contribute a factor of:
(1 + x
2
)(1 + x
4
)(1 + x
6
)· · ·
to the overall generating function.
The odd parts can be treated in the usual fashion. One contributes a factor of
1 + x + x
2 + · · · =
1
1 − x
.
Three contributes a factor of
1 + x
3 + x
6 + x
9 + · · · =
1
1 − x
3
.
And so on with the odd parts.
Thus the generating function for the number of partitions of a number into parts which have
at most one of each distinct even part is:
(1 + x
2
)(1 + x
4
)(1 + x
6
)· · ·
(1 − x)(1 − x
3)(1 − x
5)· · ·
.
The generating function for the number of partitions of a number in which each part can appear
at most three times is:
(1 + x + x
2 + x
3
)(1 + x
2 + x
4 + x
6
)(1 + x
3 + x
6 + x
9
)· · · =
(1 − x
4
)(1 − x
8
)(1 − x
12)· · ·
(1 − x)(1 − x
2)(1 − x
3)· · ·
.
We now wish to show that these two are equal, i.e.:
(1 + x
2
)(1 + x
4
)· · ·
(1 − x)(1 − x
3)· · ·
=
(1 − x
4
)(1 − x
8
)(1 − x
12)· · ·
(1 − x)(1 − x
2)(1 − x
3)· · ·
.
Common to the denominator on both sides is (1 − x)(1 − x
3
)· · · so we can cancel that out:
(1 + x
2
)(1 + x
4
)(1 + x
6
)· · · =
(1 − x
4
)(1 − x
8
)(1 − x
12)· · ·
(1 − x
2)(1 − x
4)(1 − x
6)· · ·
Now, using the fact that a
2 − b
2 = (a + b)(a − b) the (1 + x
2
)(1 + x
4
)(1 + x
6
)· · · cancels on the
right hand side. Thus we are left with:
(1 + x
2
)(1 + x
4
)(1 + x
6
)· · · = (1 + x
2
)(1 + x
4
)(1 + x
6
)· · · .
This is indeed true so we are done.