Here is a slightly simpler iterator than

jethro's, but surely it's doing the same thing. It avoids the recursion blowup of

JavaFan.

`sub iterator {
return unless @_;
my @p = ($_[0]);
return sub {
## collect all the trailing 2s, and last trailing 3 if present
my $temp = 0;
$temp += pop @p while @p and $p[-1] == 2;
$temp += pop @p if @p and $p[-1] == 3; ## updated
return if ! @p;
## reduce the last guy by 1, avoid total of 1 leftover
$p[-1]--, $temp++;
$p[-1]--, $temp++ if $temp == 1;
## redistribute collected amount, as large as possible
## (largest increments can be $p[-1])
if ($temp % $p[-1] == 0) {
push @p, ($p[-1]) x ($temp/$p[-1]);
## special case to avoid 1 leftover
} elsif ( $temp % $p[-1] == 1 ) {
my $m = int ($temp/$p[-1]) -1;
push @p, ($p[-1]) x $m, $p[-1]-1, 2;
} else {
my $m = int ($temp/$p[-1]) ;
push @p, ($p[-1]) x $m, ($temp - $p[-1]*$m);
}
@p;
}
}
my $iter = iterator(shift || 50);
while (my @part = $iter->()) {
print "@part\n";
}
`

Like

jethro's, there could be some efficiencies gained by keeping track of some pointers, but the main problem is that there are just so many partitions.
So I think the problem statement should be clarified, especially if 10^6 is involved.

**Update:** updated the marked line according to BrowserUk's suggestion. (added "`@p and`").

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