# 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

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