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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.

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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.]

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    $\begingroup$ Oh, now I feel a bit dumb for asking. Hopefully some enjoyment can be derived from the question, anyway. $\endgroup$ – augurar Jan 3 '16 at 3:12
  • $\begingroup$ @augurar I didn't think the question "dumb"-- in fact it took me a few days to come up with the idea to divide each side, and several pieces of paper to check it on some small numbers. Note that the even numbers used in a sum (triangular) + sum of evens must be positive, but it seems if one allows $0$ as a summand there would be infinitely many ways. $\endgroup$ – coffeemath Jan 3 '16 at 4:12
  • $\begingroup$ Thinking about this a bit more, I realize that the identity only works properly for odd triangular numbers, otherwise it double-counts sums containing even triangular numbers. $\endgroup$ – augurar Jan 13 '16 at 4:27
  • $\begingroup$ For $n=6$ one cannot use 6 itself as the triangular number, else the remaining part 0 has no partitions into positive even numbers. But for $n=8$ which has six partitions into odd numbers, we do need the possibility of using the triangular number 6 and then add the partition of 2 into itself, in order to get six in all in the "triangular + even partition into positive evens" to also come out six. $\endgroup$ – coffeemath Jan 14 '16 at 0:03
  • $\begingroup$ @augurar I will try to get some lists of numbers of partitions into even/odd parts and so maybe verify for more examples. But keep in mind for a given $n$ one can only use triangular numbers which are strictly less than $n$ since one must leave room for a partition of the rest into positive even numbers. [It seems partitions of an even $n$ are the same as unrestricted partitions of $n/2,$ so that may help, and I'm pretty sure I can get hold of a sequence at o.e.i.s. for partitions into odd parts. $\endgroup$ – coffeemath Jan 14 '16 at 0:10

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