# About a certain signed sum of consecutive integers ($\sum_{k=1}^n k\cdot \varsigma(k)=\sum_{k=n}^{2n} k\cdot \varsigma(k)$)

I hope you will excuse my vague title. Let's define

$$\varsigma \ \colon\ \mathbb N\to \mathbb Z\ \colon\ k\mapsto (-1)^{\lfloor \frac{k-1}{2}\rfloor}$$ and $$S(m,n)=\displaystyle\sum_{k=m}^n k\cdot \varsigma(k)$$

1. Find every integer $n\ge 1$ such that $S(1,\lfloor n/2\rfloor)=S(\lfloor n/2\rfloor+1,n)$

2. Show that $-(n+1)\le \delta(n)\le 2(n+1)$, when $\delta(n)=S(1,\lfloor n/2\rfloor)-S(\lfloor n/2\rfloor+1,n)$

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what's the definition of delta? – Qiaochu Yuan Sep 21 '10 at 20:09
Also, please mention the source if possible. This sounds like homework. – Aryabhata Sep 21 '10 at 20:12
Excusez-moi: Edited. The source: my fantasy :P – tetrapharmakon Sep 21 '10 at 22:17
I suggest we tag it as arithmetic... and remove number-theory and combinatorics. – Aryabhata Sep 21 '10 at 22:28
Have you found any values of $n$ for which part 1. holds? – Weltschmerz Sep 21 '10 at 23:54

The easiest way to attack this is elementary; if we define $S_0(n) = S(1,n)$, then $S(m,n) = S_0(n)-S_0(m)$, so all that's needed is to analyze $S_0$. That most easily happens by cases; a bit of quick calculation and some simple algebra shows that $S_0(4k+1) = 1$, $S_0(4k+3) = 0$, $S_0(4k+2) = 4k+3$, and $S_0(4k) = -4k$. From there it's straightforward to find the possible cases for $\delta(n)$ based on $n$ mod $8$; in particular, it seems like $\delta(n)$ will be zero whenever $n = 8k$ (where $S_0(n/2) = S_0(n)/2$) or $n=8k+7$ (where $S_0(n) = S_0(\lfloor n/2 \rfloor) = 0$).