First of all powerset is wrong cause equal factors will produce duplicates, e.g.

P{2,2} = { (,),(2,),(,2),(2,2)}

Then I doubt you can multiply in a way which produces all divisors in a sorted way, so final sorting can't be avoided.

I'd try a recursive/iterative approach,

  1. initialize the result set D with 1
  2. shift the smallest prime p' from P
  3. multiply all divisors in the temporary result set D with p' to get D'
  4. push the new divisors D' into D
  5. continue with step 2 till P empty
  6. sort D
but note you'll need a test if p' and p'' are duplicates to avoid the aforementioned problem! :)

update

looks good! :)

use warnings;
use strict;

my @P = qw/2 2 3 5 11 277412413/;
@P= sort { $a<=>$b } @P;              # sort to be sure

my @D     = (1);
my @D_new = ();
my $p_old = 0;

for my $p (@P) {
    if ( $p == $p_old ) {
        @D_new = map { $_* $p } @D_new;
    } else {
        @D_new = map { $_* $p } @D;
    }
    push @D,@D_new;
    $p_old = $p;
}

@D= sort { $a<=>$b } @D;

print "Factors @P\n";
print "Divisors: @D\n";

[download]

out 

Factors 2 2 3 5 11 277412413
Divisors: 1 2 3 4 5 6 10 11 12 15 20 22 30 33 44 55 60 66 110 132 165 
+220 330 660 277412413 554824826 832237239 1109649652 1387062065 16644
+74478 2774124130 3051536543 3328948956 4161186195 5548248260 61030730
+86 8322372390 9154609629 12206146172 15257682715 16644744780 18309219
+258 30515365430 36618438516 45773048145 61030730860 91546096290 18309
+2192580

[download]

Cheers Rolf

(addicted to the Perl Programming Language and ☆☆☆☆ :)

In reply to Re: Finding divisors from factors by LanX
in thread Finding divisors from factors by danaj

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