Kevin Ryde on Mon, 23 Jul 2018 09:37:35 +0200

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Re: generating function solution to poly

Karim Belabas <> 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 <> 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 <> 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.