Polite Dirichlet Series

Let $n$ be a positive integer. First of all, consider the following result, a proof of which is added below: $\text{politeness of} \; n \; = \; \text{num. of odd factors of} \; n \; \text{greater than}\; 1$ Hence, we can rewrite the series as $\displaystyle S= \sum_{n=1}^{\infty} \sum_{ \begin{aligned} & d | n, \; d > 1 \\ & d \; \text{is odd} \end{aligned}}$ $\dfrac{1}{n^2}$ and we see that each term $\dfrac{1}{n^2}$ appears exactly as many times as the number of odd divisors of $n$ that are greater than $1$ .

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Now, let $T$ be the infinite series that arises from the multiplication: $\left ( \dfrac{1}{1^2}+ \dfrac{1}{2^2}+ \dfrac{1}{3^2}+ \dfrac{1}{4^2}+...\right) \left( \dfrac{1}{3^2}+ \dfrac{1}{5^2}+ \dfrac{1}{7^2}+\dfrac{1}{9^2}+... \right)$ We observe that each term of $T$ is the reciprocal of a square and that, for each $n$ positive integer, $\dfrac{1}{n^2}$ appears as a term of $T$ uniquely under all possible forms of the type $\dfrac{1}{(n/k)^2} \dfrac{1}{k^2}$ , where $k>1$ is odd and $k|n$ , one time for each value of $k$ . And so, in the sum, the frequency of $\dfrac{1}{n^2}$ equals the number of odd divisors of $n$ greater than $1$ . Therefore, $T=S$ .

Then, knowing that

$\displaystyle \begin{aligned} & \sum_{n=1}^{\infty} \dfrac{1}{n^2} = \dfrac{\pi^2}{6} \\ & \sum_{n=1}^{\infty} \dfrac{1}{(2n-1)^2} = \sum_{n=1}^{\infty} \dfrac{1}{n^2} - \sum_{n=1}^{\infty} \dfrac{1}{(2n)^2} = \dfrac{\pi^2}{8} \end{aligned}$

we find $S = \dfrac{\pi^2}{6} \left( \dfrac{\pi^2}{8} -1\right) = \dfrac{\pi^4 - 8\pi^2 }{48}$

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so that the answer is $4+8+2+48= \boxed{62}$

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Characterization of politeness: An interesting proof of the statement above, which links the politeness of $n$ to its odd divisors, can be found in this Wikipedia page . The following, instead, can be an alternative one:

We recover from this solution the definition of odd-polite and even-polite number, along with the propositions $(i)$ and $(ii)$ (the first five lines). We will show the existence of a bijection between odd divisors of $n$ greather than one and representations of $n$ as a sum of at least two consecutive positive integers.

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Let $d>1$ be an odd integer such that $d|n$ . If the $RHS$ of $(i)$ is satisfied by $d$ (i.e. $\color{#3D99F6} \dfrac{n}{d} \ge \dfrac{d+1}{2}$ ), then $n$ is an odd-polite number and, precisely, it is the sum of all $\color{#3D99F6} d$ consecutive positive integers starting from $\color{#3D99F6} \dfrac{n}{d}-\dfrac{d-1}{2}$ . On the other hand, suppose that the $RHS$ of $(i)$ is not satisfied, i.e. we suppose $\color{#D61F06} \dfrac{n}{d} < \dfrac{d+1}{2}$ . Now, if we consider that $n=(2h+1)x$ , where $2h+1=d$ and $x \ge 1$ , we get $x<h+1$ , and so $x \le h$ . However, $n=2xh+x$ implies that $n \equiv_{2x} x$ , while the condition $x \le h$ is equivalent to $\frac{(2h+1)x-x}{2x} \ge x$ and consequently to $\frac{n-x}{2x} \ge x$ . Hence, it follows that the $RHS$ of $(ii)$ is satisfied by the even integer $p=2x=2\dfrac{n}{d}>1$ . So, $n$ is an even-polite integer and, precisely, it is the sum of all $\color{#D61F06} 2\dfrac{n}{d}$ consecutive positive integers starting from $\color{#D61F06} \dfrac{d+1}{2} - \dfrac{n}{d}$ .

To sum up:

$d>1 \,\land \, d \; \text{odd}, \; d|n \; \implies n= \begin{cases} \left| \color{#3D99F6} \dfrac{n}{d}-\dfrac{d-1}{2}\color{#333333}+...\right|_{\color{#3D99F6} d}\; \; , \, \color{#333333} \text{if} \; \color{#3D99F6} \dfrac{n}{d} \ge \dfrac{d+1}{2} \\ \left| \color{#D61F06} \dfrac{d+1}{2} - \dfrac{n}{d}\color{#333333}+...\right|_{\color{#D61F06} 2\frac{n}{d}} \color{#333333}, \, \text{if} \; \color{#D61F06} \dfrac{n}{d} < \dfrac{d+1}{2} \end{cases}$ $\;$ (where $|n+...|_k$ means the sum of $k$ consecutive integers starting from $n$ ).

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Conversely, let $n$ be a polite number. If $n$ is the sum of an odd number $d>1$ of consecutive positive integers, then $n$ is an odd-polite number, and so $(i) \implies d|n$ . If, instead, $n$ is the sum of an even number $2x>1$ of consecutive positive integers, then $n$ is an even-polite number, and hence, as we have seen above, $(ii) \implies n\equiv_{2x} x \, \land \, h \ge x$ , so that $n=(2h+1)x \, \land \, 2h+1>1$ . In both cases, we associated an odd divisor of $n$ greater than $1$ to the generic polite representation of $n$ .

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The association laws described in the last two paragraphs form a bijection between the sets $\{ \text{polite representations of} \; n \}$ and $\{d \in \mathbb{Z_{>1}}: \, d \; \text{odd} \, \land \, d|n\}$ . Hence, $p(n) = \Bigl|\{d \in \mathbb{Z_{>1}}: \, d \; \text{odd} \, \land \, d|n \}\Bigr|$ .