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For given $N$, $x$ and $k$ such that $0\leq x and $2\leq k\leq \left\lfloor \frac{N+1-2x}{2}\right\rfloor $, does it exist $p,$ $2\leq p\leq \left\lfloor \frac{N+1}{2}\right\rfloor $ such that

\begin{align} & \frac{(N+1-p)(N-p)}{2}\leq \frac{(N+1-x-k)(N-x-k)}{2}+\frac{(x+1)x}{2}<\frac{(N+1-p)(N-p)}{2}+\frac{p(p-1)}{2}+p+1-x \end{align}

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    seems that $p=k+x$ makes the trick.2012-09-07

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