# mod 5 partition identity proof

I am looking for a proof that:

$$\prod_\limits{m=0}^\infty \dfrac{1}{(1-x^{5m+1})}=\sum_\limits{i=0}^\infty \dfrac{x^i}{\prod_\limits{j=1}^i (1-x^{5j})}$$

The left hand side expands into:

$$\dfrac{1}{(1-x)(1-x^6)(1-x^{11})\dots}$$

which gives the number of partitions into parts $1\mod5$.

The right hand side expands into:

$$1+\dfrac{x}{(1-x^5)}+\dfrac{x^2}{(1-x^5)(1-x^{10})}+\dfrac{x^3}{(1-x^5)(1-x^{10})(1-x^{15})}+\dots$$.

The only proof I have is a fairly brute force method.

EDIT: I think we have:

$$\dfrac{1}{(1-x)(1-ax)(1-a^2x)\dots}\\=1+\dfrac{x}{(1-a)}+\dfrac{x^2}{(1-a)(1-a^2)}+\dfrac{x^3}{(1-a)(1-a^2)(1-a^3)}+\dots$$

• Isn't each term of the right side the generating function of the number of partitions into $n$ terms of size 1 modulo 5? – Anthony Quas Feb 5 '16 at 16:33
• @AnthonyQuas; I think so, so there is a 1-1 relation, my proof consists of fairly bluntly matching each partition to its LHS counterpart – JMP Feb 5 '16 at 16:39
• I have a strong feeling this is in Stanleys book... – Per Alexandersson Feb 5 '16 at 20:53

More generally, there is an equality of power series in two variables $$\prod_\limits{m=0}^\infty \dfrac{1}{(1-t x^{5m+1})}=\sum_\limits{i=0}^\infty \dfrac{t^i x^i}{\prod_\limits{j=1}^i (1-x^{5j})}.$$ The coefficient of $t^i x^n$ on the left hand side is the number of partitions of $\frac{n-i}{5}$ into at most $i$ pieces (by convention the number of partitions of a non-integer in $0$). The coefficient of $t^i x^n$ on the right hand side is the number of partitions of $\frac{n-i}{5}$ into pieces of size at most $i$. Conjugation gives a bijection between the two kinds of partitions.

It is not about five at all, as may be seen from Julian Rosen's answer, but let me say it in more explicit form.

Simply denote $x^5=y$ and get more general identity $$\prod_{m=0}^\infty \frac{1}{1-xy^m}=\sum_{i=0}^\infty \frac{x^i}{(1-y)(1-y^2)\dots (1-y^i)}.\,\,\,(*)$$ Coefficient of $x^a y^b$:

on the left - is the number of partitions of $b$ into at most $a$ parts;

on the right - is the number of partitions of $b$ into parts not exceeding $a$.

Yes, it is the same by conjugation.

If you ask for purely algebraic solution (conjugation is rather combinatorial bijection), you may note that each part of $(*)$ satisfies initial condition $F(0,0)=1$ and the functional equation $F(x,y)(1-x)=F(xy,y)$, which determine unique formal power series.

• разбиение на n частей = partition into n parts. – KConrad Feb 5 '16 at 18:50
• Спасибо, Keith! – Fedor Petrov Feb 5 '16 at 19:07