An identity related to the second-order Eulerian numbers. Recently, some of the remarkable properties of second-order
Eulerian numbers $ \left\langle\!\!\left\langle n\atop k\right\rangle\!\!\right\rangle$ A340556 have been proved on MSE [  a ,
 b ,  c  ].
But there are also other notable identities to which these numbers lead.
For instance, we also noticed the following identity, which we haven't seen
elsewhere (reference?):
$$ \sum_{j=0}^{k} \binom{n-j}{n-k} 
\left\langle\!\! \left\langle n\atop  j\right\rangle\!\! \right\rangle \,=\, 
\sum_{j=0}^k (-1)^{j+k} \binom{n+k}{n+j} \left\{ n+j \atop  j\right \} \quad ( n \ge 0) $$
If we use the notation $ \operatorname{W}_{n, k} $ for these numbers,
we can also introduce the corresponding polynomials.
$$ \operatorname{W}_{n}(x) = \sum_{k=0}^n \operatorname{W}_{n, k} x^k \quad ( n \ge 0) $$
So far so routine, but then came the surprise:
$$ 3^n W_n\left(-\frac13\right) \, = \, 2^n \left\langle\!\!\left\langle - \frac{1}{2} \right\rangle\!\!\right\rangle_n  \quad ( n \ge 0) $$
On the right side are the numbers we recently asked about their
combinatorial significance!
Should I trust this strange equation?
 A: In trying to verify the identity
$$\sum_{j=0}^{k} {n-j \choose n-k} 
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle 
=  \sum_{j=0}^k (-1)^{j+k} {n+k \choose n+j} 
\left\{ n+j \atop  j\right \}$$
we quote from MSE
the following identity
$$\left\langle\!\! \left\langle n\atop k 
\right\rangle\!\! \right\rangle =
\sum_{j=0}^k (-1)^{k-j} {2n+1\choose k-j} {n+j\brace j}.$$
We get for the LHS
$$\sum_{j=0}^{k} {n-j \choose n-k} 
\sum_{p=0}^j (-1)^{j-p} {2n+1\choose j-p} {n+p\brace p}
\\ = \sum_{p=0}^k {n+p\brace p} 
\sum_{j=p}^k (-1)^{j-p} {2n+1\choose j-p} {n-j\choose n-k}.$$
The inner sum is
$$\sum_{j=0}^{k-p} (-1)^j {2n+1\choose j} 
{n-j-p\choose n-k}
\\ = \sum_{j=0}^{k-p} (-1)^j {2n+1\choose j} 
{n-j-p\choose k-p-j}
\\ =  [z^{k-p}] (1+z)^{n-p}
\sum_{j=0}^{k-p} (-1)^j {2n+1\choose j} 
\frac{z^j}{(1+z)^j}.$$
Here the coefficient extractor enforces the upper limit of the sum
and we may extend $j$ to infinity:
$$[z^{k-p}] (1+z)^{n-p}
\sum_{j\ge 0} (-1)^j {2n+1\choose j} 
\frac{z^j}{(1+z)^j}
\\ =  [z^{k-p}] (1+z)^{n-p}
\left(1-\frac{z}{1+z}\right)^{2n+1}
= [z^{k-p}]
\frac{1}{(1+z)^{n+p+1}}
\\ = (-1)^{k-p} {k-p+n+p\choose n+p}
= (-1)^{k-p} {n+k\choose n+p}.$$
Introducing the leading term,
$$\sum_{p=0}^k (-1)^{k+p}
{n+k\choose n+p} {n+p\brace p}$$
This is the claim.
 As for the polynomials we find
$$\sum_{k=0}^n x^k 
\sum_{j=0}^{k} {n-j \choose n-k} 
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle
= \sum_{j=0}^n 
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle
\sum_{k=j}^n {n-j\choose n-k} x^k
\\ = \sum_{j=0}^n 
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle x^j
\sum_{k=0}^{n-j} {n-j\choose n-j-k} x^k
= \sum_{j=0}^n 
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle x^j
\sum_{k=0}^{n-j} {n-j\choose k} x^k
\\ = \sum_{j=0}^n 
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle x^j
(1+x)^{n-j}
= (1+x)^n \sum_{j=0}^n 
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle 
\frac{x^j}{(1+x)^j}.$$
Multiplying by $3^n$ and evaluating at $x=-1/3$ we obtain
$$3^n \frac{2^n}{3^n}
\sum_{j=0}^n 
\left\langle\!\! \left\langle n\atop  j
\right\rangle\!\! \right\rangle 
\frac{(-1/3)^j}{(2/3)^j}
= 2^n 
\sum_{j=0}^n 
\left\langle\!\! \left\langle n\atop j
\right\rangle\!\! \right\rangle 
\left(-\frac{1}{2}\right)^j$$
as claimed. 
Remark. Maybe we can pause here for a few days. Thanks! 
A: A sketch in three steps: First, we need a definition for $ \operatorname{W}_{n}(x)$. With this, we do not want to assume the
second-order Eulerian numbers. So we take the RHS of the identity.
The next and primary step is to show that in general
$$ \operatorname{W}_{n}(x) = (1 + x)^n 
\left\langle \! \left\langle \frac{x}{1+x} \right\rangle \! \right\rangle _n. $$
Finally, plug in $ x = -\frac13 $, et voilà,
$$ \operatorname{W}_{n}\left( - \frac13 \right) = \left( \frac23 \right)^n 
\left\langle \! \left\langle -\frac12 \right\rangle \! \right\rangle _n .$$
The choice of the notation $ \operatorname{W}_{n}(x) $ was not accidental.
We are speaking here about the Ward polynomials. Their coefficients are
the Ward numbers which be found at A269939.
