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Let $K\geq 3$ and consider $K$ real numbers $\mu_1<\mu_2<...<\mu_K$.

Let $\delta_j\equiv \mu_{j+1}-\mu_j$ $\forall j \in \{1,...,K-1\}$.

For $h\in \{2,...,K-1\}$, consider the following set of conditions $$ \begin{aligned} &\delta_1+\delta_2+...+\delta_{h-3}+\delta_{h-2}+ \delta_{h-1} \neq \delta_{K-1},\\ &\delta_1+\delta_2+...+\delta_{h-3}+\delta_{h-2}\hspace{1.2cm} \neq \delta_{K-1}+\delta_{K-2},\\ &\delta_1+\delta_2+...+\delta_{h-3}\hspace{2.4cm} \neq \delta_{K-1}+\delta_{K-2}+\delta_{K-3},\\ &...\\ &\delta_1 \hspace{5.4cm}\neq \delta_{K-1}+\delta_{K-2}+\delta_{K-3}+...+\delta_{K-h-1}. \end{aligned} $$

For example, for $K=3$ we have $$ \begin{aligned} &\delta_1\neq \delta_2. \end{aligned} $$

For $K=4$ we have $$ \begin{aligned} &\delta_1\neq \delta_3,\\ & \delta_1+\delta_2 \neq \delta_3,\\ &\delta_1\neq \delta_3+\delta_2. \end{aligned} $$

For $K=5$ we have $$ \begin{aligned} &\delta_1\neq \delta_4,\\ & \delta_1+\delta_2 \neq \delta_4,\\ &\delta_1+\delta_2+\delta_3\neq \delta_4,\\ &\delta_1+\delta_2\neq \delta_4+\delta_3,\\ &\delta_1\neq \delta_4+\delta_3+ \delta_2,\\ & \delta_1\neq \delta_4+\delta_3\\ \end{aligned} $$ I would like to know whether these conditions can be linked to some geometric/linear algebra/graph theory concept. Somehow, they rule out some specific linear combinations with a sort of geometric path, but if I was asked to explain the above conditions in English I would not be able to do so. Could you help?

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Note that $$\sum_{j=m}^n\delta_j = \mu_{n+1} -\mu_m,$$ since the middle terms cancel. We can then rewrite a condition $$\delta_1+\dots+\delta_h\neq \delta_i+\dots+\delta_{K-1}$$ with $1\leq h<i\leq K-1 $ as $$\mu_{h+1} - \mu_1 \neq\mu_K-\mu_i.$$ Rearranging yields that the sum of the first and last $\mu$ is different from the sum of any other pair of $\mu$'s (since $h$ and $i$ can be chosen arbitrarily), and different from any other $\mu$ doubled (take $h=i-1$).

This is a massive simplification IMO, but I don't know where this sort of thing would arise.

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    $\begingroup$ Thanks. Note that $\sum_{j=m}^n \delta_j=\mu_{n+1}-\mu_m$. Does this change your conclusions? $\endgroup$ – STF Nov 12 '19 at 18:49
  • $\begingroup$ Just taking your answer it looks like that my condition can be rewritten as: for $h=1,...,K-2$ and for $i=2,...,K-1$, $\mu_{h+1}-\mu_1\neq \mu_K-\mu_i$. Correct? Does this way of rewriting rings any bell of theorems, notions of graph theory, etc? $\endgroup$ – STF Nov 12 '19 at 19:05
  • $\begingroup$ Sorry, could you clarify whether your answer should be interpreted as: my condition is equivalent to $\mu_{h+1}-\mu_1\neq \mu_K-\mu_i$ for each $h,i$ such that $1\leq h<i\leq K-1$? $\endgroup$ – STF Nov 12 '19 at 19:21
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    $\begingroup$ I recalculated, and I think this should be correct now. This kind of index juggling can be a bit error prone. I think your second to last calculation is correct. We can get around the $h<i$ with a bit of regrouping. I'll try to think of any bells this might ring, but I don't see anything now. $\endgroup$ – Aldoggen Nov 12 '19 at 19:24
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Adloggan's answer has the arithmetic of the situation. So, here's one possible intuitive interpretation. Imagine a stick of length $\mu_K - \mu_1$ with notches carved into it at intervals representing $\mu_j$ in the natural way. Then, the set of inequalities imply that the stick cannot be broken along those notches into $>1$ pieces so that the leftmost piece has the same length as the rightmost piece.

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