Fullerenes 1

Fullerene counting
Rewrite relations (1), (2), and (3) as:
• 3vv = 2 2ee (1’)
•
•
5ff 5 + 6 6ff 6 = 2 2ee
v v + ff = 2 + ee
(2’)
(3’)
• Substituting v in (3’) by its value from (1’) one can write:
• (2/3) (2/3)ee + f f = 2 + ee
• 2ee + 3 3ff = 6 + 3 3ee
• ee = 3 3f f – 6 (4)
• Expressing ff by its composition: ( (ff = 5 ff + f 6), ), relation (4) becomes:
• ee = 3( 3(ff 5 + 6) ff ) – 6 (5)
• Substituting e from (5) in (2’) one obtains:
• 5ff 5 + 6 6ff 6 = 6( 6(ff 5 + 6) ff ) – 12
• 5 ff = 12 (6)
• From (1’), (2’), and (6) the expression for 6 ff is obtined:
• 5ff 5 + 6 6ff 6 = 3 3vv
• 6 ff = vv/2 /2 – 10 (7)
19