# pentagonal number theorem

Theorem :

$$\prod _{k=1}^{\mathrm{\infty}}(1-{x}^{k})=\sum _{n=-\mathrm{\infty}}^{\mathrm{\infty}}{(-1)}^{n}{x}^{n(3n+1)/2}$$ | (1) |

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

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

$$\prod _{k=1}^{\mathrm{\infty}}(1-{x}^{k})=\sum _{n=0}^{\mathrm{\infty}}f(n){x}^{n}.$$ |

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
$\{1,2,\mathrm{\dots},s\}\to {\mathbb{N}}^{+}$ 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

$$1+\sum _{n=1}^{\mathrm{\infty}}{(-1)}^{n}{x}^{n(3n+1)/2}+\sum _{n=1}^{\mathrm{\infty}}{(-1)}^{n}{x}^{n(3n-1)/2}.$$ |

Therefore what we want to prove is

$e(n)$ | $=$ | $d(n)+{(-1)}^{m}\mathit{\hspace{1em}\hspace{1em}}\text{if}n=m(3m\pm 1)/2\text{for some}m$ | (2) | ||

$e(n)$ | $=$ | $d(n)\mathit{\hspace{1em}\hspace{1em}}\text{otherwise.}$ | (3) |

For $m\ge 1$ we have

$m(3m+1)/2$ | $=$ | $2m+(2m-1)+\mathrm{\dots}+(m+1)$ | (4) | ||

$m(3m-1)/2$ | $=$ | $(2m-1)+(2m-2)+\mathrm{\dots}+m$ | (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

$g(j)$ | $=g(1)-j+1\mathit{\hspace{1em}\hspace{1em}}\text{for}j\in [1,k],g(j)$ | $$ |

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

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

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

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

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 $(m,{g}_{0})$:

$${g}_{0}(x)=2m+1-x\mathit{\hspace{1em}\hspace{1em}}\text{for}x\in [1,m]$$ |

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

$${g}_{1}(x)=2m-x\mathit{\hspace{1em}\hspace{1em}}\text{for}x\in [1,m]$$ |

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^{} |
---|---|

Canonical name | PentagonalNumberTheorem |

Date of creation | 2013-03-22 13:57:51 |

Last modified on | 2013-03-22 13:57:51 |

Owner | bbukh (348) |

Last modified by | bbukh (348) |

Numerical id | 7 |

Author | bbukh (348) |

Entry type | Theorem |

Classification | msc 11P81 |

Classification | msc 14K25 |