Re: Silly question - Request for quick answer

In article <1168794936.210036.203730@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"babu" <nasif4003@xxxxxxxxx> wrote:

Can anybody give information about is there any theorem/formula for the
following..
if number of vertex is n then how many 3 connected cubic planar
graphs are possible?

For example, if n=4 then number of graphs possible=1
if n=6 then number of graphs possible=1
I found a link given below:

http://people.csse.uwa.edu.au/gordon/remote/cubicplanar/viewgraph.html
But here no formula is provided for n vertex.

Maybe nobody knows one.

Have you tried searching the online encyclopedia of integer
sequences (http://www.research.att.com/~njas/sequences/)?

I see they have the sequence

A002851
Number of unlabeled trivalent (or cubic) connected graphs with 2n nodes.
1, 2, 5, 19, 85, 509, 4060, 41301, 510489, 7319447, 117940535,
2094480864, 40497138011, 845480228069, 18941522184590, 453090162062723,
11523392072541432, 310467244165539782, 8832736318937756165

and the sequence

A058931
Number of 3-connected claw-free cubic graphs with 2n nodes.
0, 1, 60, 0, 0, 19958400, 0, 0, 622452999168000, 0, 0,
258520167388849766400000, 0, 0, 675289572271869736778268672000000, 0, 0,
7393367369949286697176489031997849600000000, 0, 0

(together with lots of information on the sequences); maybe
if you look hrad enough at that site you'll find what you want.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)
.

Relevant Pages