Mertens function

From Rosetta Code
Revision as of 17:37, 25 January 2020 by Thundergnat (talk | contribs) (→‎{{header|Perl 6}}: Clean up some editing debris)
Mertens function is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The Mertens function M(x) is the count of square-free integers up to x that have an even number of prime factors, minus the count of those that have an odd number.

It is an extension of the Möbius function. Given the Möbius function μ(n), the Mertens function M(x) is the sum of the Möbius numbers from n == 1 through n == x.


Task
  • Write a routine (function, procedure, whatever) to find the Mertens number for any positive integer x.
  • Use that routine to find and display here, on this page, at least the first 99 terms in a grid layout. (Not just one long line or column of numbers.)
  • Use that routine to find and display here, on this page, the number of times the Mertens number is equal to zero in the range M(1) through M(1000).
  • Use that routine to find and display here, on this page, the number of times the Mertens number crosses zero in the range M(1) through M(1000). (Crossing defined as this term equal to zero but preceding term not.)


See also

This is not code golf. The stackexchange link is provided as a algorithm reference, not as a guide.


Related tasks


Perl 6

Works with: Rakudo version 2019.11

Mertens number is not defined for n == 0. Perl 6 arrays are indexed from 0 so store a blank value at position zero to keep x and M(x) aligned.

<lang perl6>use Prime::Factor;

sub μ (Int \n) {

   my @p = prime-factors(n);
   +@p == +Bag(@p).keys ?? +@p %% 2 ?? 1 !! -1 !! 0

}

my @mertens = lazy [\+] flat ' ', 1, (2..*).map: -> \n { μ(n) };

put "Mertens sequence - First 199 terms:\n",

   @mertens[^200]».fmt('%3s').rotor(20).join("\n"),
   "\n\nEquals zero ", +@mertens[1..1000].grep( !* ),
   ' times between 1 and 1000', "\n\nCrosses zero ",
   +@mertens[1..1000].kv.grep( {!$^v and @mertens[$^k - 1]} ),
   ' times between 1 and 1000';</lang>
Output:
Mertens sequence - First 199 terms:
      1   0  -1  -1  -2  -1  -2  -2  -2  -1  -2  -2  -3  -2  -1  -1  -2  -2  -3
 -3  -2  -1  -2  -2  -2  -1  -1  -1  -2  -3  -4  -4  -3  -2  -1  -1  -2  -1   0
  0  -1  -2  -3  -3  -3  -2  -3  -3  -3  -3  -2  -2  -3  -3  -2  -2  -1   0  -1
 -1  -2  -1  -1  -1   0  -1  -2  -2  -1  -2  -3  -3  -4  -3  -3  -3  -2  -3  -4
 -4  -4  -3  -4  -4  -3  -2  -1  -1  -2  -2  -1  -1   0   1   2   2   1   1   1
  1   0  -1  -2  -2  -3  -2  -3  -3  -4  -5  -4  -4  -5  -6  -5  -5  -5  -4  -3
 -3  -3  -2  -1  -1  -1  -1  -2  -2  -1  -2  -3  -3  -2  -1  -1  -1  -2  -3  -4
 -4  -3  -2  -1  -1   0   1   1   1   0   0  -1  -1  -1  -2  -1  -1  -2  -1   0
  0   1   1   0   0  -1   0  -1  -1  -1  -2  -2  -2  -3  -4  -4  -4  -3  -2  -3
 -3  -4  -5  -4  -4  -3  -4  -3  -3  -3  -4  -5  -5  -6  -5  -6  -6  -7  -7  -8

Equals zero 92 times between 1 and 1000

Crosses zero 67 times between 1 and 1000