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I would like your help to write the following set of constraints using a notation that works for any generic $K\in \{3,4,5,6,...\}$.

Consider $K$ real numbers $\mu_1<\mu_2<...<\mu_K$. Let $\alpha_j\equiv \mu_{j+1}-\mu_j$ for $j=1,...,K-1$.

Let me report below the constraints that I want to impose. RHS stays for right-hand-side. LHS stays for left-hand-side.


For $K=3$: $$ \begin{cases} \alpha_1\neq \alpha_2 & \text{(1): [first difference $\neq$ from last difference]}\\ \end{cases} $$


For $K=4$: $$ \begin{cases} \alpha_1\neq \alpha_3 & \text{(1): [first difference $\neq$ from last difference]}\\ ------\\ \alpha_1+\alpha_2\neq \alpha_3 & \text{(2): [add one adjacent difference to the LHS of (1)]}\\ \alpha_1\neq \alpha_2+ \alpha_3& \text{(3): [add one adjacent difference to the RHS of (1)]}\\ \end{cases} $$


For $K=5$:

$$ {\small \begin{cases} \alpha_1\neq \alpha_4 & \text{(1): [first difference $\neq$ from last difference]}\\ ------\\ \alpha_1+\alpha_2\neq \alpha_4 & \text{(2): [add one adjacent difference to the LHS of (1)]}\\ \alpha_1\neq \alpha_3+ \alpha_4& \text{(3): [add one adjacent difference to the RHS of (1)]}\\ ------\\ \alpha_1+\alpha_2+\alpha_3\neq \alpha_4 & \text{(4): [add one adjacent difference to the LHS of (2)]}\\ \alpha_1+\alpha_2\neq \alpha_4+\alpha_3 & \text{(5): [add one adjacent difference to the RHS of (2)]}\\ ------\\ \alpha_1+\alpha_2\neq \alpha_3+ \alpha_4& \text{ [add one adjacent difference to the RHS of (3)] [redundant]}\\ \alpha_1\neq \alpha_2+ \alpha_3+ \alpha_4& \text{(6): [add one adjacent difference to the RHS of (3)]}\\ \end{cases}} $$

For $K=6$: $$ \begin{cases} \alpha_1\neq \alpha_5 & \text{(1): [first difference $\neq$ from last difference]}\\ -------\\ \alpha_1+\alpha_2\neq \alpha_5 & \text{(2): [add one adjacent difference to the LHS of (1)]}\\ \alpha_1\neq \alpha_5+\alpha_4 & \text{(3): [add one adjacent difference to the RHS of (1)]}\\ -------\\ \alpha_1+\alpha_2+\alpha_3\neq \alpha_5 & \text{(4): [add one adjacent difference to the LHS of (2)]}\\ \alpha_1+\alpha_2\neq \alpha_5+\alpha_4 & \text{(5): [add one adjacent difference to the LHS of (2)]}\\ -------\\ \alpha_1+\alpha_2\neq \alpha_5+\alpha_4 & \text{ [add one adjacent difference to the RHS of (3)] [redundant]}\\ \alpha_1\neq \alpha_5+\alpha_4+\alpha_3 & \text{(6): [add one adjacent difference to the RHS of (3)]}\\ -------\\ \alpha_1+\alpha_2+\alpha_3+\alpha_4\neq \alpha_5 & \text{(7): [add one adjacent difference to the LHS of (4)]}\\ \alpha_1+\alpha_2+\alpha_3\neq \alpha_5+\alpha_4 & \text{(8): [add one adjacent difference to the LHS of (4)]}\\ -------\\ \alpha_1+\alpha_2+\alpha_3\neq \alpha_5+\alpha_4 & \text{[add one adjacent difference to the LHS of (5)] [redundant]}\\ \alpha_1+\alpha_2\neq \alpha_5+\alpha_4+\alpha_3 & \text{(9): [add one adjacent difference to the LHS of (5)]}\\ -------\\ \alpha_1+\alpha_2\neq \alpha_5+\alpha_4+\alpha_3& \text{[add one adjacent difference to the RHS of (6)] [redundant]}\\ \alpha_1\neq \alpha_5+\alpha_4+\alpha_3+\alpha_2 & \text{(10): [add one adjacent difference to the RHS of (6)]}\\ \end{cases} $$


Question: There is a mechanic behind these constraints that is replicable for any $K$. I can see some kind of circular argument, but I find it hard to formalise. Could you suggest a notation to represent these constraints that works for any $K$?


Update:

I found this way of writing the desired constraints: for $h=2,...,K$ $$ \begin{aligned} &\alpha_1+\alpha_2+...+\alpha_{h-3}+\alpha_{h-2}+ \alpha_{h-1} \neq \alpha_{K-1}\\ &\alpha_1+\alpha_2+...+\alpha_{h-3}+\alpha_{h-2}\hspace{1.2cm} \neq \alpha_{K-1}+\alpha_{K-2}\\ &\alpha_1+\alpha_2+...+\alpha_{h-3}\hspace{2.4cm} \neq \alpha_{K-1}+\alpha_{K-2}+\alpha_{K-3}\\ &...\\ &\alpha_1 \hspace{5.2cm}\neq \alpha_{K-1}+\alpha_{K-2}+\alpha_{K-3}+...+\alpha_{K-h-1}\\ \end{aligned} $$ Is it correct? Are there simpler ways?

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We consider the case $K=6$, rearrange the constraints conveniently and derive a formula which describes all the constraints. From this setting we derive a formula for the general case easily.

Case: $K=6$

We have

\begin{align*} \alpha_1&\ne\alpha_5&\\ &\ne \alpha_4+\alpha_5&\\ &\ne \alpha_3+\alpha_4+\alpha_5&\\ &\ne \alpha_2+\alpha_3+\alpha_4+\alpha_5&\color{blue}{\alpha_1\ne \alpha_N+\cdots+\alpha_5}\\ &&\qquad\qquad \color{blue}{ (2\leq N\leq 5)}\\ \alpha_1+\alpha_2&\ne\alpha_5&\\ &\ne \alpha_4+\alpha_5&\\ &\ne \alpha_3+\alpha_4+\alpha_5&\color{blue}{\alpha_1+\alpha_2\ne \alpha_N+\cdots+\alpha_5}\\ &&\qquad\qquad \color{blue}{ (3\leq N\leq 5)}\\ \alpha_1+\alpha_2+\alpha_3&\ne\alpha_5&\\ &\ne \alpha_4+\alpha_5&\color{blue}{\alpha_1+\alpha_2+\alpha_3\ne \alpha_N+\cdots+\alpha_5}\\ &&\qquad\qquad \color{blue}{ (4\leq N\leq 5)}\\ \alpha_1+\alpha_2+\alpha_3+\alpha_4&\ne\alpha_5 &\color{blue}{\alpha_1+\alpha_2+\alpha_3+\alpha_4\ne \alpha_N+\cdots+\alpha_5}\\ &&\qquad\qquad \color{blue}{ (5\leq N\leq 5)}\\ \end{align*}

We see in case $K=6$ we have $4$ groups, each group has constraints with equal terms at the left-hand side. On the right-hand side we have blue marked formulas, one for each group.

We can now derive a formula which describes all the constraints, namely \begin{align*} \alpha_1+\cdots+\alpha_M\ne \alpha_N+\cdots+\alpha_5\qquad\qquad 1\leq M<N\leq 5\tag{1} \end{align*}

We obtain from (1) a formula describing all constraints for general $K\geq 3$, namely \begin{align*} \color{blue}{\alpha_1+\cdots+\alpha_M\ne \alpha_N+\cdots+\alpha_{K-1}\qquad\qquad 1\leq M<N\leq K-1} \end{align*} or using sigma notation: \begin{align*} \color{blue}{\qquad\qquad\sum_{m=1}^M\alpha_m\ne\sum_{n=N}^{K-1} \alpha_n\qquad\qquad\qquad\qquad 1\leq M<N\leq K-1} \end{align*}

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