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F
m
=
m
1
r
=
2
(
a
(
n
)
−
a
(
n
)
11
a
(
n
)
12
a
(
n
)
a
(
n
)
rr
a
(
n
)
mm
a
(
n
)
n
+
+
+
+
)+
+
rr
mm
,
rr
−
1
+
1
−
1
m
−
1
r
=
2
a
(
n
+
1
)
a
(
n
+
1
)
11
a
(
n
+
1
)
Trace
of
M
(
n
+
1
)
=
Trace
of
B
n
=
+
+
=
rr
mm
m
k
=
1
(
1
−
2cosβ
k
)
m
h
=
1
sin
2
2
n
=
(
h
β
k
)
+
m
1
and from Lemma
2
it follows
F
n
,
m
=
m
k
=
1
(
1
−
2cosβ
k
)
2
n
,
(7.21)
which establishes the formula.
n
,
m
and
n
,
m
7.3 The Cardinality of
F
F
n
,
m
and
n
,
m
. The elements of
n
,
m
are the func-
Let us consider now the sets
F
F
F
tions
A
∈ F
n
,
m
satisfying
Δ
A
(
i
)
∈{
0
,
1
,−
1
,
m
−
1
,
1
−
m
},
for
i
=
1
,
2
,...,
n
−
1,
n
,
m
is the subset of
n
,
m
set up by the lattice paths satisfying the additional
and
F
F
condition
Δ
A
(
n
)=
A
(
1
)
−
A
(
n
)
∈{
0
,
1
,−
1
,
m
−
1
,
1
−
m
},
3
n
,
m
1
n
,
m
so the relation
F
⊂ F
⊂ F
n
,
m
is satisfied.
n
m
is equal to
3
n
−
1
mifm
3
, and to
2
n
Theorem 3.
The cardinality of the set
F
,
for m
=
2
.
n
,
m
we define the
n
values
Proof.
First let us assume
m
3. For every
A
∈ F
a
,
x
1
,
x
2
,...,
x
n
−
1
to be
⎧
⎨
Δ
A
(
i
)
,
if
|
Δ
A
(
i
)
|≤
1
,
a
=
A
(
1
)
,
x
i
=
−
1
,
if
Δ
A
(
i
)=
m
−
1
,
⎩
1
,
if
Δ
A
(
i
)=
1
−
m
,
so
a
∈{
1
,
2
,...,
m
}
and
x
i
∈{
0
,−
1
,
1
},
i
=
1
,
2
,...,
n
−
1
.
n
There is a bijection between the set of sequences
a
,
x
1
,
x
2
,...,
x
n
−
1
and
F
m
,and
,
clearly the number of different possibilities for the sequences
a
,
x
1
,
x
2
,...,
x
n
−
1
is
equal to
m
3
n
−
1
,
which is the desired conclusion.
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