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THE OMEGA POLYNOMIAL OF THE CORCOR DOMAIN OF GRAPHENE<br />

One can see that the number of rows inside and outside big<br />

hexagons are equal to<br />

⎛ ⎡ n ⎤ ⎡n<br />

⎤ ⎞<br />

⎡n<br />

⎤<br />

⎜6 + 4(n − ) ⎟ − 4 = 4n + 2 − 4 and<br />

3<br />

3<br />

3<br />

⎝ ⎢ ⎥ ⎢ ⎥ ⎠<br />

⎢ ⎥<br />

⎡n<br />

⎤<br />

⎡n<br />

⎤<br />

6 n − 4 − (4n + 2<br />

− 4) = 2n − 2 , respectively. From Figure 1, one can see<br />

⎢ 3⎥<br />

⎢3<br />

⎥<br />

that the molecular graph of CorCor[n] can be partitioned into six equal parts<br />

with the same number of hexagons. If we consider one half of this graph<br />

then three cases of these six parts must be considered. Define three<br />

matrices and as follows:<br />

• A is an<br />

⎛ ⎡n⎤⎞<br />

⎛ ⎡n⎤⎞<br />

⎜n −<br />

⎟× ⎜2n<br />

− 2 +<br />

⎟ matrix with 0 & 1 entries.<br />

⎝ ⎢3⎥⎠<br />

⎝ ⎢3⎥⎠<br />

The entries corresponding to the hexagons of CorCor[n] are<br />

equal to 1, and other entries are zero, see Figure 2. As an example,<br />

the matrix A 6 is as follows:<br />

⎡0<br />

0 0 0 0 0 0 0 0 1 1 0⎤<br />

⎢<br />

⎥<br />

= ⎢<br />

0 0 0 0 0 0 1 1 1 1 1 0<br />

A6<br />

⎥<br />

⎢0<br />

0 0 1 1 1 1 1 1 1 1 1⎥<br />

⎢<br />

⎥<br />

⎣1<br />

1 1 1 1 1 1 1 1 1 1 1⎦<br />

• Suppose A = [a ij ], A′ = [b ij ] is an<br />

⎛ ⎡n⎤⎞<br />

⎛ ⎡n⎤⎞<br />

⎜2n − 2 +<br />

⎟× ⎜n<br />

−<br />

⎟ matrix<br />

⎝ ⎢3⎥⎠<br />

⎝ ⎢3⎥⎠<br />

defined by<br />

⎧<br />

⎡n⎤<br />

⎪a<br />

⎛ ⎡n⎤<br />

⎞<br />

i + j ≤ 2n<br />

−1+<br />

⎪ ⎜n−<br />

( )<br />

⎢ ⎥<br />

⎟ + −<br />

⎝<br />

− j+<br />

1 i j 1<br />

3<br />

⎢3⎥<br />

⎠<br />

A″ = [c<br />

=<br />

ij ] is an<br />

bij ⎨<br />

⎪<br />

⎡n⎤<br />

a<br />

+ > − +<br />

⎪ ⎛ ⎡n⎤<br />

⎞⎛<br />

⎡ ⎤⎞<br />

i j 2n<br />

1<br />

⎜ − + ⎟⎜<br />

+ − + − ⎟<br />

⎢ ⎥<br />

⎪<br />

⎩ ⎝ ⎢ ⎥<br />

−<br />

n<br />

n j 1 i j 2n<br />

1<br />

3<br />

3 ⎠⎝<br />

⎢3⎥⎠<br />

⎛ ⎡n⎤⎞<br />

⎛ ⎡n⎤⎞<br />

⎜2n − 2 +<br />

⎟× ⎜n<br />

−<br />

⎟ matrix defined by c ij = a j ⎛ ⎡ n ⎤ ⎞ .<br />

⎜ 2n−1+<br />

⎟<br />

⎝ ⎢3⎥⎠<br />

⎝ ⎢3⎥⎠<br />

⎝ ⎢ ⎥<br />

− i<br />

3 ⎠<br />

It is easy to see that the number of hexagons in the central row of<br />

⎛<br />

⎞<br />

G[n] is 2⎜<br />

⎡n⎤<br />

⎛ ⎡n⎤⎞<br />

⎡n⎤<br />

3 2 ⎟<br />

⎟ − 3 = 4 + 2 − 3<br />

3<br />

+ ⎜n<br />

−<br />

3<br />

n . Suppose S′ ⎢3<br />

i and S i′ denote the<br />

⎝ ⎢ ⎥ ⎝ ⎢ ⎥⎠⎠<br />

⎥<br />

summation of all entries in the i th row of the matrices A′ and A″, respectively.<br />

Then<br />

S′ = ∑n − − ⎡ ⎤<br />

= 1 n / 3<br />

i j 0 a( n − j−⎡n<br />

/ 3⎤)<br />

( i+<br />

j ) and S ′ = ∑ n = − ⎡n<br />

/ 3<br />

i<br />

⎤ j 1<br />

a j 2n−i−1<br />

+ n / 3<br />

( ⎡ ⎤)<br />

235

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