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2
Sum [kn^(1/3)]
Sum [kn^(1/3)]
Source: Romanian 2018 TST Problem 2 Day 3
May 25, 2020
number theory
floor function
series summation
Problem Statement
Given a square-free integer
n
>
2
n>2
n
>
2
, evaluate the sum
∑
k
=
1
(
n
−
2
)
(
n
−
1
)
⌊
(
k
n
)
1
/
3
⌋
\sum_{k=1}^{(n-2)(n-1)} \lfloor ({kn})^{1/3} \rfloor
∑
k
=
1
(
n
−
2
)
(
n
−
1
)
⌊(
kn
)
1/3
⌋
.
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