Divisor Functions and Pentagonal Numbers

Abstract

Let p(n, ni) be the number of partitions of n with at most m summands¹, $\omega \left( n \right) = \frac{1}{2}\left( {3{n^2} - n} \right),n \in \mathbb{Z}$ be the pentagonal numbers² and ${\sigma _j}\left( n \right) = \sum\nolimits_{d\left| n \right.} {{d^j}},j \in \mathbb{N},$ be the divisor functions. Then σ(n) – the number of the divisors of n – satisfies

$$ {\sigma _0}\left( n \right) = g\left( {n,0} \right) + g\left( {n,1} \right) +, \ldots, + g\left( {n,n - 1} \right) $$

(1)

Where

$$ g\left( {n,m} \right) = p\left( {n,m} \right) - \sum\limits_{i = 1}^\infty {{{\left( { - 1} \right)}^{i - 1}}} \left( {p\left( {n - \omega \left( i \right),m} \right) + p\left( {n - \omega \left( { - i} \right),m} \right)} \right). $$

Pentagonal numbers are given by a well-known identity due to Euler

$$ {\left( q \right)_\infty }\mathop = \limits^{def} \prod\limits_{n = 1}^\infty {\left( {1 - {q^n}} \right)} = \sum\limits_{n = - \infty }^\infty {{{\left( { - 1} \right)}^n}} {q^{n\left( {3n - 1} \right)/2}}.$$

(2)

They are correlated with the number of partition p(n) of $n \in \mathbb{N},$ generated by

$$\sum\limits_{{n = 0}}^{\infty } {p\left( n \right)} {{q}^{n}} = \frac{1}{{{{{\left( q \right)}}_{\infty }}}}, $$

(3)

through the identity

$$ 1 = \frac{{{{\left( q \right)}_\infty }}}{{{{\left( q \right)}_\infty }}} = \left( {1 - {q^1} - {q^2} + {q^5} + {q^7} - \cdot \cdot \cdot } \right) \cdot \left( {p\left( 0 \right){q^0} + p\left( 1 \right){q^1} + \cdot \cdot \cdot } \right) $$

or equivalently

$$ 0 = p\left( n \right) - \sum\limits_{j = 1}^\infty {{{\left( { - 1} \right)}^{j - 1}}} \left( {p\left( {n - \omega \left( j \right)} \right) + p\left( {n - \omega \left( { - j} \right)} \right)} \right).$$

(4)

On the other hand, the pentagonal numbers are connected with the divisor function σ ¹ (n), for instance, by³

$$ {\sigma _1}\left( n \right) = {\sum\limits_{i = 1}^\infty {\left( { - 1} \right)} ^{i - 1}}\left( {\omega \left( i \right)p\left( {n - \omega \left( i \right)} \right) + \omega \left( { - i} \right)p\left( {n - \omega \left( { - i} \right)} \right)} \right). $$

Our statement (1) is a similar identity for ao(n). By way of illustration, for n = 5 we obtain $\mathop {\lim }\limits_{x \to \infty } \left( {n - {a_n}} \right) = \sum\limits_{i = 1}^\infty {{\sigma _0}} \left( i \right){q^i}$ where the sequence aⁿ is defined by a⁰ = 0 and

$$ {a_n} = 1 + \left( {1 - {q^{n - 1}}} \right){a_{n - 1}}.$$

(6)

Iterating the recurrence leads to

$$ {a_n} = \prod\limits_{i = 1}^{n - 1} {\left( {1 - {q^i}} \right)} \sum\limits_{j = 0}^{n - 1} {\frac{1}{{\left( {1 - q} \right) \ldots \left( {1 - {q^j}} \right).}}}$$

(7)

Now, theroduct ${\left( {\left( {1 - q} \right) \ldots \left( {1 - {q^m}} \right)} \right)^{ - 1}}\mathop = \limits^{def} \left( q \right)_m^{ - 1}$ is well-known as generating function of the numbers p(n, m), hence

$$ \sum\limits_{n = 0}^\infty {p\left( {n,m} \right){q^n}} = \frac{1}{{\left( {1 - q} \right)\left( {1 - {q^2}} \right) \ldots \left( {1 - {q^m}} \right)}},m \geqslant 0. $$

(8)

With (8) the equation (7) can be written as

$${{a}_{n}} = \sum\limits_{{m = 0}}^{{n - 1}} {\underbrace{{{{{\left( q \right)}}_{{n - 1}}}\sum\limits_{{h = 0}}^{\infty } {p\left( {h,m} \right){{q}^{h}}.} }}_{{\mathop{ = }\limits^{{def}} {{H}_{{n,m}}}\left( q \right)}}} $$

(9)

For $\sum\limits_{n = 0}^\infty {g\left( {n,m} \right){q^n}} = \mathop {\lim }\limits_{x \to \infty } {H_{n,m}}\left( q \right)\mathop = \limits^{\left( 2 \right)} \sum\limits_{i = - \infty }^\infty {{{\left( { - 1} \right)}^i}{q^{i\left( {3i - 1} \right)/2}}} \sum\limits_{j = 0}^\infty {p\left( {j,m} \right){q^j}}$ with g(0, m) = 1 and

$$ \begin{gathered} g\left( {n,m} \right) = p\left( {n,m} \right) - p\left( {n - 1,m} \right) - p\left( {n - 2,m} \right) + p\left( {n - 5,m} \right) + \cdots ,m \geqslant 1. \hfill \\ \hfill \\ \end{gathered} $$

Note that p(n, m) = p(n), for m ≥ n. This implies $g\left( {n,m} \right)\mathop = \limits^{\left( 4 \right)} 0,form \geqslant n \geqslant 1.$ Therefore

$$ \mathop {\lim }\limits_{n \to \infty } (n - a_n ) = \sum\limits_{j = 1}^\infty {\left( {g\left( {j,0} \right) + \cdots + g\left( {j,j - 1} \right)} \right)q^j } $$

which, together with (5), completes the proof of (1).