Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I wonder if the function $(1+y)(1+y^2)(1+y^3)(1+y^4)\cdots, 0< y<1$, converges to some well-known function.

If we let $ (1+y)(1+y^2)(1+y^3)(1+y^4)\cdots = \prod_{i=1}^\infty (1+y^i) = \sum_{i=0}^\infty a_i y^i$ then $a_i$ satisfies the following relation: $$a_0=1, i > 0$$ $a_i$ is the the number of partitions $(b_1, \cdots , b_s) $ of natural number $i$ such that $\sum_{t=1}^s b_t =i$ and $b_t < b_{t+1}$. For instance $a_{10} = 9$

$$(1,2,3,4) \; (1,2,7) \; (1,3,6) \; (1,4,5) \; (1,9) \; (2,8) \; (3,7) \; (4,6) \; (10)$$

$a_{10}$ is like Ramanujan's function $p(n)$. Is there anything in number theory related with $a_i$? At any rate the infinite product converges the function we know well? That is to say, it is a


I found the material :

Let $p(n) $ is a Ramanujan function. For instance p(4) =4 :

1+1+1+1 = 1+ 3  =2+2 = 2+ 1+1

$\sum^\infty_{n=0} p(n) x^n = \Pi^\infty_{n=1} \frac{1}{1-x^n}$

Here the convenience is $p(0)=0$

share|cite|improve this question
2 – user17762 Oct 16 '12 at 6:53
up vote 2 down vote accepted

There is not a simple closed function for $$\prod_{i=1}^\infty (1+y^i)$$ but there are the alternatives $$ 1\left/\prod_{m =0}^\infty \left(1-y^{2m+1}\right)\right. $$ and $$ \sum_{k= 0}^\infty \prod_{j=1}^k \frac{y^j}{1-y^j}.$$

OEIS A000009 has more information.

share|cite|improve this answer
+1 I had some difficulty reading your sentence, you are saying the three expressions are all equal, but none of them qualifies as closed form. – Marc van Leeuwen Oct 16 '12 at 7:54
It is nice to notice that the equality of the first two expressions here is precisely the theorem of Euler. – Mariano Suárez-Alvarez Oct 17 '12 at 3:21

One extraordinarily beautiful result, due to Euler, is that the number of partitions of a number with odd parts is always equals the number of partitions of that number with distinct parts. Your $a_i$ is the latter.

This is the beginning of a beautiful part of combinatorics.

share|cite|improve this answer
I do not understand the number of partitions of a number with odd parts. What is odd parts ? – HK Lee Oct 16 '12 at 13:32
@user37116: You can decompose 7=1+1+5=1+3+3 and other ways, but not 7=1+2+4 because 2 and 4 are even. You might start with Wikipedia on partitions – Ross Millikan Oct 16 '12 at 13:47
It is a partition with odd parts, not a number with odd parts. Maybe clearer to say partitions of a number into odd parts. – GEdgar Oct 16 '12 at 14:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.