chemia - Studia

G. H. FATH-TABAR, F. GHOLAMI-NEZHAAD
To compute the Tutte polynomial of Ns[n], we proceed inductively
but at first, we compute T(Ns[0], x, y) in the following
6
⎛ x − x ⎞
3
Lemma 2. T(Ns[0], x, y) = ⎜ + y x .
x 1
⎟
⎝ − ⎠
Proof. Suppose H 1 , H 2 and H 3 are hexagons in Ns[0]; then
5
T(Ns[0], x, y) = x T(Ns[0] - H
1, x, y) + T(Ns[0] - H1 + C
5, x, y)
= x T(Ns[0] - H , x, y) + x T(Ns[0] - H , x, y) +
5 4
1 1
T(Ns[0] - H + C , x, y)
1 4
= x T(Ns[0] - H , x, y) + x T(Ns[0] - H , x, y) +
5 4
1 1
3
3
x T(Ns[0] - H
1, x, y)+T(Ns[0] - H1 + C
3, x, y)
6
⎛ x − x ⎞
= ⎜ + y⎟T( Ns[0] −H1, x, y),
⎝ x −1
⎠
where Ns[0]-H 1 + C i is constructed from Ns[0] by removing H 1 and replacing
C i . As we did in the above,
6
⎛ x − x ⎞
T(Ns[0], x, y)= ⎜ + y T ( Ns[0]
− H1
− H
2
, x,
y).
x 1
⎟
⎝ − ⎠
3
6
⎛ x − x ⎞
Thus, T(Ns[0], x, y) = ⎜ + y T ( Ns[0]
− H1
− H
2
− H
3,
x,
y).
x 1
⎟
This
⎝ − ⎠
implies that
6
⎛ x − x ⎞
3
T(Ns[0], x, y) = ⎜ + y x .
x 1
⎟
⎝ − ⎠
6
15⎛
x − x ⎞
Lemma 3. T(Ns[1], x, y) = x
⎜ + y .
1
⎟
⎝ x − ⎠
Proof. By a similar proof as Lemma 2, we can see that
3
6
⎛ x − x ⎞
12
T(Ns[1], x, y) = ⎜ + y x T ( Ns[0],
x,
y).
x 1
⎟
⎝ − ⎠
6
15⎛
x − x ⎞
Thus, T(Ns[1], x, y) = x
⎜ + y .
1
⎟
⎝ x − ⎠
6
n + 1
2×
4 + 7 ⎛ x − x ⎞
Theorem 4. T( Ns[n], x, y ) = x
⎜ + y .
1
⎟
⎝ x − ⎠
134
2
12
12
9
n
9×
2 −6