Conjugate m-dimensional partitions

[A.N.W.Hone]

Dear Franklin and Paul,

The functional equation (1), that is

A(x) = 1/A(-x A(x)),

seems really interesting. As I have pointed out, it has a formal
power series solution

A(x)= 1 +\sum_{n=1}^\infty a_j x^j

containing infinitely many free parameters, namely the odd
index coefficients a_1,a_3,... (with the even index ones all being
uniquely determined by these).

If a_1 = b is the only non-zero parameter, then one gets the solution

A(x) = 1/(1 + b x),

while if only a_1 and a_3 are non-zero one finds

A(x) = (1 + c x)/(1 + b x + c(b-c) x^2).

I guess there must be an infinite sequence of rational solutions

A_k(x) = P_k(x) / Q_k(x)

depending on k parameters, with P_k(0)=1=Q_k(0) and
deg P_k + 1 = k =  deg Q_k, and presumably one can choose the
parameters suitably to get an analytic function A_\infty(x) in some
disc centred at x=0.

It is also interesting to note that this functional equation has
(at least) a one parameter family of solutions that are not analytic at x=0,
but with a simple pole there, namely

A(x) = 1 - d/x  (d arbitrary).

Do any other seqfans know this functional equation?

I'm afraid I've lost the original thread on "conjugate
m-dimensional partitions", so I don't recall the significance
of the sequence 1,1,1,2,4,7,13,24,42,...

Anyway, it would be interesting to see if this sequence
has a generating function of the above form A_k(x) for finite k,
or whether it corresponds to the A_\infty case, and in that case
does it have a closed form expression?

Andy

On Wed, 10 May 2006 A.N.W.Hone at kent.ac.uk wrote:

> Not so unlikely! The two conditions are completely equivalent.
>
> Condition (2) can be rewritten as
>
> x*A(x)=f^(-1)(x),
>
> where f(x)=x*A(-x). Applying f to both sides gives
>
> f(x*A(x))=x
>
> or
>
> x*A(x)*A(-x*A(x))=x,
>
> which immediately gives condition (1):
>
> A(x) = 1/A(-x*A(x)).
>
> Actually, substituting a general series A(x)=a_0 + a_1*x + a_2*x^2 +...
> into either functional equation (1) or (2), one finds that the even terms are
> uniquely determined at each order by the previous ones, but the odd terms are
> arbitrary. So any formal series of the form
>
> A(x)=1 + a_1*x + a_1^2*x^2 + a_3*x^3  + (3*a_1*a_3-2*a_1^4)*x^4 + a_5*x^5
> +  (4*a_1*a_5+2*a_3^2-18*a_1^3*a_3+13*a_1^6)*x^6 + a_7*x^7 + ...
>
> satsifies the functional equation.
>
> The case you are interested in corresponds to a_1=1, a_3=2, a_5=7, a_7=24,...
> so some other consistent functional relation would be needed to uniquely
> determine the particular series
>
> 1+x+x^2+2*x^3+4*x^4+7*x^5+13*x^6+24*x^7+42*x^8+...
>
> Andy
>
>
> ----- Original Message -----
> From: franktaw at netscape.net
> Date: Tuesday, May 9, 2006 7:24 pm
> Subject: Re: Conjugate m-dimensional partitions
> To: seqfan at ext.jussieu.fr
>
> > It does seem unlikely.
> >
> > Is there a unique sequence that satisfies these two
> > conditions?  If so, I would think it a worthwhile thing to
> > add to the OEIS.  (If it's not there already - it might be
> > A002547.)
> > Franklin T. Adams-Watters
> >
> > -----Original Message-----
> > From: Paul D Hanna pauldhanna at juno.com
> >
> >
> > Franklin,
> >      BTW, should *both* of those conditions
> > continue to hold
> > (I would be surprised),  the next term is required to
> > be:  42.
> >
> > Paul
> >
> > -- "Paul D Hanna" <pauldhanna at juno.com> wrote:
> > Franklin (and seqfans),
> >     Perhaps this is just coincidence,
> > but so far the g.f. A(x) of the terms
> > > 1,1,1,2,4,7,13,24
> >
> > satisfies these 2 relations:
> >
> > (1) A(x) = 1/A(-x*A(x))
> >
> > (2) x*A(x) = series_reversion(x*A(-x))
> >
> > It would be curious (since unlikely) if these trends continue.
> >
> > Paul
> > ___________________________________________________
> > Try the New Netscape Mail Today!
> > Virtually Spam-Free | More Storage | Import Your Contact List
> > http://mail.netscape.com
> >
>