The Universe of Disco

Mon, 05 Mar 2007

An integer partition puzzle
Last month I wrote an article about calculating Yahtzee probabilities and another one about counting permutations in which integer partitions came up. An integer partition of some integer N is an unordered sequence of positive integers that sums to N. For example, there are 5 different integer partitions of 4:

1 1 1 1
2 1 1
2 2
3 1
4
I've spent a lot of time tinkering with partitions since then.

Here's one interesting fact: it's quite easy to calculate the number of partitions of N. Let P(n, k) be the number of partitions of n into parts that are at least k. Then it's easy to see that:

$$P(n, k) = \sum_{i=k}^{n-1} P(n-i, k)$$

And there are simple boundary conditions: P(n, n) = 1; P(n, k) = 0 when k > n, and so forth. And P(n), the number of partitions of n into parts of any size, is just P(n, 1). So a program to calculate P(n) is very simple:

        my @P;
        sub P {
          my ($n, $k) = @_;
          return 0 if $n < 0;
          return 1 if $n == 0;
          return 0 if $k > $n;
          my $r = $P[$n] ||= [];
          return $r->[$k] if defined $r->[$k];
          return $r->[$k] = P($n-$k, $k) + P($n, $k+1);
        }

        sub part {
          P($_[0], 1);
        }

        for (1..100) {
          printf "%3d %10d\n", $_, part($_);
        }

I had a funny conversation once with someone who ought to have known better: I remarked that it was easy to calculate P(n), and disagreed with me, asking why Rademacher's closed-form expression for P(n) had been such a breakthrough. But the two properties are independent; the same is true for lots of stuff. Just because you can calculate something doesn't mean you understand it. Calculating ζ(2) is quick and easy, but it was a major breakthrough when Euler discovered that it was equal to π²/6. Calculating ζ(3) is even quicker and easier, but nobody has any idea what the value represents.

Similarly, P(n) is easy to calculate, but harder to understand. Ramanujan observed, and proved, that P(5k+4) is always a multiple of 5, which had somehow escaped everyone's notice until then. And there are a couple of other similar identities which were proved later: P(7k+5) is always a multiple of 7; P(11k+6) is always a multiple of 11. Based on that information, any idiot could conjecture that P(13k+7) would always be a multiple of 13; this conjecture is wrong. (P(7) = 15.)

Anyway, all that is just leading up to the real point of this note, which is that I was tabulating the number of partitions of n into exactly k parts, which is also quite easy. Let's call this Q(n, k). And I discovered that Q(13, 4) = Q(13, 5). There are 18 ways to divide a pile of 13 beans into 4 piles, and also 18 ways to divide the beans into 5 piles.

1 1 1 10
1 1 2 9
1 1 3 8
1 1 4 7
1 1 5 6
1 2 2 8
1 2 3 7
1 2 4 6
1 2 5 5
1 3 3 6
1 3 4 5
1 4 4 4
2 2 2 7
2 2 3 6
2 2 4 5
2 3 3 5
2 3 4 4
3 3 3 4

1 1 1 1 9
1 1 1 2 8
1 1 1 3 7
1 1 1 4 6
1 1 1 5 5
1 1 2 2 7
1 1 2 3 6
1 1 2 4 5
1 1 3 3 5
1 1 3 4 4
1 2 2 2 6
1 2 2 3 5
1 2 2 4 4
1 2 3 3 4
1 3 3 3 3
2 2 2 2 5
2 2 2 3 4
2 2 3 3 3

The question I'm trying to resolve: is this just a coincidence? Or is there something in the structure of the partitions that would lead us to suspect that Q(13, 4) = Q(13, 5) even if we didn't know the value of either one?

So far, I haven't turned anything up; it seems to be a coincidence. A simpler problem of the same type is that Q(8, 3) = Q(8, 4); that seems to be a coincidence too:

1 1 6
1 2 5
1 3 4
2 2 4
2 3 3

1 1 1 5
1 1 2 4
1 1 3 3
1 2 2 3
2 2 2 2

Looking at this, one can see all sorts of fun correspondences. But on closer inspection, they turn out to be illusory. For example, any partition into 4 parts can be turned into a partition into 3 parts by taking the smallest of the 4 parts, dividing it up into 1's, and distributing the extra 1's to the largest parts. But there's no reason why that should always yield different outputs for different inputs, and, indeed, it doesn't.

Oh well, sometimes these things don't work out the way you'd like.

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