[seqfan] Re: Functional Equation Challenge

[Paul D Hanna]

Seqfans,        The answer is easier than I expected.  SOLUTION. If     A( +sqrt( A(x^2*F(x)^2) ) ) = xthen  A( -sqrt( A(x^2*F(x)^2) ) ) = G(-x)where   G(x) = B(x*F(-x))such that    B(x*F(x)) = x.   Thank you, Olivier, for your thoughtful response.     Paul                On Wed, Jun 8, 2016 at 5:20 AM, Paul D Hanna <pauldhanna at juno.com> wrote: > SeqFans,>        Here is a challenge for you, the solution being as yet unknown to> me, and yet does not seem to be intractable.> And there are sequences of coefficients for certain A(x) that may be> appropriate for submission to the OEIS. 
 > OBJECTIVE.> Given F(x) with F(0)=1, suppose A(x) satisfies>     A( +A(x)^2 * F(A(x)) ) = x^2> then find G(x) such that>     A( -A(x)^2 * F(A(x)) ) = G(-x^2).> > OBSERVATION:> For the same A(x), F(x), and G(x), we have>     A( +sqrt( A(x^2 * F(x)) ) ) = x> and>     A( -sqrt( A(x^2 * F(x)) ) ) = G(-x).> > Below I give 6 simple examples.> > Note that we may generate A(x) for a given F(x) by iterating the relation>    A(x) = Series_Reversion( sqrt( A(x^2 * F(x)) ) ).> > > How does one determine G(x) from F(x)?> > Solution, anyone?>       Paul> >