chemia - Studia

MAHBOUBEH SAHELI, ALI REZA ASHRAFI, MIRCEA V. DIUDEA
A 6
Matrix
0 0 0 0 0 0 0 0 0 1 1 0
0 0 0 0 0 0 1 1 1 1 1 0
0 0 0 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1
0 0 1 1
1 1 1 1
A'' 6
Matrix
1 1 1 1
0 1 1 1
0 1 1 1
0 1 1 1
0 0 1 1
0 0 1 1
0 0 1 1
0 0 0 1
0 0 0 1
0 0 0 1
1 0 0 0
1 0 0 0 A' 6
Matrix
1 1 0 0
1 1 0 0
1 1 1 0
1 1 1 0
1 1 1 1
1 1 1 1
1 1 1 0
1 1 0 0
1 1 0 0
1 0 0 0
236
Figure 2. Construction of the Matrices A 6 , A′
6 and A′
6′
and so,
Ω(
G[
n],
x)
= 6 n ⎡n
/ 3⎤
3 j 1 ⎡ j / 2
∑ − ⎤
j 1
x
− +
=
2 2 ⎡ / 3 1
⎤
( 2 2 ⎡ / 3⎤)
6 n n S S n j n
j 1
x
j j
∑ − + + ′ + ′′ + + − +
+
=
+ 3x
4n−2+
2⎡n
/ 3⎤
Thus for computing the omega polynomial of G[n], it is enough to
compute S′
i and S i′ . By a simple calculations, one can see that Ω(G[1]) = 6x 3
+ 3x 4 and Ω(G[2]) = 6x 3 + 6x 6 + 15x 8 . So, we can assume that n ≥ 3. Our main
proof consider three cases that n ≡ 0 (mod 3), n ≡ 1 (mod 3) and n ≡ 2 (mod 3).
We first assume that n ≡ 0 (mod 3). In this case the number of rows
in the big hexagons is 7n/3 – 2. By definition of A n , if 1 ≤ j ≤ 4n/3 – 2 then
we have S′ j = ⎡j
/ 2⎤.
If 4n/3 – 1 ≤ j ≤ 7n/3 – 2 then we can define
j = 4 n / 3 − 2 + k , where 1 ≤ k ≤ n. Thus,
‣ S i
′ = 2n
/ 3 − (2k
− 2),
where k ≡ 1 or 2 (mod 3),
‣ S i
′ = 2n
/ 3 − (2k
−1),
where k ≡ 0 (mod 3).
To compute S ′
j
, we consider four cases that 1 ≤ j ≤ n/3 – 1, j = n/3,
j = n/3 + 1 and n/3 + 2 ≤ j ≤ 7n/3 – 2. In the first case S′ j = 2 j,
and for the
second and third cases we have S ′ j = 2n / 3.
For the last case, we assume
that j = n/3 + k + 1, 1 ≤ k ≤ 2n – 3. Then S′
j = n − n / 3 − ⎡k
/ 3⎤=
2n
/ 3 − ⎡k
/ 3⎤.
To compute the omega polynomial, we define the following polynomials:
1) ε / 3 1 3 1 7 / 3 ⎡ / 2⎤ 1 = ∑n − x
j − + n + j ,
j = 1