MathDB
\sum_{i=1}^{k} b_i = k \cdot b_k + \sum_{j=b_{k+1}}^{n} a_j, sequences

Source: 2008 Dutch IMO TST P3

January 10, 2020
combinatoricsSequenceSum

Problem Statement

Let m,nm, n be positive integers. Consider a sequence of positive integers a1,a2,...,ana_1, a_2, ... , a_n that satisfies m = a_1 \ge a_2\ge ... \ge a_n \ge 1. Then define, for 1\le  i\le  m, bi=b_i = # \{ j \in \{1, 2, ... , n\}: a_j \ge i\}, so bib_i is the number of terms aja_j of the given sequence for which a_j  \ge i. Similarly, we define, for 1\le   j \le  n, cj=c_j= # \{ i \in \{1, 2, ... , m\}: b_i \ge j\} , thus cjc_j is the number of terms bi in the given sequence for which b_i \ge j. E.g.: If aa is the sequence 5,3,3,2,1,15, 3, 3, 2, 1, 1 then bb is the sequence 6,4,3,1,16, 4, 3, 1, 1. (a) Prove that aj=cja_j = c_j for 1  \le j  \le n. (b) Prove that for 1\le  k \le m: i=1kbi=kbk+j=bk+1naj\sum_{i=1}^{k} b_i = k \cdot b_k + \sum_{j=b_{k+1}}^{n} a_j.