Am I a Fibonacci Number?

Question

Your Task:

Write a program or function to check if a number that is inputted is a Fibonacci number. A Fibonacci number is a number contained in the Fibonacci sequence.

The Fibonacci Sequence is defined as: F(n) = F(n - 1) + F(n - 2)

With the seeds being F(0) = 0 and F(1) = 1.

Input:

A non-negative integer between 0 and 1,000,000,000 that may or may not be a Fibonacci number.

Output:

A truthy/falsy value indicating whether or not the input is a Fibonacci number.

Examples:

0-->truthy
1-->truthy
2-->truthy
12-->falsy

Scoring:

This is code-golf, lowest byte count wins.

Answer 1

Neim, 2 bytes

f𝕚

Explanation:

f       Push an infinite fibonacci list
 𝕚      Is the input in that list?

Works the same as my It's Hip to be Square answer, but uses a different infinite list: f, for fibonacci.

Try it!

Answer 2

Regex (ECMAScript), 392 358 328 224 206 165 164 162 161 bytes

-1 byte by using ECMAScript non-participating capture group behavior
-2 bytes by by calculating \$F_{2n+3}\$ differently
-1 byte by moving around a remainder

The techniques that need to come into play to match Fibonacci numbers with an ECMAScript regex (in unary) are a far cry from how it's best done in most other regex flavors. The lack of forward/nested backreferences or recursion means that it is impossible to directly count or keep a running total of anything. The lack of lookbehind makes it often a challenge even to have enough space to work in.

Many problems must be approached from an entirely different perspective, and seem unsolvable until the arrival of some key insight. It forces you to cast a much wider net in finding which mathematical properties of the numbers you're working with might be able to be used to make a particular problem solvable.

In March 2014, this is what happened for Fibonacci numbers. Looking at the Wikipedia page, I initially couldn't figure out a way, though one particular property seemed tantalizingly close. Then the mathematician teukon outlined a method that made it quite clear it would be possible to do, using that property along with another one. He was reluctant to actually construct the regex. His reaction when I went ahead and did it:

You're crazy! ... I thought you might do this.

As with my other ECMAScript unary math regex posts, I'll give a warning: I highly recommend learning how to solve unary mathematical problems in ECMAScript regex. It's been a fascinating journey for me, and I don't want to spoil it for anybody who might potentially want to try it themselves, especially those with an interest in number theory. See that post for a list of consecutively spoiler-tagged recommended problems to solve one by one.

So do not read any further if you don't want some unary regex magic spoiled for you. If you do want to take a shot at figuring out this magic yourself, I highly recommend starting by solving some problems in ECMAScript regex as outlined in that post linked above.

The challenge I initially faced: A positive integer x is a Fibonacci number if and only if 5x² + 4 and/or 5x² - 4 is a perfect square. But there is no room to calculate this in a regex. The only space we have to work in is the number itself. We don't even have enough room to multiply by 5 or take the square, let alone both.

teukon's idea on how to solve it (originally posted here):

The regex is presented with a string of the form ^x*$, let z be its length. Check whether or not z is one of the first few Fibonacci numbers by hand (up to 21 should do). If it is not:

1. Read a couple of numbers, a Use forward look-aheads to build a², ab, and b².
2. Assert that either 5a² + 4 or 5a² - 4 is a perfect square (so a must be F_n-1 for some n).
3. Assert that either 5b² + 4 or 5b² - 4 is a perfect square (so b must be F_n).
4. Check that z = F_2n+3 or z = F_2n+4 by using the earlier built a², ab, and b², and the identities:
   • F_2n-1 = F_n² + F_n-1²
   • F_2n = (2F_n-1 + F_n)F_n

In brief: these identities allow us to reduce the problem of checking that a given number is Fibonacci to checking that a pair of much smaller numbers are Fibonacci. A little algebra will show that for large enough n (n = 3 should do), F_2n+3 > F_n + 5F_n² + 4 so there should always be enough space.

And here is a mockup of the algorithm in C which I wrote as a test before implementing it in regex. My final golfed regex is rather different. There turned out to be no need to find both \$a\$ and \$b\$, because \$b\$ is the Lucas number \$L_n\$, which along with \$F_n\$ can be used to derive \$F_{2n+2}\$ and \$F_{2n+3}\$ as needed:

• \${L_n}^2 = 5{F_n}^2±4\$
• \$L_{2n} = ({5{F_n}^2 + {L_n}^2}) / 2 = 5{F_n}^2±2 = {L_n}^2∓2\$
• \$F_{2n} = L_n F_n\$
• \$F_{2n+1} = ({L_{2n} + F_{2n}})/2\$
• \$F_{2n+2} = F_{2n} + F_{2n+1}\$
• \$F_{2n+3} = F_{2n+1} + F_{2n+2}\$

So with no further ado, here is the regex:

^(?=(x*).*(?=x{4}(x{5}(\1{5}))(?=\2*$)\3+$)(|x{4})(?=xx(x*)\5)\4(x(x*))(?=(\6*)\7+$)(?=\6*$\8)\6*(?=x\1\7+$)(x*)(\9x?)|)(\9\10(\5\10)\12?|xx?x?|x{5}|x{8}|x{21})$

Try it online!

And the pretty-printed, commented version:

^                          # tail = N = input number
(?=
    (x*)                   # \1+1 = potential number for which 5*(\1+1)^2 ± 4 is a
                           # perfect square; this is true iff \1+1 is a Fibonacci number,
                           # which we shall call F_n. Outside the surrounding lookahead
                           # block, \1+1 is guaranteed to be the largest number for which
                           # this is true such that \1 + 5*(\1+1)^2 + 4 <= N.
    .*
    (?=                    # tail = (\1+1) * (\1+1) * 5 + 4
        x{4}
        (                  # \2 = (\1+1) * 5
            x{5}
            (\1{5})        # \3 = \1 * 5
        )
        (?=\2*$)
        \3+$
    )
    (|x{4})                # \4 = parity - determined by whether the index of Fibonacci
                           #               number \1+1 is odd or even;
    (?=
        xx                 # tail = arithmetic mean of (\1+1) * (\1+1) * 5 and \6 * \6
                           #      = ((F_n * F_n * 5) + (L_n * L_n)) / 2 = L_{2n}
        (x*)\5             # \5 = floor(tail / 2) = floor(L_{2n} / 2)
    )
    \4
    # require that the current tail is a perfect square
    (x(x*))                # \6 = potential square root, which will be the square root
                           #      after the following two lookaheads; \7 = \6-1
    (?=(\6*)\7+$)          # \8 = must be zero for \6 to be a valid square root
    (?=\6*$\8)
    # \6 is now the Lucas number L_n corresponding to the Fibonacci number F_n.
    \6*
    (?=x\1\7+$)            # tail = F_{2n} = L_n * F_n = \6 * (\1+1), where \6 is larger
    (x*)(\9x?)             # \9 = floor(tail / 2);
                           # \10 = ceil(tail / 2); the remainder tail % 2 will always be
                           #       the same as the remainder discarded by \5, because
                           #       F_{2n} is odd iff L_{2n} is odd, thus this ceil()
                           #       can complement the floor() of \5 when adding \5 + \10
|                          # Allow everything above to be skipped, resulting in all
                           # capture groups being unset.
)
(
    # Note that if the above was skipped using the empty alternative in the lookahead,
    # the following will evaluate to 0. This relies on ECMAScript NPCG behavior.
    \9\10(\5\10)           # \12  = F_{2n+1} = (L_{2n} + F_{2n})/2 = \5 + \10;
                           # head = F_{2n+2} = F_{2n}   + F_{2n+1}
                           #                 = \9+\10   +   \12
    \12?                   # head = F_{2n+3} = F_{2n+2} + F_{2n+1}    (optionally)
                           #                 = head     +   \12
|
    xx?x?|x{5}|x{8}|x{21}  # The Fibonacci numbers 1, 2, 3, 5, 8, 21 cannot be handled
                           # by our main algorithm, so match them here; note, as it so
                           # happens the main algorithm does match 13, so that doesn't
                           # need to be handled here.
)$

The multiplication algorithm is not explained in those comments, but is briefly explained in a paragraph of my abundant numbers regex post.

I was maintaining six different versions of the Fibonacci regex: four that ratchet from shortest length to fastest speed and use the algorithm explained above, and two others that use a different, much faster but much more lengthy algorithm, which as I found can actually return the Fibonacci index as a match (explaining that algorithm here is beyond the scope of this post, but it's explained in the original discussion Gist). I don't think I would maintain that many very-similar versions of a regex again, because at the time I was doing all my testing in PCRE and Perl, but my regex engine is fast enough that concerns of speed are not as important anymore (and if a particular construct is causing a bottleneck, I can add an optimization for it) – though I'd probably again maintain one fastest version and one shortest version, if the difference in speed were big enough.

The "return the Fibonacci index minus 1 as a match" version (not heavily golfed):

Try it online!

All of the versions are on github with the full commit history of golf optimizations:

regex for matching Fibonacci numbers - short, speed 0.txt (the shortest but slowest one, as in this post)
regex for matching Fibonacci numbers - short, speed 1.txt
regex for matching Fibonacci numbers - short, speed 2.txt
regex for matching Fibonacci numbers - short, speed 3.txt
regex for matching Fibonacci numbers - fastest.txt
regex for matching Fibonacci numbers - return index.txt

Answer 3

JavaScript (ES6), 34 bytes

f=(n,x=0,y=1)=>x<n?f(n,y,x+y):x==n

Recursively generates the Fibonacci sequence until it finds an item greater than or equal to the input, then returns item == input.

Answer 4

Retina, 23 bytes

^$|^(^1|(?>\2?)(\1))*1$

Try it online!

Input in unary, outputs 0 or 1.

Explanation

The Fibonacci sequence is a good candidate for a solution with forward references, i.e. a "backreference" that refers either to a surrounding group or one that appears later in the regex (in this case, we're actually using both of those). When matching numbers like this, we need to figure out a recursive expression for the difference between sequence elements. E.g. to match triangular numbers, we usually match the previous segment plus one. To match square numbers (whose differences are the odd numbers), we match the previous segment plus two.

Since we obtain the Fibonacci numbers by adding the second-to-last element to the last one, the differences between them are also just the Fibonacci numbers. So we need to match each segment as the sum of the previous two. The core of the regex is this:

(         # This is group 1 which is repeated 0 or more times. On each
          # iteration it matches one Fibonacci number.
  ^1      # On the first iteration, we simply match 1 as the base case.
|         # Afterwards, the ^ can no longer match so the second alternative
          # is used.
  (?>\2?) # If possible, match group 2. This ends up being the Fibonacci
          # number before the last. The reason we need to make this optional
          # is that this group isn't defined yet on the second iteration.
          # The reason we wrap it in an atomic group is to prevent backtracking:
          # if group 2 exists, we *have* to include it in the match, otherwise
          # we would allow smaller increments.
  (\1)    # Finally, match the previous Fibonacci number and store it in
          # group 2 so that it becomes the second-to-last Fibonacci number
          # in the next iteration.
)*

Now this ends up adding the Fibonacci numbers starting at 1, i.e. 1, 1, 2, 3, 5, .... Those add up to 1, 2, 4, 7, 12, .... I.e. they are one less than the Fibonacci numbers, so we add a 1 at the end. The only case this doesn't cover is zero, so we have the ^$ alternative at the beginning to cover that.

Answer 5

Python 3, 48 bytes

lambda n:0in((5*n*n+4)**.5%1,abs(5*n*n-4)**.5%1)

Try it online!

Answer 6

Python 2, 48 44 bytes

f=lambda n,a=0,b=1:n>a and f(n,b,a+b)or n==a

Try it online

_{Thanks to Jonathan Allan for saving 4 bytes}

Answer 7

05AB1E, 8 7 bytes

>ÅF¹å¹m

Explanation:

>ÅF       # Generate Fibbonacci numbers up to n+1
   ¹å     # 0 if original isn't in the list, 1 if it is
     ¹m   # 0**0 = 1 if input was 0 (hotfix for 0).
          # or 0**n if not fibb / 1**n if it is a fibb.

Try it online!

-1 thanks to Jonathan Allan for the 0-not-a-fibonacci-number workaround.

Answer 8

Perl 6, 23 bytes

{$_∈(0,1,*+*...*>$_)}

Answer 9

Seriously, 3 bytes

,fu

Try it online!

Returns the index +1 in the list of Fibonacci numbers if truthy, otherwise returns falsy.

Explanation:

,fu
,   read input
 f  0-indexed index of that number in the fibonacci sequence (-1 if not in the sequence)
  u increment. (Makes the -1 value falsy and the 0-value truthy)

Answer 10

Jelly,  8 7  6 bytes

-r‘ÆḞċ

Try it online!

How?

-r‘ÆḞċ - Link: non negative number, n
-      - literal -1      = -1
 r     - inclusive range = [-1,0,1,2,3,4,5,...,n]
  ‘    - increment n     = [ 0,1,2,3,4,5,6,...,n+1]
   ÆḞ  - Fibonacci       = [ 0,1,1,2,3,5,8,...,fib(n+1)]
     ċ - count occurrences of n (1 if n is a Fibonacci number, 0 otherwise)

Notes:

• the increment, ‘, is needed so this works for 2 and 3, since they are the 3^rd and 4^th Fibonacci numbers - beyond 3 all Fibonacci numbers are greater than their index.
• the - is needed (rather than just ‘R) so this works for 0 since 0 is the 0^th Fibonacci number;

Answer 11

ZX81 BASIC 180 151 100 ~94 tokenized BASIC bytes

With thanks to Moggy on the SinclairZXWorld forums, here is a much neater solution that saves more bytes.

 1 INPUT I
 2 FOR F=NOT PI TO VAL "1E9"
 3 LET R=INT (VAL ".5"+(((SQR VAL "5"+SGN PI)/VAL "2")**I)/SQR VAL "5")
 4 IF R>=I THEN PRINT F=R
 5 IF R<I THEN NEXT F

This will output 1 if a Fibonacci number is entered, or zero if not. Although this saves bytes, it's much slower than the old solutions below. For speed (but more BASIC bytes) remove the VAL wrappers around the string literal numbers. Here is the old(er) solutions with some explanations:

 1 INPUT A$
 2 LET A=SGN PI
 3 LET B=A
 4 LET F=VAL A$
 5 IF F>SGN PI THEN FOR I=NOT PI TO VAL "1E9"
 6 LET C=A+B
 7 LET A=B
 8 LET B=C
 9 IF B>=F THEN GOTO CODE "£"
10 IF F THEN NEXT I
12 PRINT STR$(SGN PI*(B=F OR F<=SGN PI)) AND F>=NOT PI;"0" AND F<NOT PI

The amendments above saves further BASIC bytes to condensing the IF statements into a single PRINT in line 12; other bytes were saved by using the VAL keyword and, and using GOTO CODE "£", which in the ZX81 character set is 12. Strings save more bytes over numbers as all numeric values are stored as floats so therefore take more space on the VAR stack.

Answer 12

C#, 109 bytes

bool f(int n){int[]i=new[]{0,1,0};while(i[0]<n||i[1]<n){i[i[2]%2]=i[0]+i[1];i[2]++;}return n==i[0]||n==i[1];}

Definitely could be improved, but I didn't have time.

Answer 13

><>, 21 19+3 = 24 22 bytes

i1\{=n;
?!\:@+:{:}(

Input is expected to be on the stack at program start, so +3 bytes for the -v flag.

Try it online!

This works by generating Fibonacci numbers until they are greater than or equal to the input number, then checking the last generated number for equality with the input. Outputs 1 if it was a Fibonacci number, 0 otherwise.

To ensure that 0 is handled correctly, the seed is -1 1 - the first number generated will be 0 rather than 1.

Thanks to @cole for pointing out that i can be used to push -1 onto the stack when STDIN is empty. Very clever!

Previous version:

01-1\{=n;
}(?!\:@+:{:

Answer 14

Mathematica, 37 bytes

!Fibonacci@n~Table~{n,0,#+1}~FreeQ~#&

-4 bytes from @ngenisis

Answer 15

Dyalog APL, 17 bytes

0∊1|.5*⍨4 ¯4+5××⍨

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Answer 16

MATL (16 bytes)

2^5*4-t8+hX^tk=s

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Not the golfiest solution, but wanted to use the direct method of checking if "5*x^2+/-4" is a perfect square.

Explanation:

2^5*    % 5 times the input squared
4-      % push the above minus 4
t8+     % push the above plus 8 (+4 overall)
hX^     % concatenate them into an array and then take the sqrt().
tk      % push a copy of the array that is rounded using floor().
=       % test if the sqrt's were already integers
s       % sum the results, returns 0 if neither was a perfect square.

Note:

In the "0" case it returns "2" because both 4 and -4 are perfect squares, same with 1 which produces "1 1". Consider any non-zero output as "truthy", and 0 as "falsy".

Answer 17

Pari/GP, 32 bytes

n->!prod(x=0,n+1,fibonacci(x)-n)

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Answer 18

PHP, 44 bytes

for(;0>$s=$x-$argn;)$x=+$y+$y=$x?:1;echo!$s;

Try it online!

PHP, 58 bytes

for($x=0,$y=1;$x<$argn;$x=$y,$y=$t)$t=$x+$y;echo$x==$argn;

Try it online!

Answer 19

C# (.NET Core), 51 bytes

bool f(int n,int a=0,int b=1)=>a<n?f(n,b,a+b):a==n;

Try it online!

-6 bytes thanks to @Oliver!

This solution uses a pretty straightforward recursive function.

• The variable n is the number to be tested.
• The variables a and b are the 2 most recent numbers in the sequence.
• Checks if the first of the 2 most recent number is less than input. When this is the case, a recursive call is made to the next numbers in the series.
• Otherwise, check whether the first number is equal to input and return the result.

The TIO link demonstrates this working for 1134903170 which exceeds the maximum value required by the challenge.

Answer 20

Alchemist, 205 134 bytes

^{Big thanks to ASCII-only for rather clever merging of states, saving 71 bytes!!}

_->In_x+c+u
u+b->u+a+d
u+0b->v
v+c->v+b+d
v+0c->w
w+a+x->w+y
w+0a+0x->Out_"1"
w+a+0x->Out_"0"
w+0a+x+y->w+2x
w+0a+0y+d->w+c
w+0d+0a->u

Try it online or verify in batch!

Ungolfed

# read input, initialize (c = 1)
_ -> In_x + c + s0

# a,d <- b
s0 +  b -> s0 + a + d
s0 + 0b -> s1

# b,d <- c
s1 +  c -> s1 + b + d
s1 + 0c -> s2

s2 +  a +  x -> s2 + y            # y <- min(a,x)
s2 + 0a + 0x -> Out_"1"           # if (a == x): was Fibonacci
s2 +  a + 0x -> Out_"0"           # if (a >  x): stop (we exceeded target)
s2 + 0a +  x +  y -> s2 + 2x      # if (a <  x): x += a (since y = a) / restore x
s2 + 0a      + 0y +  d -> s2 + c  # once that's done; c <- d
s2 + 0a           + 0d->s0        # and finally loop

Answer 21

Java (JDK), 37 bytes

s->s.matches(".?|(\\2?+(\\1|^.))*..")

Try it online!

This works for numbers up to 1,836,311,903 (47th fibonacci number) included. Above that, the result is undefined (including a potential infinite loop).

Credits

• Thanks to Kevin Cruijssen and David Conrad for helping golfing :)
• Thanks to Kevin Cruijssen for porting the regex version, saving 12 bytes in the process!

Answer 22

Vyxal r, 4 bytes

ÞFce

Try it Online! - show all boolean results
Try it Online! - show only numbers returning truthy

Thanks to Steffan. This based on the alternative 5 byte answer below, removing the need for the $.

Vyxal, 5 bytes

ÞF0pc

Try it Online! - show all boolean results
Try it Online! - show only numbers returning truthy

ÞF   # All 1-indexed Fibonacci numbers as a LazyList.
0p   # Prepend the list with 0
c    # Is the input contained in the list?

Alternative 5 bytes:

ÞFc$e

Try it Online! - show all boolean results
Try it Online! - show only numbers returning truthy

ÞF   # All 1-indexed Fibonacci numbers as a LazyList.
c    # Is the input contained in the list?
$    # Swap above result with the input (this creates a copy of the input)
e    # Exponentiate. For an input of zero this will yield 0**0 == 1; for other
     # Fibonacci numbers, 1**{n>0} == 1; otherwise 0**{n>0} == 0.

The above two solutions work around the fact that there's no 0-indexed version of ÞF. Alternatively using ∆F requires a different workaround, but it can also be done in 5 bytes:

Þn∆Fc

Try it Online! - show all boolean results
Try it Online! - show only numbers returning truthy

Þn  # All integers in an infinite list (0, 1, -1, 2, -2, ...)
∆F  # nth Fibonacci number, 0-indexed
c   # Contains - Check if one thing contains another

Using ʀ (Inclusive range [0..input]) instead of Þn wouldn't work, because the Fibonacci numbers \$2\$ and \$3\$ are not members of the lists \$[0,1,1]\$ and \$[0,1,1,2]\$ respectively. It would be necessary to use ›ʀ (the range [0..input+1]), but for non-Fibonacci numbers that is exponentially slower than using Þn (and Þn∆Fc is in turn a bit slower than ÞFc$e).

Note that Þn includes negative numbers, and ∆F on a negative number returns a negative Fibonacci number. However, Þn∆Fc only returns truthy for nonnegative Fibonacci numbers, probably due to a bug in Vyxal.

---

Without using any Fibonacci builtins (7 bytes):

k≈⁽+dḞc

Try it Online! - show all boolean results
Try it Online! - show only numbers returning truthy

The is based on lyxal's answer to Fibonacci function or sequence.

k≈   # the list [0, 1]
⁽+d  # lambda x, y: x + y
Ḟ    # Create an infinite sequence based on the function and the initial list.
c    # Does the list contain the input?

Answer 23

Jelly, 5 bytes

ȷḶÆḞi

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Returns 0 for non-Fibonacci numbers, and the 1-indexed position of the number in the Fibonacci sequence for Fibonacci numbers.

Explanation:

ȷḶÆḞi
ȷ        The literal number 1000
 Ḷ       Range [0,1,...,999]
  ÆḞ     Get the ith Fib number; vectorizes [1,1,2,3,5,...,<1000th Fib number>]
    i    Get the first index of element in list, or 0 if not found

Answer 24

R, 43 40 bytes

pryr::f(x%in%DescTools::Fibonacci(0:45))  

pryr::f creates a function:

function (x) 
x %in% DescTools::Fibonacci(0:45)

Uses DescTools::Fibonacci to create the first x+1 fibonacci numbers and checks for inclusion. x+1 because the third fibnum is 2, and that wouldn't be enough to check for inclusion of 3.

Luckily Desctools::Fibonacci(0)=0, so that is a nice freebee.

-3 bytes thanks to MickyT

Answer 25

Haskell, 31 bytes

f=0:scanl(+)1f
(`elem`take 45f)

Try it online! This exploits the fact that the input will be in the range from 0 to 1,000,000,000, hence we need to check only the first 45 Fibonacci numbers. f=0:scanl(+)1f generates an infinite list of Fibonacci numbers, take 45f is the list of the first 45 Fibonacci numbers and elem checks whether the input is in this list.

---

Unrestricted version: 36 bytes

f=0:scanl(+)1f
g n=n`elem`take(n+3)f

Try it online! For any n, taking the first n+3 Fibonacci numbers will guarantee that n will be in this list if it's a Fibonacci number.

Note that this approach is incredible inefficient for high numbers that are not Fibonacci numbers, because all n+3 Fibonacci numbers need to be computed.

Answer 26

Javascript (ES6 without the ** operator), 44 bytes

f=(x,c=x*(Math.sqrt(5)-1)/2%1)=>x*(c-c*c)<.5

Relies on the ratio between successive Fibonacci numbers approaching the golden ratio. The value of c is the fractional part of the input divided by the golden ratio - if the input is Fibonacci then this will be very close to 1 and the value of c-c² will be very small.

Not as short as some of the other JS answers but runs in O(1) time.

Answer 27

Julia 0.4, 29 bytes

!m=in(0,sqrt(5*m*m+[4,-4])%1)

Try it online!

---

If this isn't how you do a Julia answer, let me know. I'm unsure of how input works on TIO.

Answer 28

R, 32 31 bytes

Takes input from stdin, returns TRUE or FALSE as appropriate.

any(!(5*scan()^2+-1:1*4)^.5%%1)

Answer 29

Common Lisp, 61 54 bytes

(defun f(x)(do((a 0 b)(b 1(+ a b)))((>= a x)(= a x))))

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The reduction in size with respect to the previous version:

(defun f(x)(do((a 0 b)(b 1 c)(c 1(+ b c)))((>= a x)(= a x))))

was triggered by the idea that to generate the sequence of the Fibonacci numbers only two variables are necessary, not three.

Answer 30

D, 98 79 bytes 77 bytes

-19 bytes: changed square number algorithm (which also had the side effect of not needing import std;) and realized the non-UFCS way was 1 byte shorter

golfed:

int p(real z){return z^^.5%1==0;}int f(int z){return p(5*z*z+4)||p(5*z*z-4);}

readable:

int p(real z){
    return z^^.5%1==0
}
int f(int z){
    return p(5*z*z+4)||p(5*z*z-4);
}

where f is the function that tests if a number is a fibonacci number.

This uses the fact that a positive integer z is a Fibonacci number if and only if one of 5z^2 + 4 or 5z^2 − 4 is a perfect square. wikipedia

Also real instead of float for one less byte, and returning an int instead of bool to save another byte (since ints are truthy if they're nonzero, and D implicitly casts expressions to the function's return value if possible, in this case casting a bool to an int).

There's probably some way to avoid having the extra function p (used to check if a number is a perfect square) but I can't find it out.

edit: replaced z^^2 with z*z