TITLE: How to show that this identity holds?
QUESTION [5 upvotes]: By comparing the first terms of the Taylor expansion at $0$, it seems that
for $|x|<4/27$ the following identity holds:
$$\ln\left(\sum_{n=1}^{\infty}
\binom{3n}{n}\frac{x^{n-1}}{2n+1}\right)=
\sum_{n=1}^{\infty}
\binom{3n}{n}\frac{x^n}{n}.$$
I tried by differentiating both sides, but I am completely lost in a terrible mess. I wonder if there is a better strategy to handle it. Any idea?

REPLY [3 votes]: Suppose we seek to show that
$$\sum_{n\ge 0} {3n+3\choose n+1} \frac{z^n}{2n+3}
= \exp\left(\sum_{n\ge 1} {3n\choose n} \frac{z^n}{n}\right).$$
This is the same as showing that
$$q_n = [z^n] f(z) = {3n+3\choose n+1} \frac{1}{2n+3}
\quad\text{where}\quad
f(z) = 
\exp\left(\sum_{n\ge 1} {3n\choose n} \frac{z^n}{n}\right).$$
and it  can be  shown by induction.  Basic arithmetic verifies  it for
$q_0 = 1.$ Differentiating we get
$$f'(z) = \exp\left(\sum_{n\ge 1} {3n\choose n} \frac{z^n}{n}\right)
\left(\sum_{n\ge 1} {3n\choose n} z^{n-1} \right)
\\ = f(z) \left(\sum_{n\ge 0} {3n+3\choose n+1} z^{n} \right).$$
Extracting coefficients we obtain the recurrence
$$[z^n] f'(z) = (n+1) q_{n+1}
= \sum_{k=0}^n q_k {3n+3-3k\choose n+1-k}$$
or
$$q_{n+1} = \frac{1}{n+1}
\sum_{k=0}^n q_k {3n+3-3k\choose n+1-k}.$$
This means to verify the claim we have to show that
$${3n+6\choose n+2} \frac{1}{2n+5}
= \frac{1}{n+1}
\sum_{k=0}^n {3k+3\choose k+1} \frac{1}{2k+3} 
{3n+3-3k\choose n+1-k}$$
or
$${3n+6\choose n+2} \frac{n+1}{2n+5}
= \sum_{k=0}^{n} {3k+3\choose k+1} \frac{1}{2k+3} 
{3n+3-3k\choose n+1-k}$$
which is
$${3n+6\choose n+2} \frac{n+2}{2n+5}
= {3n+6\choose n+1}
\\ = \sum_{k=0}^{n+1} 
{3k+3\choose k+1} \frac{1}{2k+3} 
{3n+3-3k\choose n+1-k}.$$
What we  have here  is a straightforward  convolution of  two ordinary
generating functions.  With some help from  the OEIS we  learn that we
require  the ternary  unlabeled tree  function $T(z)$  with functional
equation
$$T(z) = 1 + z T(z)^3
\quad\text{or}\quad
z = \frac{T(z)-1}{T(z)^3}.$$
We then obtain
$$[z^n] T(z) =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} T(z) \; dz.$$
Using  the functional  equation and  letting  $w=T(z)$ we  have $dz  =
(3-2w)/w^4 \; dw$ and get for the integral
$$\frac{1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{w^{3n+3}}{(w-1)^{n+1}} w \frac{3-2w}{w^4} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{3w^{3n}-2w^{3n+1}}{(w-1)^{n+1}}  \; dw.$$
Now with $w^q = \sum_{p=0}^q {q\choose p} (w-1)^p$ this will produce
$$3\times {3n\choose n} - 2\times {3n+1\choose n}
= \left(3 - 2\frac{3n+1}{2n+1}\right) {3n\choose n}
= \frac{1}{2n+1} {3n\choose n}.$$
This generates  the left term of  the convolution. For  the right term
consider
$$[z^n] \frac{1}{1-3zT(z)^2}
= [z^n] \frac{1}{1-3(T(z)-1)/T(z)} = [z^n] \frac{T(z)}{3-2T(z)}$$
to get
$$\frac{1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{w^{3n+3}}{(w-1)^{n+1}} \frac{w}{3-2w} \frac{3-2w}{w^4} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{w^{3n}}{(w-1)^{n+1}}  \; dw = {3n\choose n}.$$
Recall the convolution that we are working with which now becomes
$$\sum_{k=0}^{n+1} [z^{k+1}] T(z) 
[z^{n+1-k}] \frac{T(z)}{3-2T(z)}
\\ = \sum_{k=0}^{n+1} [z^k] \frac{T(z) - 1}{z}
[z^{n+1-k}] \frac{T(z)}{3-2T(z)}.$$
We are left with
$$[z^{n+1}] \frac{1}{z} \frac{T(z)(T(z)-1)}{3-2T(z)}
= [z^{n+2}] \frac{T(z)(T(z)-1)}{3-2T(z)}.$$
Using the sáme integral as before this becomes
$$\frac{1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{w^{3n+9}}{(w-1)^{n+3}} \frac{w(w-1)}{3-2w} \frac{3-2w}{w^4} \; dw
\\ = \frac{1}{2\pi i}
\int_{|w-1|=\gamma} 
\frac{w^{3n+6}}{(w-1)^{n+2}} \; dw
\\ = {3n+6\choose n+1}.$$
This concludes the argument.
The   sequences    where   this   material    resides   are   OEIS
A005809              and              OEIS
A001764.
The structure  of the initial equation suggests a counting argument
involving sets of labeled directed cycles where each item may have one
of three colors and the reader is invited to investigate this further.