# pentagonal number theorem

 $\prod_{k=1}^{\infty}(1-x^{k})=\sum_{n=-\infty}^{\infty}(-1)^{n}x^{n(3n+1)/2}$ (1)

where the two sides are regarded as formal power series over $\mathbb{Z}$.

Proof: For $n\geq 0$, denote by $f(n)$ the coefficient of $x^{n}$ in the product  on the left, i.e. write

 $\prod_{k=1}^{\infty}(1-x^{k})=\sum_{n=0}^{\infty}f(n)x^{n}\;.$

By this definition, we have for all $n$

 $f(n)=e(n)-d(n)$

where $e(n)$ (resp. $d(n)$) is the number of partitions   of $n$ as a sum of an even (resp. odd) number of distinct summands. To fix the notation, let $P(n)$ be set of pairs $(s,g)$ where $s$ is a natural number  $>0$ and $g$ is a decreasing mapping $\{1,2,\ldots,s\}\to\mathbb{N}^{+}$ such that $\sum_{x}g(x)=n$. The cardinal of $P(n)$ is thus $f(n)$, and $P(n)$ is the union of these two disjoint sets:

 $E(n)=\{(s,g)\in P(n)\mid s\text{ is even}\},$
 $D(n)=\{(s,g)\in P(n)\mid s\text{ is odd}\}.$

Now on the right side of (1) we have

 $1+\sum_{n=1}^{\infty}(-1)^{n}x^{n(3n+1)/2}+\sum_{n=1}^{\infty}(-1)^{n}x^{n(3n-% 1)/2}\;.$

Therefore what we want to prove is

 $\displaystyle e(n)$ $\displaystyle=$ $\displaystyle d(n)+(-1)^{m}\qquad\text{if }n=m(3m\pm 1)/2\text{ for some }m$ (2) $\displaystyle e(n)$ $\displaystyle=$ $\displaystyle d(n)\qquad\text{otherwise.}$ (3)

For $m\geq 1$ we have

 $\displaystyle m(3m+1)/2$ $\displaystyle=$ $\displaystyle 2m+(2m-1)+\ldots+(m+1)$ (4) $\displaystyle m(3m-1)/2$ $\displaystyle=$ $\displaystyle(2m-1)+(2m-2)+\ldots+m$ (5)

Take some $(s,g)\in P(n)$, and suppose first that $n$ is not of the form (4) nor (5). Since $g$ is decreasing, there is a unique integer $k\in[1,s]$ such that

 $\displaystyle g(j)$ $\displaystyle=g(1)-j+1\qquad\text{for }j\in[1,k],g(j)$ $\displaystyle

If $g(s)\leq k$, define $\overline{g}\colon[1,s-1]\to\mathbb{N}^{+}$ by

 $\overline{g}(x)=\begin{cases}g(x)+1,&\text{if }x\in[1,g(s)],\\ g(x),&\text{if }x\in[g(s)+1,s-1].\end{cases}$

If $g(s)>k$, define $\overline{g}\colon[1,s+1]\to\mathbb{N}^{+}$ by

 $\overline{g}(x)=\begin{cases}g(x)-1,&\text{if }x\in[1,k],\\ g(x),&\text{if }x\in[k+1,s],\\ k,&\text{ if }x=s+1.\end{cases}$

In both cases, $\overline{g}$ is decreasing and $\sum_{x}\overline{g}(x)=n$. The mapping $g\to\overline{g}$ maps takes an element having odd $s$ to an element having even $s$, and vice versa. Finally, the reader can verify that $\overline{\overline{g}}=g$. Thus we have constructed a bijection $E(n)\to D(n)$, proving (3).

Now suppose that $n=m(3m+1)/2$ for some (perforce unique) $m$. The above construction still yields a bijection between $E(n)$ and $D(n)$ excluding (from one set or the other) the single element $(m,g_{0})$:

 $g_{0}(x)=2m+1-x\qquad\text{for }x\in[1,m]$

as in (4). Likewise if $n=m(3m-1)/2$, only this element $(m,g_{1})$ is excluded:

 $g_{1}(x)=2m-x\qquad\text{for }x\in[1,m]$

as in (5). In both cases we deduce (2), completing the proof.

Remarks: The name of the theorem derives from the fact that the exponents $n(3n+1)/2$ are the generalized pentagonal numbers  .

The theorem was discovered and proved by Euler around 1750. This was one of the first results about what are now called theta functions  , and was also one of the earliest applications of the formalism of generating functions.

The above proof is due to F. Franklin, (Comptes Rendus de l’Acad. des Sciences, 92, 1881, pp. 448-450).

Title pentagonal number theorem  PentagonalNumberTheorem 2013-03-22 13:57:51 2013-03-22 13:57:51 bbukh (348) bbukh (348) 7 bbukh (348) Theorem msc 11P81 msc 14K25