This can be expanded into a prime sieve that shows what he has here.

  #prime.pl

  use Math::Complex;
  use Math::Trig;
  use POSIX;
  # P = A = 1/2 B (6 n-cos(pi n)+3) \ B = 1/2 (6 n-cos(pi n)+3)

  my $pi = pi;

  my @k;
  my $count = "100";
  my $n = "1";
  my @c;

  ### given two sets A, B returns B \ A (ie all elements that are in B but NOT in A)
  sub complement
  {
      my @set1 = @c;
      my @set2 = @k;
      my @ans=();    

      ## find intersection of A and B
      my @intersection=(@set1);

      for(my $i=0;$i<@set2;$i++)
      {
         for(my $j=0;$j<@intersection;$j++)
         {
            if($set2[$i] eq $intersection[$j])
            {
               $set2[$i]="?";
            }
         }
      }

      for(my $i=0;$i<@set2;$i++)
      {
         push(@ans,$set2[$i]) if $set2[$i]!~m/\?/;
      }
      return @ans;
  }

  for ($count = 100000 ;$count >= $n; $n++){
  	my $k = ((3 * $n) - (0.5 * cos($pi*$n))) + 1.5;
  	push (@k, $k);
  }

  my $c; 
  my $x = $#k;
  my $y = $k[$x];

  foreach (@k){
  	my $b = $_;
  	my $n = "1";
  	$c=1;
  	next if ($b >= floor(sqrt($y + 1)));
  	for ($count = $y + 1 ;$count >= $c; $n++){
  		$c = (((3 * $n) - (0.5 * cos($pi*$n))) + 1.5) * $b;
  		next unless ($c <= $y + 1);
  		push (@c, $c);
  	}
  }

  print "stophere";

  %hashTemp = map { $_ => 1 } @c;
  @c = keys %hashTemp;
  @c = sort { $a <=> $b } @c;

  #@P = &complement;

  print "stophere";