# calculating Möbius function

I wonder if there is any efficient way to calculate Möbius function for a array of number 1:1000000

http://en.wikipedia.org/wiki/M%C3%B6bius_function

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## 4 Answers

Here are a few papers that might be helpful.

Shallit and Shamir, Number-theoretic functions which are equivalent to number of divisors, Inform. Process. Lett. 20 (1985), no. 3, 151–153, MR0801982 (86k:11076). The review, by Hale Trotter, says $\mu(n)$ can be calculated from a single value $d(n^q)$ where $d$ is the divisor function and $q$ is a prime greater than $1+\log_2n$.

Lioen and van de Lune, Systematic computations on Mertens' conjecture and Dirichlet's divisor problem by vectorized sieving, in From Universal Morphisms to Megabytes: a Baayen Space Odyssey, 421–432, Math. Centrum, Centrum Wisk. Inform., Amsterdam, 1994, MR1490603 (98j:11125). The review, by Jonathan P. Sorenson, says the authors present sieving algorithms for computing $\mu(n)$.

Herman te Riele, Computational sieving applied to some classical number-theoretic problems, in Number Theory in Progress, Vol. 2 (Zakopane-Kościelisko, 1997), 1071–1080, de Gruyter, Berlin, 1999, MR1689561 (2000e:11119). The review, by Marc Deléglise, starts, "This paper is a survey about different applications of sieving in number theory. Finding all the primes belonging to a given interval and computing all the values of the Möbius function on a given interval are obvious examples."

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The key phrase here is for an array. Then try the Sieve of Eratosthenes:

• initialize an array of length N of 1's
• for each prime $p < \sqrt{N}$ iterate over the array:

• set each $x \equiv 0 \, (\mod p\,)$ to $-1$
• set each $x \equiv 0 \, (\mod p^2)$ to $\;\,0$

We stop at power $2$ since we are just looking for squarefree (OEIS: A005117).

This is certainly not the fastest but maybe easiest to implement. I downloaded a list of primes off the internet (or you can generate or find your own sieve).

import numpy as np

L = np.ones(100000).astype(int)

for p in P:
L[::p]    *= -1
L[::p**2] *=  0


The output looks good (See also OEIS:A008683)

0,  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


### The Sieve

N = np.ones(100000).astype(int)
N[:2] = 0

P = []

p = 1
while p < np.sqrt(100000):

p = np.argmax(N)
N[::p] = 0
P += [p]

P = np.hstack((P, np.where(N > 0)[0]))
P[100:105]


This is a sample implementation. The results, they look correct:

547, 557, 563, 569, 571

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I think you want to not just 'replace 1 by -1' but just 'multiply each entry $\equiv 0\pmod p$ by -1'; your code itself gets it right, but the description suggests that once a value has been set to -1 it can't be flipped back. –  Steven Stadnicki Mar 19 at 1:05
@StevenStadnicki just noticed I have reproduce Section 3 in Systematic computations on Mertens' conjecture and Dirichlet's divisor problem by vectorized sieving referred to in the checked answer. –  john mangual Mar 19 at 2:18

You don't need the bigger machinery of a segmented sieve for such a small range. Here is a simple $O(n\log\log n)$ algorithm to calculate all $\mu(i)$ up to $n$ based on the Sieve of Eratosthenes. In fact, the sieve does fully factor all the square-free numbers; it just doesn't do it one at a time.

Depending on your computer, this approach is practical up to around $2^{30}$ at which point you need to start using higher-precision arithmetic and computing a range of $\mu(i)$ values in batches.

public static int[] GetMu(int max)
{
var sqrt = (int)Math.Floor(Math.Sqrt(max));
var mu = new int[max + 1];
for (int i = 1; i <= max; i++)
mu[i] = 1;
for (int i = 2; i <= sqrt; i++)
{
if (mu[i] == 1)
{
for (int j = i; j <= max; j += i)
mu[j] *= -i;
for (int j = i * i; j <= max; j += i * i)
mu[j] = 0;
}
}
for (int i = 2; i <= max; i++)
{
if (mu[i] == i)
mu[i] = 1;
else if (mu[i] == -i)
mu[i] = -1;
else if (mu[i] < 0)
mu[i] = 1;
else if (mu[i] > 0)
mu[i] = -1;
}
return mu;
}


Running GetMu(1000000) takes about 10 msec on my computer.

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Doing it the second-most-stupid-way, as described in my answer with Mathematica takes 1sec. I did not try it using Gerhard's extra-stupid way, but it is safe to say that anything works in this range. –  Igor Rivin Jun 14 '12 at 2:04
So, how many different spellings of Eratosthenes can we accumulate in this thread? –  Gerry Myerson Jun 14 '12 at 5:21
@Gerry: Thanks. Fixed here ... and in all my other documents. –  Rick Sladkey Jun 14 '12 at 17:42
In Rick Sladkey's answer, it seems to me that after running through the sieve, mu[i] should be 0 or +/- i. So why have the extra conditions: else if (mu[i] < 0) mu[i] = 1; else if (mu[i] > 0) mu[i] = -1; ?? –  Steve Robbins Mar 23 '14 at 4:33
It's not true that all the prime factors of a number are less than its square root. But, if there is such a factor, there can only be one. –  Rick Sladkey Mar 23 '14 at 14:55

For range 1:1000000 the stupidest possible algorithm (factor each integer, check if it's square free, raise -1 to the right power) will terminate in a blink of an eye. For much larger ranges, an obvious modifican of the sieve of erathsosphenes will be very fast ( just remember to zero every $p$th number at the $p$th step.

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I want to avoid factoring it. I'm also considering Sieve of Eratosthenes –  zimbra314 Jun 13 '12 at 16:53
Calling any algorithm the "stupidest possible" underestimates the power of human ingenuity.... –  Greg Martin Jun 13 '12 at 17:34
@Greg: I double dog dare ya :) –  Igor Rivin Jun 13 '12 at 17:39
For calclulating mu(n) when n>1, pick a prime p at random. If n/p is an integer, use (-1) times mu(n/p), taking care to check if p divides n/p. Otherwise go back and try again. There are stupider variations of course. Gerhard "Dare You Double-Dog Dare Me?" Paseman, 2012.06.13 –  Gerhard Paseman Jun 13 '12 at 18:03
Is erathsosphenes what you get when you cross Eratosthenes with Aristophanes? –  Gerry Myerson Jun 13 '12 at 23:14