# Binary Fibonacci

## Challenge

You need to generate a program or function that takes in a positive integer N, calculates the first N terms of the Fibonacci sequence in binary, concatenates it into a single binary number, converts that number back to decimal, and then outputs the decimal as an integer.

For example

1 ->  -> 0 to decimal outputs 0
3 -> [0, 1, 1] -> 011 to decimal outputs 3
4 -> [0, 1, 1, 10] -> 01110 to decimal outputs 14


You do not need to output the ->, just the number (e.g. if the user types 4, just output 14). The arrows are just to help explain what the program must do.

## Test cases

1 -> 0
2 -> 1
3 -> 3
4 -> 14
5 -> 59
6 -> 477
7 -> 7640
8 -> 122253
9 -> 3912117
10 -> 250375522
11 -> 16024033463
12 -> 2051076283353
13 -> 525075528538512
14 -> 134419335305859305
15 -> 68822699676599964537
16 -> 70474444468838363686498
17 -> 72165831136090484414974939
18 -> 147795622166713312081868676669
19 -> 605370868394857726287334099638808
20 -> 4959198153890674493745840944241119317


The program must be able to output up to the limit of the language in use. No lookup tables or common workarounds allowed.

This is , so the answer with the shortest number of bytes wins!

• Added test cases from 0 to 20 from tio.run/##DYxBCoQwDAC/…. Credit to @alephalpha for the program. – Nathan Wood Mar 30 '18 at 16:46
• As it hasn't been said yet: Welcome to PPCG! Nice first challenge. – Laikoni Mar 30 '18 at 17:54
• @Laikoni Thanks! – Nathan Wood Mar 30 '18 at 17:56
• Where exactly does the language-specific limit apply? Would a C function that returns a 32-bit integer be allowed? Like int32_t binary_concat_Fib(int n), which would limit the resulting output value to 2^31-1. i.e. you get to assume all the concatenated bits fit in an integer. Or should the function work up to the point where the largest Fibonacci number doesn't fit in an integer on its own, so concatenating the bits takes extended precision? – Peter Cordes Mar 30 '18 at 20:43
• And does the "converts to decimal" have to be explicit, calling an integer->string function or writing one yourself? Concatenating the bits into a single integer gives you a representation of the final value. If I understand correctly, Dennis's Python answer is returning an integer, leaving it up to the caller to turn that value into a decimal string or do whatever with it. Integer values in computer languages that support bit-shift operators are naturally binary, not decimal, unless they're stored in strings. In languages without shifts / bitwise operators, nothing implies any base. – Peter Cordes Mar 30 '18 at 20:47

# Python, 64 bytes

f=lambda n,a=0,b=1,r=0:n and f(n-1,b,a+b,r<<len(bin(a))-2|a)or r


Try it online!

• Nice, I was just trying to achieve this style. – Jonathan Allan Mar 30 '18 at 17:46

# Jelly,  7  6 bytes

ḶÆḞBẎḄ


Try it online!

### How?

ḶÆḞBẎḄ - Link: integer, n
Ḷ      - lowered range -> [0,1,2,3,4,5,...,n]
ÆḞ    - Fibonacci (vectorises) -> [0,1,1,2,3,5...,F(n)]
B   - to binary (vectorises) -> [,,,[1,0],[1,1],[1,0,1],...,B(F(n))]
Ẏ  - tighten -> [0,1,1,1,0,1,1,1,0,1,...,B(F(n)),B(F(n)),...]
Ḅ - from binary -> answer

• Messing around with the new quicks, I found that the first n Fibonacci numbers can also be found using Ṛc’SƲƤ which could be useful for similar sequences. – miles Mar 31 '18 at 5:06

# Python 3.6, 61 bytes

f=lambda n,a=0,b=1:n and int(f'{a:b}{f(n-1,b,a+b)*2:b}',2)//2


Try it online!

# brainfuck, 397 bytes

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


Well, that was fun!

Takes ASCII input (e.g. 11), outputs result in ASCII.

Note: to try this online, make sure you set the cell size to 32 bits (on the right side of the webpage). If you do not enter an input, your browser might crash.

The interpreter cannot handle input of 11 and higher because it only supports up to 32 bits.

Try it on copy.sh

## Explanation

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


Get decimal input and add one (to mitigate off-by-one)

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


Generate fibonacci numbers on the tape.

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


Set up for the incoming binary concatenation loop

So the cells contain the value, starting from the first position,

1 | 0 | 1 | 1 | 2 | 3 | 5 | ... | f_n | 0 | 1 | 0 | 1 | 0 | f_n | 1 | 0 | 0 | 0...


Look at these cells:

f_n | 0 | 1 | 0 | 1 | 0 | f_n | 1


I'll label this:

num | sum | cat | 0 | pow | 0 | num | pow


pow is there to find the maximal power of 2 that is strictly greater than num. sum is the concatenation of numbers so far. cat is the power of 2 that I would need to multiply the num in order to concatenate num in front of the sum (so I would be able to simply add).

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


Loop: Check whether f_n is strictly less than pow.

Truthy:

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


Zero out junk. Then, add num * cat to sum. Next, load the next Fibonacci number (= f_(n-1); if it doesn't exist, exit loop) and set cat to cat * pow. Prepare for next loop (zero out more junk, shift scope by one).

Falsey:

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


Set pow to 2 * pow, restore num.

]


Repeat until there is no Fibonacci number left.

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


Clean garbage. Take each digit of the resulting number and output each (in ascii).

# Husk, 7 bytes

ḋṁḋ↑Θİf


Try it online!

### Explanation

ḋṁḋ↑Θİf                              4
İf    The Fibonacci numbers     [1,1,2,3,5,8..]
Θ      Prepends 0                [0,1,1,2,3,5..]
↑     Take n elements from list   [0,1,1,2]
ḋ        Convert to binary digits  [,,,[1,0]]
ṁ       Map function then concat    [0,1,1,1,0]
ḋ        Convert from base 2         14

• Welcome to PPCG! :) – James Mar 30 '18 at 23:17

# Japt, 9 bytes

ÆMgX ¤Ã¬Í


Run it

## Explanation:

ÆMgX ¤Ã¬Í
Æ     Ã     | Iterate X through the range [0...Input]
MgX        |   Xth Fibonacci number
¤      |   Binary
¬    | Join into a string
Í   | Convert into a base-2 number

• Bah! Beat me to it! – Shaggy Mar 30 '18 at 16:57
• @Shaggy I knew this one was going to be a race against you :P – Oliver Mar 30 '18 at 16:59

# Pyth, 22 bytes

JU2VQ=+Js>2J)is.BM<JQ2


Try it here

### Explanation

JU2VQ=+Js>2J)is.BM<JQ2
JU2                       Set J = [0, 1].
VQ       )             <Input> times...
=+Js>2J              ... add the last 2 elements of J and put that in J.
<JQ     Take the first <input> elements...
.BM        ... convert each to binary...
s           ... concatenate them...
i       2    ... and convert back to decimal.


# Perl 6, 38 bytes

{:2([~] (0,1,*+*...*)[^$_]>>.base(2))}  Try it online! • Note that this starts to get noticeably slower with inputs above 200. It takes about 8 seconds to generate the output with an input of 1000. (20 seconds if you include printing it) – Brad Gilbert b2gills Mar 30 '18 at 18:07 # JavaScript (Node.js), 70655857 55 bytes • thanks to Shaggy for reducing 2 bytes ('0b+C-0 to '0b'+C) f=(n,a=C=0,b=1)=>--n?f(n,b,a+b,C+=b.toString(2)):'0b'+C  Try it online! • You can ditch the -0 at the end to save another 2 bytes. – Shaggy Mar 30 '18 at 21:51 # 05AB1E, 6 bytes LÅfbJC  Try it online! 1-indexed. # J, 36 Bytes 3 :'#.;<@#:"0]2}.(,{:+_2&{)^:y _1 1'  ### Explanation: 3 :'#.;<@#:"0]2}.(,{:+_2&{)^:y _1 1' | Explicit function (,{:+_2&{)^:y _1 1 | Make n fibonacci numbers, with _1 1 leading 2}. | Drop the _1 1 <@#:"0] | Convert each item to binary and box ; | Unbox and join #. | Convert back from binary  # x86, 3722 21 bytes Changelog • -13 by using bsr. Thanks Peter Cordes! • -1 by using a while loop instead of loop and push/pop ecx (credit Peter Cordes). Input in edi, output in edx. .section .text .globl main main: mov$5, %edi            # n = 5

start:
dec     %edi                # Adjust loop count
xor     %ebx, %ebx          # b = 0
mul     %ebx                # a = result = 0
inc     %ebx                # b = 1

fib:
add     %ebx, %eax          # a += b
xchg    %eax, %ebx          # swap a,b
bsr     %eax, %ecx          # c = (bits of a) - 1
inc     %ecx                # c += 1
sal     %cl, %edx           # result >>= c
add     %eax, %edx          # result += a

dec     %edi                # n--; do while(n)
jnz     fib

ret


Objdump:

00000005 <start>:
5:   4f                      dec    %edi
6:   31 db                   xor    %ebx,%ebx
8:   f7 e3                   mul    %ebx
a:   43                      inc    %ebx

0000000b <fib>:
d:   93                      xchg   %eax,%ebx
e:   0f bd c8                bsr    %eax,%ecx
11:   41                      inc    %ecx
12:   d3 e2                   shl    %cl,%edx
16:   4f                      dec    %edi
17:   75 f2                   jne    b <fib>
19:   c3                      ret

• Use lea to shift-and-add in fib2. Also, extracting each bit one at a time is unnecessary. Use bsr %eax, %ecx to find the number of bits in the binary representation, and use a shift by CL / or to merge, like Dennis's Python answer is doing. – Peter Cordes Mar 30 '18 at 20:54
• You need cl for shift counts, so take your loop counter in a different reg (like %edi) and use dec %edi / jnz (3 bytes in 32-bit code, 4 bytes in 64-bit). In 32-bit code, that saves 1 byte total from dropping the push/pop ecx. Don't fall into the trap of using loop when it makes the problem harder, not easier. (Your calling convention is already custom, clobbering %ebx, so don't call your function main) You might be able to return in EAX while still taking advantage of 1-byte xchg, no need to be non-standard if you don't need to. – Peter Cordes Mar 31 '18 at 2:04
• You can replace the extra inc %ecx of the shift count with an extra left-shift as you add, using lea (%eax, %edx, 2), %edx. Neutral in bytes for 32-bit, saves one for x86-64. But saves an instruction. – Peter Cordes Mar 31 '18 at 2:08
• Every time I end up using loop in code golf, I feel dirty. Well not quite, but disappointed that I couldn't find an equally small implementation that avoided that slow instruction; outside of code golf, loop is one of my pet peeves. I wish it was fast on modern CPUs, because it would be very nice for extended-precision loops without partial-flag stalls, but it's not and should be considered only as an obscure size optimization instruction that makes your code slow. – Peter Cordes Mar 31 '18 at 2:12
• Anyway, nice job. Other than push/pop/loop -> dec/jnz, I don't see any savings, just the LEA speedup that's neutral in code-size. I'd always wondered if there was ever a real use case for the xor/mul trick to zero three registers (do you ever need that many zeroes?), but using that as part of creating a 1 makes it more sensible. – Peter Cordes Mar 31 '18 at 2:35

# APL (Dyalog), 26 22 bytes

4 bytes saved thanks to @H.PWiz

{2⊥∊2∘⊥⍣¯1¨1∧+∘÷\~⍵↑1}


Try it online!

f=0:scanl(+)1f
foldr1(\x y->y+x*2*2^floor(logBase 2.read.show$y)).(takef)  Ungolfed version: import Data.Bits fib = 0:scanl (+) 1 fib catInt :: Integer -> Integer -> Integer catInt x y = x' + y where position = floor$ succ $logBase 2$ realToFrac y
x' = shift x position

answer n = foldr1 catInt fib' where
fib' = take n fib

• Welcome to PPCG and Haskell golfing in particular! A shorter way to generate an infinite list of Fibonacci numbers is f=0:scanl(+)1f (taken from here). Functions can be anonymous, so you can drop the leading g=, see our Guide to Ggolfing Rules in Haskell. – Laikoni Mar 31 '18 at 9:47
• Thanks! That compensates for some of the longer functions used. I spent a while trying to find a way to implement bit-shifting in a more concise way, but came up short. – user9549915 Mar 31 '18 at 18:25
• You can replace $realToFrac y with .read.show$y for one byte – H.PWiz Mar 31 '18 at 18:59

# Pari/GP, 59 bytes

n->fromdigits(concat([binary(fibonacci(i))|i<-[0..n-1]]),2)


Try it online!

# APL+WIN, 55 bytes

Prompts for screen input of integer.

v←b←0 1⋄⍎∊(⎕-2)⍴⊂'v←v,c←+/¯2↑v⋄b←b,((1+⌊2⍟c)⍴2)⊤c⋄'⋄2⊥b


APL+WIN's maximum integer precision is 17 and integer limit is of the order of 10E300 therefore the maximum input number is 55 and the result is: 1.2492739026634838E300

# Python 3, 94 bytes

f=lambda n,a=[0,1]:n>len(a)and f(n,a+[sum(a[-2:])])or int(''.join(bin(v)[2:]for v in a[:n]),2)


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# Jelly, 6 bytes

ḶÆḞBFḄ


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Ḷowered range -> nth ÆḞibonacci number -> from dec to Binary -> Flatten -> from Ḅinary to dec

# MATL, 21 bytes

0li:"yy+]xx&h"@B]&hXB


Try it online!

### Explanation

0l        % Push 0, then 1 (initial terms of the Fibonacci sequence)
i:"       % Do n times, where n is the input
yy+     %   Duplicate top two numbers and push their sum
]       % End
xx        % Delete the last two results. The stack now contains the
% first n Fibonacci numbers, starting at 0
&h        % Concatenate all numbers into a row vector
"         % For each
@       %   Push current number
B       %   Convert to binary. Gives a vector of 0 and 1
]         % End
&h        % Concatenate all vectors into a row vector
XB        % Convert from binary to decimal. Implicitly display


# J, 25 bytes

2(#.;)<@#:@(1#.<:!|.)\@i.


Try it online!

## Explanation

2(#.;)<@#:@(1#.<:!|.)\@i.  Input: n
i.  Range [0, n)
\@    For each prefix
|.         Reverse
!           Binomial coefficient (vectorized)
<:            Decrement
1#.              Sum
#:                   Convert to binary
<                      Box
;                        Link. Join the contents in each box
2 #.                         Convert to decimal from base 2


# Python 3, 86 bytes

def f(N):
a,b,l=0,1,''
for _ in range(N):l+=format(a,'b');a,b=b,a+b
return int(l,2)


Try it online!

D,f,@@@@*,V$2D+G1+dAppp=0$Qp{f}p
D,k,@,¿1=,1,bM¿
D,g,@,¿1_,1_001${f},1¿{k} D,w,@,BBbR D,l,@,ßR€gp€w@0b]@¦+VcG2$Bb


Try it online!

## PHP, 124 Bytes

Try it online!

So I was looking for a way to output fibonacci numbers using the series, until I found this. It turns out you can calculate the fibonacci series via rounding, so I tried the challenge with a recursive function.

I found the approach of "rounding" really interesting, also a professor showed me this a while ago.

Code

function f($n,$i=0,$b=''){ if($n>$i){$b.=
decbin(round(pow((sqrt(5)+1)/2,$i)/sqrt(5)));f($n,$i+1,$b);}else{echo bindec($b);}}  Explanation function f($n,$i=0,$b=''){           #the function starts with $i=0, our nth-fib number if($n>$i){ #it stops once$n (the input) = the nth-fib
$b.=decbin( #decbin returns an integer as bin, concatenates round(pow((sqrt(5)+1)/2,$i)/sqrt(5))
#the formula, basically roundign the expression
);                           #it returns the (in this case) $i-th fib-number f($n,$i+1,$b);                   #function is called again for the next index
}else{                               #and the current string for fibonacci

echo bindec($b); #"echo" the result, bindec returns the base 10 #value of a base 2 number } }  Also check this stackoverflow post the best answer refers to the same article on Wikipedia. • Interesting way to do it! – Nathan Wood Apr 4 '18 at 21:13 # Stax, 9 bytes ü1∞╓♪εw≤+  Run and debug it at staxlang.xyz! ## Unpacked (10 bytes) and explanation: vr{|5|Bm|B v Decrement integer from input. Stax's Fibonacci sequence starts with 1 :( r Integer range [0..n). { m Map a block over each value in an array. |5 Push nth Fibonacci number. |B Convert to binary. |B Implicit concatenate. Convert from binary. Implicit print.  # Julia 0.6, 65 bytes f(n)=n<2?n:f(n-1)+f(n-2) n->parse(BigInt,prod(bin.(f.(0:n-1))),2)  Try it online! # Pyth, 27 bytes JU2V-Q2=aJ+eJ@J_2)is.BM<JQ2  Test suite Python 3 translation: Q=eval(input()) J=list(range(2)) for i in range(Q-2): J.append(J[-1]+J[-2]) print(int(''.join(map("{0:b}".format,J[:Q])),2))  ## 37 bytes J[Z1)W<lJQ=aJ+eJ@J_2)Ig1QZ.?ijkm.BdJ2  Test suite Python 3 translation: Q=eval(input()) J=[0,1] while len(J)<Q: J.append(J[-1]+J[-2]) if 1>=Q: print(0) else: print(int(''.join(map("{0:b}".format,J)),2))  # Ruby, 61 bytes ->n,a=0,b=1,s=""{s+="%b"%a;a,b=b,a+b;(n-=1)>0?redo:s.to_i(2)}  Try it online! # Jotlin, 59 bytes g(l(0,1)){l(a.sum(),a)}.take(this).j(""){a.s(2)}.i(2)  ## Test Program data class Test(val input: Int, val output: Long) val tests = listOf( Test(1, 0), Test(2, 1), Test(3, 3), Test(4, 14), Test(5, 59), Test(6, 477), Test(7, 7640), Test(8, 122253), Test(9, 3912117), Test(10, 250375522) ) fun Int.r() = g(l(0,1)){l(a.sum(),a)}.take(this).j(""){a.s(2)}.i(2) fun main(args: Array<String>) { for (r in tests) { println("${r.input.r()} vs \${r.output}")
}
}


It supports up to 10, changing .i(2) for .toLong(2) would support up to 14 if needed

# Python 2, 88 bytes

def f(n):
a,b,r=0,1,"0"
for _ in range(n-1):a,b=b,a+b;r+=bin(a)[2:]
print int(r,2)


# R, 244180 179 bytes

i=ifelse;g=function(n)i(n<3,1,g(n-1)+g(n-2))
a=scan(,"");i(a==1,0,sum(2^(which(rev(unlist(sapply(g(2:a-1),function(x)(y=rev(as.numeric(intToBits(x))))[which(!!y):32]))>0))-1)))


Try it online!

Saved some bytes by concatenating numeric vectors, not strings. Bloody special case for 0!

• Functions are acceptable. Also it is much more efficient to shift the result left by the number of bits then to bother with numeric vectors. See my or Dennis's python answer. This has the added benefit of handling the 0 case. – qwr Mar 31 '18 at 14:08
• codegolf.stackexchange.com/questions/4024/tips-for-golfing-in-r R golfing is too much for me :/ – qwr Mar 31 '18 at 14:17
• @qwr The answer is not a function; I am creating a helper function because it must be sapply’d to a vector due to the fact that it is recursive. It cannot be all wrapped into one line. As you see, the programme prompts for user’s input and then returns the answer. One byte can be saved by creating a shortcut for ifelse. And... we can remove ,"" from scan, yes. – Andreï Kostyrka Mar 31 '18 at 16:07