It seemed wasteful to me to first create all partitions and then filter out the wanted ones. And I wanted to say that it should be simple to draw a partition algorithm that directly does what you want. But it took me longer than I thought...
use strict;
use warnings;
sub partition {
my( $n, $level, $max, @part ) = @_;
$max //= $n;
$level //= $n;
my @solutions = $n <= $max ? [ @part, $n ]: ();
$max = $n-1 if $max > $n-1;
push @solutions, map { partition( $n-$_, $level-1, $_-1, @part, $_)
+} reverse 1..$max if $level > 1 and $n > 2;
return @solutions;
}
my( $number, $depth ) = @ARGV;
my @s = partition $number, $depth;
print "@$_\n" for @s;
Update: Changed 0..$max to 1..$max.
Update 2: You are not doing this to solve Kakuro puzzles?