Re: generating function solution to poly

[Kevin Ryde <user42_kevin@yahoo.com.au>]

Karim Belabas <Karim.Belabas@math.u-bordeaux.fr> writes:
>
> ? seralgdep(g,3,2)
> %2 = (y + 1)*x^3 - 2*x^2 + (2*y^2 - y + 1)*x - y^2

I didn't know that one.  Though actually I wanted the other way, have
the cubic, want the series :).

Bill Allombert <Bill.Allombert@math.u-bordeaux.fr> writes:
>
> ? g=truncate(g)-subst(P,y,truncate(g))/subst(P',y,g)
> %3 =
> %X^2+X^3+X^4+3*X^5+6*X^6+12*X^7+29*X^8+67*X^9+157*X^10 + ...

Thanks, and starting from say just 3 terms of series too.  It comes out
all terms correct in each doubled series length is it?  Nice.

Peter Pein <peter.pein@dordos.de> writes:
>
> ?  P=substvec((1+x)*g^3-2*g^2+(1+x+2*x^2)*g-x^2,[x,g],[X,y])
                                ^^^^^^^^^^^
Cut and paste gone astray, should be 1-x+2*x^2.  :)

> I'm just curious: are you sure the coefficients in g are correct?

I think correct.  They're counts (of some trees) so are to be integers
and >=0 in any case.  There might be other series "solutions" to the
cubic, starting series low terms 1*x^0 -1*x^1 and above that not
integers by my reckoning, so not what I had sought.