MathDB
Problems
Contests
National and Regional Contests
Netherlands Contests
Dutch IMO TST
2008 Dutch IMO TST
3
3
Part of
2008 Dutch IMO TST
Problems
(1)
\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
1/10/2020
Let
m
,
n
m, n
m
,
n
be positive integers. Consider a sequence of positive integers
a
1
,
a
2
,
.
.
.
,
a
n
a_1, a_2, ... , a_n
a
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,
b
i
=
b_i =
b
i
=
# \{ j \in \{1, 2, ... , n\}: a_j \ge i\}, so
b
i
b_i
b
i
is the number of terms
a
j
a_j
a
j
of the given sequence for which a_j \ge i. Similarly, we define, for 1\le j \le n,
c
j
=
c_j=
c
j
=
# \{ i \in \{1, 2, ... , m\}: b_i \ge j\} , thus
c
j
c_j
c
j
is the number of terms bi in the given sequence for which b_i \ge j. E.g.: If
a
a
a
is the sequence
5
,
3
,
3
,
2
,
1
,
1
5, 3, 3, 2, 1, 1
5
,
3
,
3
,
2
,
1
,
1
then
b
b
b
is the sequence
6
,
4
,
3
,
1
,
1
6, 4, 3, 1, 1
6
,
4
,
3
,
1
,
1
. (a) Prove that
a
j
=
c
j
a_j = c_j
a
j
=
c
j
for 1 \le j \le n. (b) Prove that for 1\le k \le m:
∑
i
=
1
k
b
i
=
k
⋅
b
k
+
∑
j
=
b
k
+
1
n
a
j
\sum_{i=1}^{k} b_i = k \cdot b_k + \sum_{j=b_{k+1}}^{n} a_j
∑
i
=
1
k
b
i
=
k
⋅
b
k
+
∑
j
=
b
k
+
1
n
a
j
.
combinatorics
Sequence
Sum