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Indonesia TST
2012 Indonesia TST
4
a_n = sum of a_floor(n/k) + 1
a_n = sum of a_floor(n/k) + 1
Source: 2012 Indonesia Round 2.5 TST 2 Problem 4
May 21, 2012
floor function
number theory unsolved
number theory
Problem Statement
The sequence
a
i
a_i
a
i
is defined as
a
1
=
1
a_1 = 1
a
1
=
1
and
a
n
=
a
⌊
n
2
⌋
+
a
⌊
n
3
⌋
+
a
⌊
n
4
⌋
+
⋯
+
a
⌊
n
n
⌋
+
1
a_n = a_{\left\lfloor \dfrac{n}{2} \right\rfloor} + a_{\left\lfloor \dfrac{n}{3} \right\rfloor} + a_{\left\lfloor \dfrac{n}{4} \right\rfloor} + \cdots + a_{\left\lfloor \dfrac{n}{n} \right\rfloor} + 1
a
n
=
a
⌊
2
n
⌋
+
a
⌊
3
n
⌋
+
a
⌊
4
n
⌋
+
⋯
+
a
⌊
n
n
⌋
+
1
for every positive integer
n
>
1
n > 1
n
>
1
. Prove that there are infinitely many values of
n
n
n
such that
a
n
≡
n
m
o
d
2012
a_n \equiv n \mod 2012
a
n
≡
n
mod
2012
.
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