What is the significance of this identity relating to partitions? I was watching a talk given by Prof. Richard Kenyon of Brown University, and I was confused by an equation briefly displayed at the bottom of one slide at 15:05 in the video.
$$1 + x + x^3 + x^6 + \dots + x^{n(n-1)/2} + \dots = \left(\frac{1-x^2}{1-x}\right) \left(\frac{1-x^4}{1-x^3}\right) \left(\frac{1-x^6}{1-x^5}\right) \dots$$
On the left we have the power series $\sum_{n=0}^{\infty}x^{T_n}$.  On the right we have some sort of infinite product.  Can anyone explain what the meaning of this identity is, in relation to integer partitions?

Background:  The speaker starts by discussing the generating function of the partition function,
$$P(x) = \prod_{k=1}^\infty \left(\frac {1}{1-x^k} \right)$$
He then uses the idea behind this generating function to derive a fun identity:
$$(1+x)(1+x^2)(1+x^3)\dots = \frac{1}{(1-x)(1-x^3)(1-x^5)\dots}$$
which shows that the number of partitions into unequal parts equals the number of partitions into odd parts.
This is the context for the above identity which I fell short of understanding.

Also: I did a bit of searching and came across a 1991 paper by Ono, Robins & Wahl concerning partitions using triangle numbers, which might be related.
This paper proves that
$$ \sum_{n=1}^{\infty}{x^{T_n}} = \prod_{n=1}^{\infty}{\frac{(1-x^{2n})^2}{1-x^n}}$$
which shows that the identity is true.
 A: If the sides are divided by the numerator of the right, the formula is
$$\frac{\sum x^{T_n}}{(1-x^2)(1-x^4)\cdots}=\frac{1}{(1-x)(1-x^3)\cdots}. \tag{1}$$
Here the left side with numerator replaced by $1$ represents partitions into even parts, while the right side partitions into odd parts. Putting the numerator back, the left side represents representations of a number by a single triangular number plus a sum of even parts, while the right side again represents representations by sums of odd parts. 
So in this form, the identity says the number of ways to write $n$ as a triangular number plus a sum of even parts is the same as the number of ways to write $n$ as a sum of odd parts. Note the single triangular number involved here may be $0$ (which is $T_0$). I didn't know the equality of these two counts, but tried it on some small numbers and it seems to be so. 
A slight correction and better explanation of the left side count: Since the taylor series of $1/[(1+x^2)(1+x^4)\cdots$ starts out with the term $1\cdot x^0,$ it is clear that series considers that $0$ is indeed the (only) partition of $0$ into even parts. [The same happens in the generating function for unrestricted partitions.] So when this series is multiplied by the numerator in (1), the result is counting, for a given $n,$ ordered pairs consisting of a triangular number $T$ (which may be zero) followed by a partition of $n-T$ into even parts, and notation such as $(6),2$ (for $n=8$) means it is the entity for which $T$ has been taken to be 6 and then $n-T=8-6=2$ is to be partitioned into even part(s), here the extra 2 after the (6) of $(6),2.$ Because one must "tag" these entitities by the triangular number used, this is a different entity than the $(0),2,6$ in the count. They come from different powers of $x$ in the numerator of (1). [I think somewhere in the answer or in comments I had erroneously insisted that the partition into even parts which follows the triangular number had to be positive even parts, but this is so only when $n$ itself is not triangular.]
