For each $k$ , find the least $n$ in terms of $k$ st the following holds: There exists $n$ real numbers $a_1 , a_2 ,\cdot \cdot \cdot , a_n$ st for each $i$ : $$0 < a_{i+1} - a_{i} < a_i - a_{i-1}$$And , there exists $k$ pairs $(i,j)$ st $a_i - a_j = 1$.
Problem
Source: Iran MO 3rd round 2023 , Day 1 P3
Tags: combinatorics
alinazarboland
16.08.2023 17:39
2$k$
First , note that each of the elements $a_i$ can be written as sum of some consecutive terms of numbers of the form $b_i = a_{i+1} - a_i$ So if $t$ is the largest number st $b_t+b_{t+1}+...+b_s = 1$ , by the condition , $s-t$ is the maximal among all other consecutive terms with sum 1 since $t$ was maximal. So $s = (s-t) + t > k-1 + k$ which means $n \geq 2k$.
Let $a_1=1 , a_2=2 $ at $i$-th step , consider another number that has difference one with the largest number you have(and larger than the current maximum , of course) and put their mean $+10^{-i}$ between them as another element of the sequence where we have add to new numbers to the sequence. One can easily check that this works and we're done.