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s
1
=
s
2
=
s
1
−
1
,
s
2
+
1
s
i
=
s
i
=
0for
i
=
3
,
4
,...,
n
and from Lemma
3
, with
h
=
1and
k
=
2 it is obtained that
f
(
s
1
,
s
2
,...,
n
)
−
s
1
,
s
2
,...,
s
n
)=(
(
−
−
)
>
.
(
,
,...,
)
is not a minimum.
From the above considerations we obtain that the minimum of
f
is either
f
s
1
s
2
1
0
Therefore
f
s
1
s
2
s
n
q
−
1
n
−
q
−
1
(
,
,...,
)=(
−
(
+
))(
−
−
)
(
−
)
,
f
s
1
s
2
s
n
ms
np
p
1
m
p
1
m
p
m
n
−
2
with the values
s
1
,
s
2
,...,
s
n
defined by (
3.18
), or
f
(
s
1
,
s
2
,...,
s
n
)=(
ms
−
2
ab
)
,
where the values
s
1
,
s
n
are defined by (
3.22
). The second part of the theorem
follows straightforwardly from Lemma
4
, and the proof is complete.
s
2
,...,
Corollary 1.
Let s, n, m, be positive integers, n
≥
3
and suppose that
2
m
3
[
s
/
2
]
≤
(3.25)
Then the minimum of function f defined by (
3.14
) for integers s
1
,
s
2
,...,
s
n
, satisfy-
ing (
3.15
) is K defined by (
3.23
) achieved with the values s
i
=
s
i
defined by (
3.18
).
Proof.
From Theorem
4
it follows that the minimum of function
f
is achieved with
the values
s
1
,
s
2
,...,
s
n
defined by (
3.18
), or with the values
s
1
,
s
2
,...,
s
n
defined
by (
3.22
). But if (
3.25
) is satisfied it follows that
n
−
1
i
=
2
s
i
b
[
s
/
2
]
s
i
=
b
=
]
≤
2
m
−
m
−
m
−
[
s
/
2
(
,
,...,
)
is satisfied. Then, from Lemma
4
, it follows that
f
is not a minimum,
therefore the minimum is
K
defined by (
3.23
) achieved with the values
s
i
s
1
s
2
s
n
=
s
i
de-
fined by (
3.18
) and the proof is complete.
Now we can solve the OEIG. We will write
{
(
i
,
j
)
:1
≤
j
≤
p
+
1
,
for
i
≤
q
,
1
≤
j
≤
p
,
for
i
>
q
},
(3.26)
1
m
(
np
(
p
+
1
)
p
+
1
p
m
)
q
−
1
n
−
q
−
1
H
=
s
−
)(
1
−
)
(
1
−
(3.27)
m
m
where
p
=[
s
/
n
]
and
q
=
s
−
np
,
{
(
i
,
j
)
:1
≤
j
≤
a
,
for
i
=
1
,
1
≤
j
≤
b
,
for
i
=
2
}
(3.28)
1
m
2
(
H
=
ms
−
2
ab
)
(3.29)
where
b
=[
s
/
2
]
and
a
=
s
−
b
.
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