Integer Partitions and the Twelvefold Way — Examples

The partitions of $6$, grouped by largest part, are

$6,$

$5+1,$

$4+2,4+1+1,$

$3+3,3+2+1,3+1+1+1,$

$2+2+2,2+2+1+1,2+1+1+1+1,$

$1+1+1+1+1+1.$

So $p(6)=11$.

For the partitions of $6$, conjugation gives the pairings

$6\leftrightarrow 1+1+1+1+1+1,$

$5+1\leftrightarrow 2+1+1+1+1,$

$4+2\leftrightarrow 2+2+1+1,$

$4+1+1\leftrightarrow 3+1+1+1,$

$3+3\leftrightarrow 2+2+2,$

and

$3+2+1\leftrightarrow 3+2+1.$

The fixed partition $3+2+1$ is self-conjugate.

The partition

$\lambda =(5,3,3,1,1)$

is self-conjugate. Its diagonal cells are $(1,1)$, $(2,2)$, and $(3,3)$, and their hook lengths are

$9,3,1.$

Thus the diagonal-hook construction sends $\lambda$ to the distinct odd partition

$9+3+1.$

Start with the odd-part partition

$5+5+3+1+1.$

Its multiplicities are

$2=2^1\text{ for the part }5,1=2^0\text{ for the part }3,2=2^1\text{ for the part }1.$

Glaisher's rule therefore replaces the two $5$'s by one $10$, keeps the $3$, and replaces the two $1$'s by one $2$. The image is the distinct-part partition

$10+3+2.$

For

$\lambda =(5,3,3,2,1),$

the Durfee length is $3$, so the upper-left $3\times 3$ square is the Durfee square.

The cells to the right of that square form the partition

$\alpha =(2),$

and the cells below it form the partition

$\beta =(2,1)\mathrm{.}$

Thus $\lambda$ is rebuilt from the $3\times 3$ square together with one partition having at most $3$ parts and one partition with parts at most $3$.

The generalized pentagonal offsets not exceeding $10$ are

$1,2,5,7.$

So the recurrence gives

$p(10)=p(9)+p(8)-p(5)-p(3)\mathrm{.}$

Using

$p(9)=30,p(8)=22,p(5)=7,p(3)=3,$

one gets

$p(10)=30+22-7-3=42.$

For $n=3$ labelled balls and $k=2$ labelled boxes:

• arbitrary maps: $2^3=8$;
• injective maps: $0$, because $3>2$;
• surjective maps: $2!S(3,2)=2\cdot 3=6$.

So the labelled-to-labelled row is $(8,0,6)$.

For $3$ indistinguishable balls and $2$ labelled boxes, the occupancy vectors are:

• arbitrary: $(3,0)$, $(2,1)$, $(1,2)$, $(0,3)$, so the count is $4$;
• injective: none, because a $0$-$1$ vector of length $2$ cannot sum to $3$;
• surjective: $(2,1)$ and $(1,2)$, so the count is $2$.

Thus the unlabelled-to-labelled row is $(4,0,2)$.

For $3$ labelled balls and $2$ unlabelled boxes:

• arbitrary placements are set partitions of $\{1,2,3\}$ into at most two blocks, namely one one-block partition and three two-block partitions, so the count is $4$;
• injective placements do not exist, because three singleton fibres would use three boxes but only two are available;
• surjective placements are exactly the three two-block set partitions.

So the labelled-to-unlabelled row is $(4,0,3)$.

For $3$ indistinguishable balls and $2$ indistinguishable boxes:

• arbitrary placements correspond to the partitions $3$ and $2+1$, so the count is $2$;
• injective placements do not exist because $3>2$;
• surjective placements correspond only to $2+1$, so the count is $1$.

Thus the unlabelled-to-unlabelled row is $(2,0,1)$.

Applying the recurrence once gives

$p_3(5)=p_2(4)+p_3(2)=2+0=2.$

The two partitions are

$3+1+1,2+2+1.$

The partitions of $7$ into distinct parts are

$7,6+1,5+2,4+3,4+2+1.$

The partitions of $7$ into odd parts are

$7,5+1+1,3+3+1,3+1+1+1+1,1+1+1+1+1+1+1.$

Glaisher's map pairs them as

$7\leftrightarrow 7,5+1+1\leftrightarrow 5+2,3+3+1\leftrightarrow 6+1,$

$3+1+1+1+1\leftrightarrow 4+3,1+1+1+1+1+1+1\leftrightarrow 4+2+1.$

The partition

$(4,2,1)$

is self-conjugate.

Conjugating the distinct-part partition

$(4,2,1)$

produces a partition into odd parts.

One may compute the coefficient of $x^3$ in the Euler product by the naive rule

$[x^3]{\prod }_{m\ge 1}(1-x^m{)}^{-1}={\sum }_{m\ge 1}[x^3](1+x^m+x^{2m}+\cdots )\mathrm{.}$