proofs

208 Identities versus bijections
thought should convince you that this is just the number of ways to write n
as a sum
n = n 1 · 1 + n 2 · 2 + n 3 · 3 + . . .
= 1 + . . . + 1 + 2 + . . . + 2 + 3 + . . . + 3 + . . . .
} {{ } } {{ } } {{ }
n 1 n 2 n 3
So the coefficient is nothing else but the number p(n) of partitions of n.
Since the geometric series 1+x k +x 2k 1
+. . . equals
1−x
, we have proved
k
our first identity:
∏ 1
1 − x k = ∑ p(n)x n . (4)
k≥1
n≥0
6 = 5 + 1
6 = 3 + 3
6 = 3 + 1 + 1 + 1
6 = 1 + 1 + 1 + 1 + 1 + 1
Partitions of 6 into odd parts: p o(6) = 4
7 = 7
7 = 5 + 1 + 1
7 = 3 + 3 + 1
7 = 3 + 1 + 1 + 1 + 1
7 = 1 + 1 + 1 + 1 + 1 + 1 + 1
7 = 7
7 = 6 + 1
7 = 5 + 2
7 = 4 + 3
7 = 4 + 2 + 1.
The partitions of 7 into odd resp. distinct
parts: p o(7) = p d (7) = 5.
1
What’s more, we see from our analysis that the factor
1−x
accounts for
k
1
the contribution of k to a partition of n. Thus, if we leave out
1−x
from k
the product on the left side of (4), then k does not appear in any partition
on the right side. As an example we immediately obtain
∏
i≥1
1
1 − x 2i−1 = ∑ n≥0
p o (n)x n , (5)
where p o (n) is the number of partitions of n all of whose summands are
odd, and the analogous statement holds when all summands are even.
By
∏
now it should be clear what the n-th coefficient in the infinite product
k≥1 (1 + xk ) will be. Since we take from any factor in (3) either 1 or x k ,
this means that we consider only those partitions where any summand k
appears at most once. In other words, our original product (1) is expanded
into
∏
+ x
k≥1(1 k ) = ∑ p d (n)x n , (6)
n≥0
where p d (n) is the number of partitions of n into distinct summands.
Now the method of formal series displays its full power. Since 1 − x 2 =
(1 − x)(1 + x) we may write
∏
k≥1(1 + x k ) = ∏ k≥1
1 − x 2k
1 − x k = ∏ 1
1 − x 2k−1
k≥1
since all factors 1 − x 2i with even exponent cancel out. So, the infinite
products in (5) and (6) are the same, and hence also the series, and we
obtain the beautiful result
p o (n) = p d (n) for all n ≥ 0. (7)
Such a striking equality demands a simple proof by bijection — at least that
is the point of view of any combinatorialist.