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Let $a_1,...,a_n $ positive integers s.t. $a_1+...+a_n=2011$, where $n $ is a positive integer, $n\geq3$ and $a_1\leq |a_2-a_3|$, $a_2\leq|a_3-a_4|$,..., $a_n\leq|a_1-a_2|$.

What can I say about the minimum and the maximum value of $n $?

I notice that the minimum value of $n $ is greater then 3. They are equal?

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  • $\begingroup$ I think that the sequence will not exist. Let $a_i$ be the largest element in this sequence, then $a_i \geq a_{i+1}$ so $a_i > a_{i+1} - 1 \geq |a_{i+1} - a_{i+2}|$. $\endgroup$ – Hw Chu Feb 16 '18 at 0:31
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(Expanding on Hw Chu's comment): There is no solution at all, not even with real numbers $a_i > 0$.

Without loss of generality (by rotating the indices if necessary) assume that $a_1$ is the largest number in $a_1, \ldots, a_n$.

Then $$ a_1 \le |a_2 - a_3| = \max(a_2, a_3) - \min(a_2, a_3) < a_1 - 0 = a_1 $$ gives a contradiction.

Remark: There are solutions if the condition “positive integers” is relaxed to “non-negative integers”, see Solving $a_1+...+a_n=2011$ , $a_1\leq |a_2-a_3|, \ldots a_n\leq|a_1-a_2|$ with non-negative integers.

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