# The Möbius Function

The Möbius function is an important number theoretic function.

Your submission should accept a positive integer n and return the value of the Möbius function evaluated at n.

### Definition

The Möbius function μ(n) is defined as follows:

       |  1 if n is squarefree and has an even number of distinct prime factors
μ(n) = | -1 if n is squarefree and has an odd number of distinct prime factors
|  0 otherwise


n is called squarefree if the exponents of the prime factorization of n are all strictly less that two. (Alternatively: No prime to the power of two divides n).

### Test cases

Here you can see the first 50 values of μ: Public Domain Image from Wikipedia

The Möbius function is sequence number A008683 in the OEIS.

These are the first 77 values:

1, -1, -1, 0, -1, 1, -1, 0, 0, 1, -1, 0, -1, 1, 1, 0, -1, 0, -1, 0, 1, 1, -1, 0, 0, 1, 0, 0, -1, -1, -1, 0, 1, 1, 1, 0, -1, 1, 1, 0, -1, -1, -1, 0, 0, 1, -1, 0, 0, 0, 1, 0, -1, 0, 1, 0, 1, 1, -1, 0, -1, 1, 0, 0, 1, -1, -1, 0, 1, -1, -1, 0, -1, 1, 0, 0, 1


Larger values can also easily be checked in Wolframalpha.com or in the b-file of OEIS, as suggested by @MartinBüttner.

## Python 2, 48 bytes

m=lambda n,d=1:d%n and-m(d,n%d<1)+m(n,d+1)or 1/n


Earlier 51 byte version:

m=lambda n:1/n-sum(m(k)for k in range(1,n)if n%k<1)


Möbius-inverts the sequence 1,0,0,0,0....

The Möbius function has the property that for any n>1, the Möbius functions of n's divisors sum to 0. So, for n>1, μ(n) is computed by negating the sum of μ(k) for all proper divisors k of n. For n=1, the output is 1.

The code handles the base case by adding a floor-division 1/n, which gives 1 for n==1 and 0 otherwise.

Thanks to Dennis for saving 3 bytes with better recursive handling inspired by a similar structure in this challenge.

# Jelly, 7 bytes

Code:

ÆF>1’PS


Explanation:

ÆF       # This computes the prime factorization as well as the exponent
>1     # Compares each element if it's greater than 1, resulting in 1's and 0's
’    # Decrement on each element
P   # Compute the product
S  # Compute the sum of the list


For example, the number 10:

ÆF       # [[2, 1], [5, 1]]
>1     # [[1, 0], [1, 0]]
’    # [[0, -1], [0, -1]]
P   # [0, 1]
S  # 1


And results in 1.

• -1 byte: ÆFỊNPS (not sure if Ị was a built-in back then, but should be fine now). Oct 28 '18 at 10:47

## Mathematica, 9 bytes

MoebiusMu


Of course, Mathematica has a built-in. (And will probably be is beaten by Jelly anyway.)

## CJam, 18 15 bytes

WrimFz~\1&+f/:*


The fact that CJam returns 1 in factorisation builtins for n = 1 makes things a bit tricky.

Thanks to @PeterTaylor for the neat 1&+ trick for handling the 1 case.

## Explanation

W                 Push -1
ri               Push input as int
mF             Factorise input into [base exponent] pairs
z~           Zip and unwrap pairs, leaving stack like [-1 bases exponents]
\1&        Setwise intersect bases with 1, giving  for 1 and [] otherwise
+       Append to exponent array
f/     Divide the previously pushed -1 by each element in the array
This gives -1 for 1s and 0 otherwise, since / is int division
:*   Take product


For n > 1, the modified array is just the exponents array. If n is squarefree then the array is all 1s, which become all -1s after division. Otherwise, if n has a prime square divisor, then there will be a 0 somewhere after division, giving a product of 0.

For n = 1 the modified array is  + , which becomes [-1 -1] after division, giving a product of 1.

Alternative 16:

rimF{1#2+3%(}%:*


This uses # (find) on each [base exponent] array to look for a 1, then maps -1 -> 0, 0 -> 1, 1 -> -1.

## Ruby, 65+7=72 62+7=69 bytes

->x{((d=x.prime_division).all?{|_,n|n<2}?1:0)*(d.size%2*-2+1)}


+7 bytes for the -rprime flag.

We're doing this the very naïve way:

->x{
(
(d=x.prime_division)  # ex. input 20 results in [[2,2],[5,1]] here
.all?{|_,n|n<2}?      # are all factors' exponents under 2?
1:0                   # if so, result in a 1; otherwise, a 0
)
*                      # multiply that 1 or 0 by...
(d.size%2*-2+1)       # magic
}


The "magic" part results in 1 if the number is even and -1 otherwise. Here's how:

Expression       Even  Odd
--------------------------
d.size%2         0     1
d.size%2*-2      0     -2
d.size%2*-2+1    1     -1


## Labyrinth, 87 bytes

1
:
}
?   @ "}){/{=:
""({! "      ;
} )   :}}:={%"
* _}}){     ;
{      #}):{{
")%#/{+


Try it online!

### Short explanation

Here's a port of the algorithm used, in Python:

divisor = 1
mobius = 1
n = int(input())

while n != 1:
divisor += 1
count = 0

while n % divisor == 0:
n //= divisor
count += 1

mobius *= (count + 3)//(count + 1)%3*-1 + 1

print(mobius)


### Long explanation

The usual primer on Labyrinth:

• Labyrinth is stack-based and two-dimensional, with execution starting at the first recognised character. There are two stacks, a main stack and an auxiliary stack, but most operators only work on the main stack.
• At every branch of the labyrinth, the top of the stack is checked to determine where to go next. Negative is turn left, zero is straight ahead and positive is turn right.

Execution for this program starts at the top-left 1.

Outer preparation
=================

1        Pop 0 (stack is bottomless and filled with 0s) and push 0*10+1 = 1
:}       Duplicate and shift to auxiliary stack
Stack is now [div=1 n | mob=1]

Top of stack positive but can't turn right. Turn left into outer loop.

Begin outer loop
================
Inner preparation
-----------------

(        Decrement top of stack

If top was 1 (and is now zero), move forward and do...
------------------------------------------------------

{!       Print mob
@        Terminate

If top of stack was greater than 1, turn right and do...
--------------------------------------------------------

)        Increment n back to its previous value
_}       Push 0 and shift to aux
This will count the number of times n can be divided by div
}){      Increment div
Stack is now [div n | count mob]

Inner loop
----------

:}}      Dup n, shift both n's to aux
:=       Dup div and swap top of main with top of aux
{%       Shift div down and take mod
Stack is now [div n%div | n count mob]

If n%div == 0, move forward and do...
-----------------------------------

;        Pop n%div
:={/     Turn n into n/div
{)}      Increment count
(continue inner loop)

If n%div != 0, turn right (breaking out of inner loop) and do...
================================================================

;        Pop n%div
{{       Pull n and count from aux
:)}      Dup and increment count, giving (count+1), and shift to aux
#+       Add length of stack to count, giving (count+3)
{/       Calculate (count+3)/(count+1)
#%       Mod by length of stack, giving (count+3)/(count+1)%3
Multiply by -1
)        Increment, giving (count+3)/(count+1)%3*-1 + 1
This is 1 if count was 0, -1 if count was 1 and 0 if count > 1
{*}      Multiply mob by this number
(continue outer loop)


## Pyth, 9 bytes

^_{IPQlPQ


Explanation:

^_{IPQlPQ    Implicit: Q=input
PQ       Prime factorization of Q
{I         Is invariant under uniquify.
{IPQ       1 if Q is squarefree; 0 otherwise.
_{IPQ       -1 if Q is squarefree; 0 otherwise.
^     lPQ    Exponentiation to length of PQ.


Try it here.

## CJam, 20 bytes

rimf1-___&=\,2%!2*(*


Run it against a range of inputs here.

Seems golfable.

• The fact that 1 is returned in mf or mF really hurts... Jan 24 '16 at 14:31
• @Sp3000 Tell me about it :-( Jan 24 '16 at 17:35

# R 3916 bytes

numbers::moebius


Requires you having the numbers package installed on your system...

Edit: Far simpler if I read the specs appropriately [thanks @AlexA.]

• This returns the Möbius function evaluated for each integer from 1 to the input, but the task for this challenge is simply to evaluate the Möbius function on the input. Jan 25 '16 at 0:44
• Along the lines of the Mathematica answer, you could even do simply numbers::moebius for 16 bytes. Jan 25 '16 at 0:45

# Pyth, 16 Bytes

?nl{PQlPQZ^_1lPQ


Try it online!

My first actual Pyth answer. Suggestions appreciated! :)

## Explanation

My solution uses the fact that a number is squarefree, if its prime-factors contain no number more than once. If the input is squarefree, the Möbius-Function takes the value -1^(number of primefactors).


?n        if not equal
l{PQ      length of the list of the distinct input-Primefactors
lPQ       length of the list of primefactors including duplicates
Z         Input is not squarefree, so output Zero
^_1lPQ  if input is squarefree, output -1^(number of prime-factors)


# MATL, 15 17 bytes

tq?YftdAwn-1w^*


This uses current release (10.1.0) of the language/compiler.

Try it online!

### Explanation

t         % implicit input. Duplicate that
q         % decrement by 1. Gives truthy (nonzero) if input exceeds 1
?         % if input exceeds 1, compute function. Otherwise leave 1 on the stack
Yf      % vector of prime factors. Results are sorted and possibly repeated
td      % duplicate and compute differences
A       % true if all differences are nonzero
w       % swap
n       % number of elements in vector of prime factors, "x"
-1w^    % -1^x: gives -1 if x odd, 1 if x even
*       % multiply by previously obtained true/false value, so non-square-free gives 0
% implicitly end "if"
% implicitly display stack contents


# 05AB1E, 8 bytes, non-competing

Dammit, once again a bug which makes my submission non-competing. Code:

.p0K1›<P


Explanation:

.p        # Get the prime factorization exponents
0K      # Remove all zeroes from the list
1›    # Compare each element if greater than 1
<   # Decrement on each element
P  # Take the product


Uses the CP-1252 encoding

• › isn't in ISO 8859-1... Jan 24 '16 at 20:39
• @ETHproductions Heh? What kind of encoding is it then? I got it from this site. Jan 24 '16 at 20:44
• I believe it's called Extended ASCII. Jan 24 '16 at 20:54
• @ETHproductions Thanks, I have edited the post :) Jan 24 '16 at 21:08
• @ThomasKwa Ahh, I have found it. It is the CP-1252 encoding. Jan 24 '16 at 21:41

# Pyth, 11

*{IPQ^_1lPQ


Test suite

This multiplies the boolean value of if the prime factors are squarefree by -1 to the power of the number of prime factors that the number has.

I is an invariance check on the preceeding operator, which here is {, which is the unique-ify operator.

# Japt, 21 bytes

V=Uk)¬¥Vâ ¬?Vl v ªJ:0


Test it online!

# Julia, 66 bytes

n->(f=factor(n);any([values(f)...].>1)?0:length(keys(f))%2>0?-1:1)


This is a lambda function that accepts an integer and returns an integer. To call it, assign it to a variable.

Ungolfed:

function µ(n::Int)
# Get the prime factorization of n as a Dict with keys as primes
# and values as exponents
f = factor(n)

# Return 0 for non-squarefree, otherwise check the length of the list
# of primes
any([values(f)...] .> 1) ? 0 : length(keys(f)) % 2 > 0 ? -1 : 1
end


# PARI/GP, 7 bytes

moebius


Sadly möbius is not available. :)

## Seriously, 19 18 bytes

,w;iX1=Mπ)l1&τD*


Try it online!

Explanation:

,w;iXDYMπ)l1&τD*
,w;                push two copies of the full prime factorization of the input
(a list of [base, exponent] pairs)
    M         map the following function:
iX1=             flatten list, discard, equal to 1
(push 1 if exponent == 1 else 0)
π)       product of list, push to bottom of stack
1&     push 1 if the # of prime factors is even else 0
τD   double and decrement (transform [0,1] to [-1,1])
*  multiply


# C# (.NET Core), 867372 65 bytes

a=>{int b=1,i=1;for(;++i<=a;)b=a%i<1?(a/=i)%i<1?0:-b:b;return b;}


Try it online!

-13 bytes: streamlined loops, added return variable (thanks to Kevin Cruijssen)
-1 byte: changed setting b to zero to a ternary from an if (thanks to Kevin Cruijssen)
-7 bytes: changed if statement in for loop to a ternary (thanks to Peter Taylor and Kevin Cruijssen)

Ungolfed:

a => {
int b = 1, i = 1;           // initialize b and i to 1

for(; ++i <= a;)            // loop from 2 (first prime) to a
b = a % i < 1 ?                     // ternary: if a is divisible by i
((a /= i) % i < 1 ? 0 : -b) :   // if true: set b to 0 if a is divisible by i squared, otherwise flip sign of b
b;                              // if false: don't change b

return b;                   // return b (positive for even numbers of primes, negative for odd, zero for squares)
}

• 73 bytes I've changed int b=1;for(int i=1; to int b=1,i=1;for(;. Removed the {}-brackets for the loop. Changed both a%i==0 to a%i<1. Changed the b*=-1; to b=-b;. And finally changed the return 0; to b=0;. Oct 30 '18 at 12:57
• Yeah, everything you suggested looked correct. I was ever so slightly worried when you said it wasn't right, because that would've meant my original code was wrong too! :) Oct 30 '18 at 13:14
• Yeah, sorry about that. :) 1 more byte to golf btw is if(a%i<1)b=0; to b=a%i<1?0:b;. Oct 30 '18 at 13:14
• Actually, that overlooks an easy improvement: b=-b;b=a%i<1?0:b; golfs to b=a%i<1?0:-b; Oct 30 '18 at 18:55
• Continuing on @PeterTaylor's golf above, you can then change if(a%i<1){a/=i;b=a%i<1?0:-b;} to b=a%i<1?(a/=i)%i<1?0:-b:b; to save 3 more bytes. Oct 30 '18 at 20:10

# GNU sed, 89 bytes

#!/bin/sed -rf
s/.*/factor &/e
s/.*://
/\b([0-9]+) \1\b/!{
s/[0-9]+//g
s/\$/1/
s/  //g
y/ /-/
}
s/ .*/0/


# J, 18 bytes

[:*/3<:@|2+2<._&q:


## Microsoft Office Excel, British version, 196 bytes

=IF(ROW()>=COLUMN(),IF(AND(ROW()=1,COLUMN()=1),1,IF(COLUMN()=1,
1)),""))),"")


Excel cell formula to be entered in cells A1 to AX50.

# Julia, 35 bytes

\(n,k=1)=k%n>0?n\-~k-k\0^(n%k):1÷n


Based on @xnor's Python answer. Try it online!

# Seriously, 11 bytes

Golfing suggestions welcome. Try it online!

;y;l0~ⁿ)π=*


Ungolfing

     Implicit input n.
;    Duplicate n.
y    Push a list of the distinct prime factors of n. Call it dpf.
;    Duplicate dpf.
l    Push len(dpf).
0~   Push -1.
ⁿ    Push (-1)**len(dpf).
)    Rotate (-1)**len(dpf) to BOS. Stack: dpf, n, (-1)**len(dpf)
π    Push product(dpf).
=    Check if product(dpf) == n.
This is only true if n is squarefree.
*    Multiply (n is squarefree) by (-1)**len(dpf).
Implicit return.

• Nice solution=) (I guess however that this language is younger that the question, is it?) Sep 30 '16 at 20:30
• @flawr Apparently the answer works just as well in Seriously, and I don't know when Actually first came online, so I changed to Seriously just to be safe. Sep 30 '16 at 20:45

# Java 8, 7268 65 bytes

n->{int r=1,i=1;for(;++i<=n;)r=n%i<1?(n/=i)%i<1?0:-r:r;return r;}


-4 bytes thanks to @PeterTaylor.

Port of @Meerkat's .NET C# answer, which I first golfed a bit further, so make sure to upvote him!

Try it online.

Explanation:

n->{                 // Method with integer as both parameter and return-type
int r=1,           //  Result-integer, starting at 1
i=1;for(;++i<=n;)  //  Loop i in the range [1, n]:
r=n%i<1?         //   If n is divisible by i:
(n/=i)        //    Divide n by i first
%i<1?        //    And if n is still divisible by i:
0           //     Change r to 0
:            //    Else:
-r          //     Swap the sign of r (positive to negative or vice-versa)
:              //    Else:
r;            //     Leave r unchanged
return r;}         //  Return r as result

• See my comment on Meerkat's answer. Oct 30 '18 at 18:55
• @PeterTaylor Smart, thanks! And then 3 more bytes can be saved by using r=n%i<1?(n/=i)%i<1?0:-r:r;. Oct 30 '18 at 20:05

## Javascript (ES6), 48 bytes

f=(n,i=1)=>n-1?n%++i?f(n,i):(n/=i)%i?-f(n,i):0:1


First of all - this is my first code golf post so I may misunderstand the rules to some extent. In this submission the last character ; could be omitted and it'll still work but I'm not even sure if the code would be valid according to the ES6 specs. Anyways, a short explanation.

First, I'll explain the idea a bit; we take n, and we try dividing it by the integer i. If it's divisible, then we do so and we check if it's divisible by i again. If that's the case, then we need to return 0. Otherwise, we can try another i. The cool thing is, we can just start at i=2 and just add increase it by 1 every time, so that we divide out all the prime factors.

So, the code works like this:

f=(n,i=1)=>                                           We will increase i by one at the start of
the function body, so default is 1
n-1?                                       Check if n==1.
n%++i?                                 If i isn't, increase i by 1, check if n
is divisible by i
f(n,i):                          if it isn't, we check the next i
(n/=i)%i?                 if it is, divide n by i, and check for
divisibility by i again
-f(n,i):         if it not, then we flip the value to
account for the 1 and -1 depending on the
amount of factors
0:       if it's divisible by i twice, done.
1      if we're at 1, divided out all factors,
then we return 1. The line with
-f(n,i) will then take care of the sign


So, that's my submission.

• Welcome to the site. I don't know js, but I can tell you that here we don't care about specs, only implementations. So if removing ; doesn't break it, it doesn't matter the specs you can remove it. Oct 28 '18 at 14:27
• Good to know! I'll remove it then ;) Oct 29 '18 at 13:13