TL;DR

On with TASK #2 from The Weekly Challenge #159. Enjoy!

The challenge

You are given a positive number $n. Write a script to generate the Moebius Number for the given number. Please refer to wikipedia page for more informations. Example 1: Input:$n = 5
Output: -1


Example 2:

Input: $n = 10 Output: 1  Example 3: Input:$n = 20
Output: 0


The questions

This challenge is a sibling of PWC150 - Square-free Integer, with the exception that this time we’re told that we have to consider only positive numbers. I mean… integer numbers, right?

The solution

We’re adapting the solution from PWC150 - Square-free Integer, just to remain on the lazy side:

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';

say $_, ' ', moebius_number($_) for (@ARGV ? @ARGV : (1 .. 10));

sub moebius_number ($n) { return 0 unless$n % 4;
($n, my$n_divisors) = $n % 2 ? ($n, 0) : ($n / 2, 1); my$divisor = 3;
while ($n >=$divisor) {
if ($n %$divisor == 0) {
++$n_divisors;$n /= $divisor; return 0 unless$n % $divisor; }$divisor += 2; # go through odd candidates only
}
return 1 - 2 * ($n_divisors % 2); }  And the same goes for Raku: #!/usr/bin/env raku use v6; sub MAIN (*@args) { put$_, ' ', möbius-number($_) for @args} sub möbius-number ($n is copy) {
return 0 if $n %% 4; ($n, my $n-divisors) =$n %% 2 ?? (($n / 2).Int, 1) !! ($n, 0);
my $divisor = 3; while$n >= $divisor { if$n %% $divisor { ++$n-divisors;
$n = ($n / $divisor).Int; return 0 if$n %% $divisor; }$divisor += 2; # go through odd candidates only
}
return 1 - 2 * ($n-divisors % 2); }  The case for divisibility by 2 is handled specially just to be able to increment the $divisor variable by 2 instead of by 1. Apart from this, we’re sticking to the definition in the Wikipedia page, returning 0 whenever we find a square in the divisors, or using the even/odd count of prime factors to figure out the right value to return.

Stay safe folks!