# plane partitions

Christian G. Bower bowerc at usa.net
Fri Jan 2 07:30:34 CET 2004

```Wouter:

> since it is well known that the plane partitions
> are counted by the generating function
> GF: Prod(k=1..oo, 1/(1-x^k)^k),

> the small step for mankind is:
> GF: Prod(k=1..oo, 1/(1-q x^k)^k)
snip
> {1},
> {2, 1},
> {3, 2, 1},
> {4, 6, 2, 1},
> {5, 10, 6, 2, 1},
> {6, 19, 14, 6, 2, 1},
> {7, 28, 28, 14, 6, 2, 1}

I've been working on an interpretation for those two triangles and this
is my best so far:

We can describe A000219 (planar partitions) as follows:

I'll define a Q-partition of a number n as a set of ordered pairs:
(a_1,b_1), (a_2,b_2), ...
such that the a's add up to n and each b satisfies: 1 <= b_k <= a_k
then the number of Q-partitions of n is the same as the number of planar
partitions. E.g. the 6 Q-partitions of 3 are:
(3,1)
(3,2)
(3,3)
(2,1),(1,1)
(2,2),(1,1)
(1,1),(1,1),(1,1)

Wouter's triangle above gives the number of Q-partitions of n into k
parts and Wouter's triangle below:

> Table[CoefficientList[ Coefficient[
> Series[Product[Product[ 1/(1 - q^j x^k), {j, k}], {k, 0, n}],
>  {x, 0, n}], x^n]/q, q], {n, 12}]

> { 1},
> { 1, 2},
> { 1, 2, 3},
> { 1, 3, 4, 5},
> { 1, 3, 6, 7, 7},
> { 1, 4, 8, 12, 12, 11},
> { 1, 4, 10, 16, 21, 19, 15},
> { 1, 5, 12, 23, 31, 36, 30, 22},
> { 1, 5, 15, 28, 45, 55, 58, 45, 30},
> { 1, 6, 17, 37, 60, 84, 94, 92, 67, 42},
> { 1, 6, 20, 44, 80, 115, 147, 153, 140, 97, 56},
> { 1, 7, 23, 55, 101, 161, 211, 249, 244, 211, 139, 77}

gives the number of Q-partitions whose b-values add up to k.

So the missing piece of the puzzle is to map between Q-partitions and
planar partitions. I'm not there yet. If anyone knows a good mapping

What I have so far is that if I restrict the planar partitions to two
rows, I get the Euler transform of 1 2 2 2 2 ... = A000990

I've been working out a scheme to match these to the corresponding
Q-partitions (those whose b-value is at most 2) and the method I have is:

place the a-value of any term with b-value 1 in the top row.  Place the
a-value of any term with b-value 2 in the bottom row if it can be done
legally (i.e. the value above it is >=) if not, put a-1 in the top row
and 1 in the bottom row.

I haven't verified that this is correct and I haven't figured out how to
do it in reverse. In any case it appears that the triangles would
correspond to non-intuitive properties of the plane partitions. E.g. the
A000990 version of Wouter's first triangle

1
2 1
2 2 1
2 5 2 1
2 6 5 2 1
2 9 10 5 2 1
2 10 15 10 5 2 1

would correspond to the number of <=2 row plane partitions of n where the
number of parts in the top row + the number of non-one parts in the
bottom row = k if my algorithm is correct.

Christian

P.S.

I suspect that the number of plane partitions with at most k rows is Euler
transform of 1 2 3 ... k-1 k k k k ...

Some searching in the EIS supports this conjecture.

Cf A000991, A002799, A001452

```