# [seqfan] Re: Seqfan Digest, Vol 20, Issue 15

Brendan McKay bdm at cs.anu.edu.au
Mon May 24 14:09:31 CEST 2010

> Date: Sun, 23 May 2010 17:51:17 -0400
> From: franktaw at netscape.net
> Subject: [seqfan]  G.f. for trees with degree at most 3
> To: seqfan at list.seqfan.eu
> Message-ID: <8CCC8D935422E1B-18B0-FD4B at webmail-m028.sysops.aol.com>
> Content-Type: text/plain; charset="utf-8"; format=flowed
>
> I asked the author of this sequence for clarification, but got no
> response. Maybe somebody here can figure out what is going on.
>
> http://www.research.att.com/~njas/sequences/A003692?
> ?
> For this sequence, a generating function is given:
>
> (1-x)(2-x-x^2) - (2-x+x^2)\sqrt{1-2x-x^2} \over 3 x^3.?
> ?
> I'm not sure if this is supposed to be?
> ?
> (1-x)*(2-x-x^2)-(2-x-x^2)*sqrt(1-2*x-x^2)/(3*x^3)?
> ?
> or?
> ?
> ((1-x)*(2-x-x^2)-(2-x-x^2)*sqrt(1-2*x-x^2))/(3*x^3),?
> ?
> but either one produces a series that includes terms with negative
> exponents; and in neither case there is any apparent relationship to
> this sequence. So what is the correct g.f.??
> ?
> Franklin T. Adams-Watters?

The number of labelled trees with d[i] vertices of degree i
for i=1,2,3 is
(n-1)! * n! / (2^d[3] * d[3]! * d[2]! * d[1]!).
Now sum over d[1]+d[2]+d[3]=n, d[1]+2*d[2]+3*d[3] = 2n-2.

Brendan.