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n
1
i
=
2
−
s
i
s
i
>
2
m
−
and let
l
be the last index for which
s
l
>
0, that is
s
i
>
0for
i
=
1
,
2
,...,
l
and
s
i
=
0
=
+
,
+
,...,
>
<
for
i
l
1
l
2
n
. Let us suppose
l
2. We distinguish two cases,
s
1
m
and
=
<
s
1
m
.If
s
1
m
, then we can define the new set of integers
s
1
=
s
l
=
1and
s
i
=
s
1
+
1
,
s
l
−
s
i
i
=
1
,
and
i
=
l
,
and applying Lemma
3
with
h
=
l
and
k
=
1 it follows that
s
i
s
1
,
s
2
,...,
s
n
)=(
−
∑
i
=
(
,
,...,
)
−
(
s
l
−
−
)
(
s
i
)
f
s
1
s
2
s
n
f
s
1
1
P
2
m
−
1
,
l
n
1
i
=
1
−
s
i
≥
(
s
l
−
s
1
−
1
)
P
(
2
−
s
i
)
.
m
−
n
−
1
s
i
Bearinginmind
s
1
≥
s
l
,
P
>
0 nd
m
−
s
i
>
2 it follows that
f
(
s
1
,
s
2
,
∑
i
=
2
s
1
,
s
2
,...,
s
n
)
>
...,
s
n
)
−
f
(
0
.
Therefore
f
(
s
1
,
s
2
,...,
s
n
)
is not a minimum.
If
s
1
=
m
, bearing in mind that
s
<
2
m
we can assert
s
2
<
m
, then we can define
the new set of integers
s
2
=
s
l
=
1and
s
i
=
s
2
+
1
,
s
l
−
s
i
i
=
2
,
and
i
=
l
,
In this case, to compute the difference
s
1
,
s
2
,...,
s
n
)
f
(
s
1
,
s
2
,...,
s
n
)
−
f
(
we can not apply Lemma
3
because neither index 2 nor index
l
is equal to 1, but it
is not difficult to compute it directly,
s
1
,
s
2
,...,
s
n
)=
∑
i
(
s
i
j
=
i
(
m
−
s
j
)
−
s
i
j
=
i
(
m
−
s
j
))
f
(
s
1
,
s
2
,...,
s
n
)
−
f
(
l
1
j
=
2
(
m
−
s
j
))(
m
−
s
l
−
m
+
s
l
)
−
j
=
1
(
m
−
s
j
)
−
s
1
j
=
1
(
m
−
s
j
)=
m
(
=
s
1
l
−
1
j
=
2
(
m
−
s
j
)
>
0
=
m
therefore
f
(
s
1
,
s
2
,...,
s
n
)
is not a minimum. Now we can assert that, if
f
(
s
1
,
s
2
,
...,
s
n
)
is the minimum of function
f
,and
s
n
=
0then
s
3
=
s
4
=
...
=
s
n
=
0. More-
over, given a set of integers
s
1
,
s
2
,...,
s
n
satisfying (
3.15
)and
s
3
=
s
4
=
...
=
s
n
=
0,
if
s
1
−
s
2
≥
2 we can define the new set of integers
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