Fullerenes 1

Fullerene counting
By substituting vv, , ee and ff (as below) in (3) one obtains:
3 v = ∑
s ⋅ f e
s s = 2
f
∑
=
s
∑
f
s s
s) f s
( 6 − = 12(
1−
g)
(1’-2’) (1’ 2’)
•For For a given genus of the surface, (5) ( ) gives the number of s-polygons. polygons.
This condition is independent of the number of hexagons, hexagons,
which is
therefore arbitrary. arbitrary
•Special Special cases are the Platonic Platonic tilings, tilings,
with a single kind of polygons, polygons,
and the Archimedean Archimedean tilings, tilings,
with two different kinds of polygons, polygons,
one
of which being here the hexagon.
•In In Platonic fullerenes (gg = 0): from (5), 5 ff =12, or
Archimedean fullerenes must always contain 12 5; ff thus
from (1’-2’): (1 ): 5ff 5 + 6 6ff 6 = 60+ 6 6 = 3 3v v ; 6 = (v/2)-10 (v/2) 10.
= 60+ 6ff 6
; ff 6
(4)
(5)
or 4 ff = 6 or 3 ff = 4.
; thus 6 ff comes out
20