Flipped to be Squared

An integer $N$ can be written as a sum of two or more consecutive positive integers precisely when we can find integers $0 \le m < n$ with $n - m \ge 2$ such that $N \; = \; (m+1) + (m+2) + \cdots + n \; = \; \tfrac12n(n+1) - \tfrac12m(m+1) \; = \; \tfrac12(n-m)(n+m+1)$ Note that $n-m,n+m+1 \ge 2$ are integers of opposite parity that multiply to $2N$ . Thus, if $N$ belongs to $S$ , then $2N$ has a nontrivial odd factor, and hence $N$ is not a power of $2$ .

On the other hand, if $N$ is not a power of $2$ , then $N$ has a nontrivial odd factor, and $2N$ has a nontrivial even factor. Thus we can write $N = ab$ where $1 < a < b$ and $a,b$ have opposite parities. One of $a,b$ could be $2^k$ where $k$ is the index of $2$ in $2N$ . We can now solve for $m,n$ such that $n-m = a$ and $n+m+1 = b$ , obtaining integers $m,n$ with $0 \le m < n$ and $n \ge m+2$ . Hence $N$ belongs to $S$ .

Thus $S$ is just the set of numbers that are not powers of $2$ , and hence $R \;= \; \tfrac16\pi^2 - \sum_{n=0}^\infty \frac{1}{(2^n)^2} \; = \; \tfrac16\pi^2 - \tfrac43 \; = \; \frac{\pi^2 - 8}{6}$ making the answer $\boxed{16}$ .

At the very least, all odd numbers can be written as a sum of two consecutive integers, since odd numbers are of the form $2n + 1$ , and $2n + 1 = n + (n+1)$ .

Now, consider a number $p$ which you would like to split into $k$ consecutive elements. If $k$ is odd, these elements will have a form of $a$ , $a\pm1$ , $a\pm2$ ... $a\pm \frac{k-1}{2}$ . Adding them all leads to $p = ka$ for some integer $a$ , which means that to divide $p$ into $k$ consecutive integers, $k | p$ . So, any number that is divisible by a prime number $k>2$ can be split into $k$ consecutive integers.

If $k$ is even on the other hand, then we will have $p$ divided as $a$ , $a\pm1$ , $a\pm2$ ... $a+ \frac{k}{2}$ for some arbitrary integer $k$ . Now, let us let $j= \frac{k}{2}$ . Adding them will lead to the statement $p = ka + j$ . and since we know that $j|k$ , this becomes $p = j(2a+1)$ .

We now have the condition such that $j|p$ for $p$ to be divided into $k$ consecutive integers. Following that, we see that $62$ can be divided into $4$ consecutive integers, but not $6$ .

So it will be quite evident at this point that every number that is not a power of 2 is not an element of $S$ .

So $R$ will have all reciprocal of squares except those squares of powers of two. OR, this can be expressed as powers of $4$ .

Knowing that $\displaystyle \sum \limits_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ , and that $\displaystyle \sum\limits_{i=0}^{\infty} \frac{1}{4^i} = \frac{4}{3}$ , it will eventually give us the sum of the elements of $R$ as

$\frac{ \pi^2 - 8 }{6}$

And there we have it. $2+8+6= \boxed {16}$ .

Let $n \in \mathbb{Z^+}$ . We can call $n$ a $k$ - polite number if $k \in \mathbb{Z_{>1}}$ and $n$ is the sum of $k$ consecutive positive integers. Also, we say that $n$ is an odd-polite number if there exists at least one odd integer $k$ for which $n$ is a $k$ -polite number, and likewise for the even case. Following then the same idea outlined in the proof by Will Van Noordt under the problem "A special property of a number" , one can notice that:

$n$ is an odd-polite number $\iff \exists$ an odd $d \in \mathbb{Z_{>1}}$ such that $d|n \, \land \, \dfrac{n}{d} \ge \dfrac{d+1}{2} \; \; \; \; (i)$

$n$ is an even-polite number $\iff \exists$ an even $p \in \mathbb{Z_{>1}}$ such that $n \equiv \dfrac{p}{2} \, \text{mod} \, p \, \land \, \dfrac{n-\frac{p}{2}}{p} \ge \dfrac{p}{2} \; \; \; \; (ii)$

For a few lines, let's work under the assumption that $n$ is not an odd-polite number. To analyze for which conditions $n$ is or is not an even-polite number, we need to verify wheter there is an integer $x \ge 1$ such that $n \equiv x \, \, \text{mod} \, 2x$ , and therefore so that $n=(2h+1)x$ , for a certain $h \in \mathbb{Z}$ . Furthermore, we set $\frac{(2h+1)x-x}{2x} \ge x$ , which leads to $h\ge x$ , because the second condition above in $(ii)$ is required too. It follows that $h\ge1$ holds true, and since accordingly $2h+1>1$ , it is a necessary condition for $n$ being an even-polite number that an odd integer different from $1$ is a factor of $n$ . This means that if $n$ is a power of $2$ , it cannot be an even-polite number.

Viceversa, if $n$ is not a power of $2$ , then there is at least one couple of integers $h,x$ so that $n=(2h+1)x$ and $2h+1>1$ and $x\ge1$ . Hence, there is an odd integer $d=2h+1$ in $\mathbb{Z_{>1}}$ which divides $n$ . By our initial assumption that $n$ is not an odd-polite number, $(i) \Rightarrow \frac{n}{d} < \frac{d+1}{2}$ , which in terms of $h$ and $x$ implies $x < \frac{2h+2}{2} \Rightarrow x < h+1 \Rightarrow h\ge x$ . Thus $n$ is an even-polite number.

We proved: $n$ is not an odd-polite number $\Rightarrow (\, n$ is a power of 2 $\iff n$ is not an even-polite number $) \; \; \; \; (P)$

Also, n is a power of 2 $\Rightarrow \nexists$ an odd $d \in \mathbb{Z_{>1}} \text{:} \; d$ divides $n \Longrightarrow^{(i)} n$ is not an odd-polite number $\Longrightarrow^{(P)} n$ is not an odd-polite number $\land \; n$ is not an even-polite number. By De Morgan, this implication is equivalent to: $n$ is an odd-polite number $\lor \; n$ is an even-polite number $\Longrightarrow n$ is not a power of 2.

On the other hand, let $n$ not be a power of 2. Assuming $n$ is not an odd-polite number, from $(P)$ we know that $n$ is an even-polite number as it is not a power of 2. Since tertium non datur , $n$ is not a power of 2 $\Longrightarrow n$ is an odd-polite number $\lor \; n$ is an even-polite number. So we can conclude that: $n \; \text{is the sum of at least two positive integers} \; \iff n \; \text{is not a power of 2}$

$\large \therefore \;$ in the problem we have $S=\mathbb{Z^+} \setminus \{2^b : b \in \mathbb{Z}, b\ge0\}$ , and as a consequence $\sum\limits_{n \in S}^{} \frac{1}{n^2} = \sum\limits_{n=1}^{\infty} \frac{1}{n^2}- \sum\limits_{b=0}^{\infty} \frac{1}{2^{2b}} =\frac{\pi^2}{6} -\frac{1}{1-\frac14} = \frac{\pi^2-8}{6}$