Colouring a cycle
Brendan McKay
bdm at cs.anu.edu.au
Mon Mar 26 05:51:26 CEST 2007
We know that the chromatic polynomial of an n-cycle is
P(x) = (x-1)^n + (-1)^n (x-1).
So P(k) is the number of proper colourings possible when
k colours are available.
But suppose I want to know how many colourings use
colour 1 n[1] times, colour 2 n[2] times, ... where
n[1]+...+n[k] = n. How many colourings are like that?
Brendan.
Joerg, Thanks for that message. All the links
to crowdog have been erased. New version of OEIS in 10 minutes.
Neil
* N. J. A. Sloane <njas at research.att.com> [Mar 26. 2007 10:29]:
>
> Peter Pein <petsie at dordos.net> reports that
>
> > > LINKS Creighton Dement, Table of n, a(n) for n = 1..184
> > >
> > > Creighton Dement, Additional Floretion Ray-traced Plots
> > >
> > Windows user should be very carefull with this link. The site tries to install
> > a trojan (avast calls it Win32:Agent-EQD [Trj], others name it Win32.Agent.aev)
>
>
> Me: this is referring to the page
>
> Creighton Dement, <a href="http://crowdog.de/45586.html">Additional Floretion Ray-traced Plots</a>
>
> which apparently no longer exits. Is that correct, Creighton?
>
> Then I should delete all your links to this file and to
> similar ones?
>
> Neil
Domain is apparently taken by a domain grabber, see
Anything at crowdog.de get redirected to
http://www.ndparking.de/crowdog.de (possibly installs MALWARE!)
I'd remove all links to that domain ASAP.
regards, jj
(This message, aside from serving its own purpose, is partially a test to
moment.)
As I said above, the OEIS is down, so I don't know if this sequence (two
Let a(1) = 2. Let a(n) be such that the continued fraction (of +-rational
terms)
[a(1);a(2),...,a(n)]
= sum{k=1 to n-1} 1/a(k), for every integer n >= 2.
So, we get the sequence of rationals beginning:
{a(k)}: 2, -2/3, 3, 7/3, -16/27, 141/49, -3023/768,...
So, for example:
1/2 -3/2 + 1/3 = 2 + 1/(-2/3 +1/(3 + 3/7)),
and
1/2 -3/2 + 1/3 + 3/7 = 2 + 1/(-2/3 +1/(3 + 1/(7/3 - 27/16))).
If I am right, then, for n >= 5,
a(n) = - (a(n-1) + a(n-2)) * (a(n-2) + a(n-3)) /(a(n-1) * a(n-2)^2).
What I wonder is, is this sequence infinite?
In other words, does a(m) not equal -a(m+1) for every positive integer m?
Thanks,
Leroy Quet
I have reported the problem to the systems people
Neil
More information about the SeqFan
mailing list