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s i (
m
s h )(
m
s k )
s i (
m
s h )(
m
s k )
i
P
P
m
s i
m
s i
i
=
h
,
k
=
h
,
k
s i
=
i
s i (
s h
s k
)
.
1
P
m
=
h
,
k
Then ( 3.17 ) can be rewritten as
s i
i
(
s h
s k
1
)
P
(
2
s i )
m
=
h
,
k
and the proof is finished.
To state the following lemma we will write
s i =
p
+
1
,
for i
q
,
(3.18)
s i =
p
for i
>
q
.
where
p
=[
s
/
n
]
and
q
=
s
pn
,
(3.19)
Lemma 4. Let s and m be integers, 0
<
s
<
2 m. The minimum of function f defined
by ( 3.14 ) for integers s 1 ,
s 2 ,...,
s n satisfying ( 3.15 ) and
n
1
i = 2
s i
s i
2
,
(3.20)
m
is achieved with the integers s i =
s i defined by ( 3.18 ) and it is equal to
q
1
(
n q
1
.
(
ms
np
(
p
+
1
))(
m
p
1
)
m
p
)
(3.21)
with p and q defined by ( 3.19 ).
Proof. Let
S
be the set defined by
2
n
i = 1 s i = s ,
n
1
i = 2
s i
S =
{
s 1
,
s 2
,...,
s n
}
: s i
integers
,
0
s i
m
,
s i
m
We want to obtain the minimum
min
f
(
s 1 ,
s 2 ,...,
s n ) .
(
s 1 ,
s 2 ,...,
s n ) ∈S
Given a set of integers
{
s 1 ,
s 2 ,...,
s n }∈S
, the symmetry of the function ( 3.14 )
allows us to assume that
s 1
s 2 ≥ ...≥
s n .
If s 1
s n =
0or s 1
s n =
1, then s i should be equal to s i (for each index i ). If
s 1 ,
s 2 ,...,
s n }∈S
s 1
s n
2, we can build the new set of integers
{
,by
 
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