Compute the Mertens function

Question

Given a positive integer n, compute the value of the Mertens function \$M(n)\$ where:

$$M(n) = \Sigma_{k=1}^{n}\mu(k)$$

and \$\mu(k)\$ is the Möbius function where \$μ(k) = 1\$ if \$k\$ has an even number of distinct prime factors, \$-1\$ if \$k\$ has an odd number of distinct prime factors, and \$0\$ if the prime factors are not distinct.

• This is code-golf so create the shortest code for a function or program that computes the Mertens function for an input integer n > 0.
• This is the OEIS sequence A002321.

Test Cases

n M(n)
1 1
2 0
3 -1
4 -1
5 -2
6 -1
7 -2
8 -2
9 -2
10 -1
117 -5
5525 5
7044 -25
8888 4
10000 -23

Answer 1

Mathematica, 22 20 bytes

Thanks to @miles for saving 2 bytes.

Tr@*MoebiusMu@*Range

Explanation

Range

Generate a list from 1 to input.

MoebiusMu

Find MoebiusMu of each number

Tr

Sum the result.

Answer 2

Python 2, 45 37 bytes

f=lambda n,k=2:n<k or f(n,k+1)-f(n/k)

Test it on Ideone.

Background

This uses the property

from A002321, which leads to the following recursive formula.

How it works

We use recursion not only to compute M for the quotients, but to compute the sum of those images as well. This saves 8 bytes over the following, straightforward implementation.

M=lambda n:1-sum(M(n/k)for k in range(2,n+1))

When f is called with a single argument n, the optional argument k defaults to 2.

If n = 1, n<k yields True and f returns this value. This is our base case.

If n > 1, n<k initially returns False and the code following or is executed. f(n/k) recursively computes one term of the sum, which is subtracted from the return value of f(n,k+1). The latter increments k and recursively calls f, thus iterating over the possible values of k. Once n < k + 1 or n = 1, f(n,k+1) will return 1, ending the recursion.

Answer 3

05AB1E, 16 15 bytes

LÒvX(ygmyyÙïQ*O

Explanation

L        # range [1 .. n]
Ò        # list of prime factors for each in list
v        # for each prime factor list
 X(ygm   # (-1)^len(factors)
 yyÙïQ*  # multiplied by factors == (unique factors)
 O       # sum

Try it online!

Answer 4

Brachylog, 22 20 bytes

yb:1a+
$p#dl:_1r^|,0

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Explanation

yb                 The list [1, 2, …, Input]
  :1a              Apply predicate 1 (second line) to each element
     +             Sum the resulting list


    $p#d               All elements of the list of prime factors of the Input are distinct
        l:_1r^         Output = (-1)^(<length of the list of prime factors>)
|                  Or
    ,0                 Output = 0

Answer 5

Jelly, 6 bytes

:Ḋ߀SC

Try it online! or verify the smaller test cases. (takes a while)

Background

This uses the property

from A002321, which leads to the following recursive formula.

How it works

:Ḋ߀SC  Main link. Argument: n

 Ḋ      Dequeue; yield [2, ..., n].
:       Perform the integer division of n by each k in [2, ..., n].
  ߀    Recursively call the main link on each result.
    S   Sum; add the results from the recursive calls.
     C  Complement; map the sum r to 1 - r.

Answer 6

Jelly, 9 bytes

RÆFỊNP€FS

Try it online! or verify all test cases.

How it works

RÆFỊNP€FS  Main link. Argument: n

R          Range; yield [1, ..., n].
 ÆF        Factor; decompose each integer in that range into prime-exponent pairs.
   Ị       Insignificant; yield 1 for argument 1, 0 for all others.
    N      Negative; map n to -n.
           This maps primes to 0, exponent 1 to -1, and all other exponents to 0.
     P€    Reduce the columns of the resulting 2D arrays by multiplication.
           The product of the prime values will always be 0; the product of the
           exponent values is 0 if any exponent is greater than, 1 if there is an
           even number of them, -1 is there is an odd number of them.
       FS  Flatten and sum, computing the sum of µ(k) for k in [1, ..., n].

Answer 7

Haskell, 29 27 bytes

f n=1-sum(f.div n<$>[2..n])

Answer 8

Jelly, 7 bytes

Ị*%ðþÆḊ

Not very efficient; determinants are hard.

Try it online! or verify the smaller test cases. (takes a while)

Background

This uses a formula from A002321:

M(n) is the determinant of the Boolean matrix A_n×n, where a_i,j is 1 if j = 1 or i | j, and 0 otherwise.

How it works

Ị*%ðþÆḊ  Main link. Argument: n

   ð     Combine the preceding atoms into a chain (unknown arity).
         Begin a new, dyadic chain with arguments a and b.
Ị        Insignificant; return 1 iff a = 1.
  %      Compute a % b.
 *       Compute (a == 1) ** (a % b).
         This yields 1 if a = 1, or if a ≠ 1 and a % b = 0; otherwise, it yields 0.
    þ    Table; construct the matrix A by calling the defined chain for every pair
         of integers in [1, ..., n].
     ÆḊ  Compute the determinant of the resulting matrix.

Answer 9

PHP, 113 bytes

for(;$i=$argv[1]--;){for($n=$j=1;$j++<$i;)if(!($i%$j)){$i/=$j;$n++;if(!($i%$j))continue 2;}$a+=$n%2?1:-1;}echo$a;

As far as I know php lacks anything like prime number functionality so this is kind of a pain. It's probably possible to do better.

use like:

 php -r "for(;$i=$argv[1]--;){for($n=$j=1;$j++<$i;)if(!($i%$j)){$i/=$j;$n++;if(!($i%$j))continue 2;}$a+=$n%2?1:-1;}echo$a;" 10000

Answer 10

Racket 103 bytes

(λ(N)(for/sum((n(range 1 N)))(define c(length(factorize n)))(cond[(= 0 c)0][(even? c)1][(odd? c)-1])))

Ungolfed:

(define f
  (λ(N)
    (for/sum ((n (range 1 N)))
      (define c (length (factorize n)))
      (cond
        [(= 0 c) 0]
        [(even? c) 1]
        [(odd? c) -1]))))

Answer 11

CJam (20 bytes)

qiM{_,:)(@@f/{j-}/}j

Online demo

Uses the formula from OEIS

sum(k = 1..n, a([n/k])) = 1. - David W. Wilson, Feb 27 2012

and CJam's memoising operator j.

Dissection

qi       e# Read stdin as an integer
M{       e# Memoise with no base cases
         e#   Memoised function: stack contains n
  _,:)(  e#   Basic manipulations to give n [2 .. n] 1
  @@f/   e#   More basic manipulations to give 1 [n/2 ... n/n]
  {j-}/  e#   For each element of the array, make a memoised recursive call and subtract
}j

Answer 12

JavaScript (ES6), 50 bytes

n=>[1,...Array(n-1)].reduce((r,_,i)=>r-f(n/++i|0))

Port of @Dennis's Python answer.

Answer 13

Julia, 26 25 bytes

!n=1-sum(map(!,n÷(2:n)))

Try it online!

Background

This uses the property

from A002321, which leads to the following recursive formula.

How it works

We redefine the unary operator ! for our purposes.

n÷(2:n) computes all required quotients, our redefined ! is mapped over them, and finally the sum of all recursive calls is subtracted from 1.

Unfortunately,

!n=1-sum(!,n÷(2:n))

does not work since dyadic sum will choke on an empty collection.

!n=n<2||1-sum(!,n÷(2:n))

fixes this, but it doesn't save any bytes and returns True for input 1.

Answer 14

C, 51 50 47 bytes

f(n,t,u){for(t=u=1;n/++u;t-=f(n/u));return t;}

Edit: Thanks to @Dennis for -3 bytes!

Answer 15

PARI/GP, 24 bytes

n->sum(x=1,n,moebius(x))

Answer 16

Scala, 53 bytes

def?(n:Int,k:Int=2):Int=if(n<k)1 else?(n,k+1)- ?(n/k)

A port of Dennis's pythin answer.

I've called the method ?, which is a token that doesn't stick to letters.

Answer 17

Pyth, 12 bytes

Defines a function y that takes in the n.

L-1syM/LbtSb

Test suite here. (Note that the trailing y here is to actually call the declared function.)

Answer 18

Actually, 18 17 16 bytes

Golfing suggestions welcome. Try it online!

R`;y;l0~ⁿ)π=*`MΣ

Ungolfing

         Implicit input n.
R        Push the range [1..n].
`...`M   Map the following function over the range. Variable k.
  ;        Duplicate k.
  y        Push the distinct prime factors of k. Call it dpf.
  ;        Duplicate dpf.
  l        Push len(dpf).
  0~       Push -1.
  ⁿ        Push (-1)**len(dpf).
  )        Move (-1)**len(dpf) to BOS. Stack: dpf, k, (-1)**len(dpf)
  π        Push product(dpf).
  =        Check if this product is equal to k.
            If so, then k is squarefree.
  *        Multiply (k is squarefree) * (-1)**(length).
            If k is NOT squarefree, then 0.
            Else if length is odd, then -1.
            Else if length is even, then 1.
           This function is equivalent to the Möbius function.
Σ        Sum the results of the map.
         Implicit return.

Answer 19

J, 19 bytes

1#.1*/@:-@~:@q:@+i.

Computes the Mertens function on n using the sum of the Möbius function over the range [1, n].

Usage

   f =: 1#.1*/@:-@~:@q:@+i.
   (,.f"0) 1 2 3 4 5 6 7 8 9 10 117 5525 7044 8888 10000
    1   1
    2   0
    3  _1
    4  _1
    5  _2
    6  _1
    7  _2
    8  _2
    9  _2
   10  _1
  117  _5
 5525   5
 7044 _25
 8888   4
10000 _23

Explanation

1#.1*/@:-@~:@q:@+i.  Input: integer n
                 i.  Range [0, 1, ..., n-1]
   1            +    Add 1 to each
             q:@     Get the prime factors of each
          ~:@        Sieve mask of each, 1s at the first occurrence
                     of a value and 0 elsewhere
        -@           Negate
    */@:             Reduce each using multiplication to get the product
1#.                  Convert that to decimal from a list of base-1 digits
                     Equivalent to getting the sum

Answer 20

Factor + math.extras math.unicode, 27 bytes

[ [1,b] [ mobius ] map Σ ]

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