2 Partitions 545
For many purposes the discussion of convergence is superfluous and Proposition 3
may be regarded simply as a relation between formal products and formal power series.
Euler also obtained an interesting counterpart to Proposition 3, which we will
derive from Jacobi’s triple product formula.
Proposition 4If|x|< 1 ,then
( 1 −x)( 1 −x^2 )( 1 −x^3 )···=
∑
m∈Z
(− 1 )mxm(^3 m+^1 )/^2.
Proof If we takeq=x^3 /^2 andz=−x^1 /^2 in Proposition XII.2, we obtain at once the
result, since
∏
n≥ 1
( 1 −x^3 n)( 1 −x^3 n−^1 )( 1 −x^3 n−^2 )=
∏
k≥ 1
( 1 −xk).
Proposition 4 also has a combinatorial interpretation. The coefficient ofxn(n≥ 1 )
in the power series expansion of
∏
k≥ 1 (^1 −x
k)is
sn=
∑
(− 1 )v,
where the sum is over all partitions ofnintounequalparts andvis the number of parts
in the partition. In other words,
sn=p∗e(n)−p∗o(n),
wherep∗e(n),resp.p∗o(n), is the number of partitions of the positive integerninto an
even, resp. odd, number of unequal parts. On the other hand,
∑
m∈Z
(− 1 )mxm(^3 m+^1 )/^2 = 1 +
∑
m≥ 1
(− 1 )m{xm(^3 m+^1 )/^2 +xm(^3 m−^1 )/^2 }.
Thus Proposition 4 says thatp∗e(n)=p∗o(n)unlessn=m( 3 m± 1 )/2forsomem∈N,
in which casepe∗(n)−po∗(n)=(− 1 )m.
From Propositions 3 and 4 we obtain
[
1 +
∑
m≥ 1
(− 1 )m{xm(^3 m+^1 )/^2 +xm(^3 m−^1 )/^2 }
][
1 +
∑
k≥ 1
p(k)xk
]
= 1.
Multiplying out on the left side and equating to zero the coefficient ofxn(n≥ 1 ),we
obtain the recurrence relation:
p(n)=p(n− 1 )+p(n− 2 )−p(n− 5 )−p(n− 7 )
+···+(− 1 )m−^1 p(n−m( 3 m− 1 )/ 2 )
+(− 1 )m−^1 p(n−m( 3 m+ 1 )/ 2 )+···,
wherep( 0 )=1andp(k)=0fork<0. This recurrence relation is quite an effi-
cient way of calculatingp(n). It was used by MacMahon (1918) to calculatep(n)for
n≤200.