the limit of infinite product $(1+y)(1+y^2)(1+y^3)(1+y^4)\cdots $

Question

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

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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 $p(0)=0$

Answer 1

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.

Answer 2

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.