How can I get the sequence of the generating function $T(z)$ given that $T(z)=z+1+(z+1)T(z)^2$

Question

given that $$T(z)=z+1+(z+1)T(z)^2$$, how can I get the sequence?

Answer 1

This is not a complete answer but some progress that may help.

The most obvious feature of the given formula, $$T_z=(z+1)+(z+1)T_z^2$$ is the repeated occurrence of $(z+1)$.

It's when I let $w=z+1$ that I finally started to make some interesting progress with this for then, $$T_z=w+wT_z^2$$ $$wT_z^2-T_z+w=0$$ Applying the formula to solve a quadratic equation to this gives, $$T_z=\frac{1\pm\sqrt{1-4w^2}}{2w}$$ and at this point the words "Catalan numbers" leap into mind.

It's well known that the Catalan numbers have the generating function and generating series,

$$\frac{1-\sqrt{1-4y}}{2y}=1+y^1+2y^2+5y^3+14y^4+42y^5+...$$ Replacing $y$ with $y^2$ inserts a single interlacing zero between the terms, $$\frac{1-\sqrt{1-4(y^2)}}{2(y^2)}=1+(y^2)^1+2(y^2)^2+5(y^2)^3+14(y^2)^4+42(y^2)^5+\dots$$ and if we multiply both sides by $y$ we have, $$\frac{1-\sqrt{1-4y^2}}{2y}=y+y^3+2y^5+5y^7+14y^9+42y^{11}+\dots$$ So from this we can deduce, as $y=w=z+1$, that, $$T_z=\frac{1\pm\sqrt{1-4(z+1)^2}}{2(z+1)}$$

At this point I couldn't face expanding this by hand and asked some software to look at what generating series this generating function expands into...

The more interesting result is,

$$T_z=\frac{1+\sqrt{1-4(z+1)^2}}{2(z+1)}$$ $$=-\frac{\sqrt{5}-1}{2}+\frac{\sqrt{5}-5}{10}x-\frac{7\sqrt{5}-25}{50}x^2+\frac{43\sqrt{5}-125}{250}x^3+\dots$$

The other possibility moves into complex numbers, $$T_z=\frac{1-\sqrt{1-4(z+1)^2}}{2(z+1)}$$ $$=-\frac{i\sqrt{3}-1}{2}-\frac{i\sqrt{3}+3}{6}x+\frac{5i\sqrt{3}+9}{18}x^2-\frac{23i\sqrt{3}+27}{54}x^3$$

The software is a "Taylor Series Expansion Calculator" at https://www.numberempire.com/taylorseriesexpansion.php

Any comments on this are welcome - I'm disappointed that such an elegant start ended up with such a dirty finish.

Answer 2

Are you expecting a sequence of complex numbers ?

If you are then Felix Marin's equation can be written as, $$T(z)=\frac{1}{2z+2}-\frac{i\sqrt{4z^2+8z+3}}{2z+2}$$ The real part of the terms are, $$\frac{1}{2z+2}=\frac{1}{2}-\frac{1}{2}z+\frac{1}{2}z^2-\frac{1}{2}z^3+\dots$$ The imaginary parts have a transient start but then also alternate, $$-\frac{i\sqrt{4z^2+8z+3}}{2z+2}=-\frac{3}{2}-\frac{5}{2}z+\frac{1}{2}z^2-\frac{1}{2}z^3+\frac{1}{2}z^4-\frac{1}{2}z^5+\dots$$ Hope this means something in the wider context of whatever it is you are doing with it.