Polynomial sequence?

[Nick Hobson]

Apologies if you've received this message twice.

Thanks everyone.

I submitted the sequence without any reference to rooted trees, which I'm  
not too familiar with.  If anyone wants to submit a comment for the  
sequence, it is A126130.

Nick


On Tue, 19 Dec 2006 17:33:54 -0000, <franktaw at netscape.net> wrote:

> Actually, that's off by one.  The number of labelled rooted trees is  
> n^(n-1), not (n+1)^n.
>
> What it actually is is something better.  n! is the number of increasing  
> rooted trees on n+1 nodes - that is, trees where each node label is  
> greater than the label of its parent.  So a(n) is the number of labelled  
> rooted trees on n+1  nodes that are not increasing.
>
> Based on this and your original description, I would recommend changing  
> the offset:  a(n) = n^(n-1) - (n-1)!.  Then a(n) = f(n) in the original  
> exposition,  not f(n+1), and a(n) is the number of labelled rooted trees  
> on n nodes that are not increasing.  You want to add an initial a(1) = 0.
>
> Franklin T. Adams-Watters
>
>
> -----Original Message-----
> From: joshua.zucker at gmail.com
>
> If I understand correctly, the observation below also means that it is 
> the number of labeled rooted trees with n nodes except for the ones 
> that are some particular tree (say the one where the nodes are all in 
> a straight line)? 
>  
> --Joshua Zucker 
>  
> On 12/17/06, Jonathan Post <jvospost3 at gmail.com> wrote: 
>> a(n) = (n+1)^n - n! = A000169(n) - A000142(n). 
>>  
>> Since A000169 is the number of labeled rooted trees with n nodes, you 
>> should be able to come up with a pure graph enumeration description
> of your 
>> sequence. When you do, I agree that it is nice.