eQ)@QJ

Test suite

0-indexed, input format: [x,y,n].

Answer 41

GolfScript, 8 bytes

~{.@+}*;

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Takes input in order x y n. Actually generates first n+1 values and pops last.

Explanation:

~{.@+}*; Full program, implicit input
~        Evaluate
 {   }*  Pop n and repeat:    Stack: x y
  .        Duplicate y               x y y
   @       Get x                     y y x
    +      Add                       y x+y
       ; Pop last

Answer 42

cQuents, 8 bytes

=A,B:z+y

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Takes input as x y n, where n is 1-indexed.

Explanation

=A,B      Set sequence start to first two inputs
    :     Mode: Sequence
     z+y  Each item in the sequence is the previous two added together

          If there is a third input, that input is `n`. Get the `n`th (1-indexed) item in the sequence
          If there is not a third input, output the whole sequence.