We seek to show that with $0\le k\le n$ the following identity holds:
$${n\brack n-k} - {n\brace n-k} =
\sum_{j=0}^k \left({n+j-1\choose 2k} - {n+k-j\choose 2k}\right)
\left\langle\!\! \left\langle k\atop j
\right\rangle\!\! \right\rangle.$$
We will use the following two representations of the Eulerian numbers
of the second order in terms of associated Stirling numbers (consult
MSE link
4037172):
$$ \sum_{j=0}^{k} (-1)^{k-j} {n-j \choose k-j}
\left\{ \!\! \left\{ n+j\atop j\right\} \!\! \right\} =
\left\langle\!\! \left\langle n\atop k\right\rangle\!\! \right\rangle
=\sum_{j=0}^{n-k+1} (-1)^{n-k-j+1} {n-j\choose k-1}
\left[\! \left[ n+j\atop j\right] \! \right] $$
We get for the first piece
$$\sum_{j=1}^k {n+j-1\choose 2k}
\sum_{p=0}^{k-j+1} (-1)^{k-j-p+1} {k-p\choose j-1}
\left[\! \left[ k+p\atop p\right] \! \right]
\\ = \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
(-1)^{k-p+1}
\sum_{j=1}^{k+1-p} (-1)^j {n+j-1\choose 2k}
{k-p\choose j-1}
\\ = \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
(-1)^{k-p} [z^{2k}] (1+z)^n
\sum_{j=1}^{k+1-p} (-1)^{j-1} (1+z)^{j-1}
{k-p\choose j-1}
\\ = \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
(-1)^{k-p} [z^{2k}] (1+z)^n
(1-(1+z))^{k-p}
\\ = \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
[z^{k+p}] (1+z)^n
= \sum_{p=0}^k
\left[\! \left[ k+p\atop p\right] \! \right]
{n\choose k+p}.$$
Now this last piece evaluates combinatorially to ${n\brack n-k}$ when
written as $\left[\! \left[ k+p\atop p\right] \! \right] {n\choose
n-k-p}$ namely we choose $n-k-p$ fixed points and split the remaining
$k+p$ elements into $p$ cycles of size at least two for a total of
$n-k$ cycles. Here we must have $k+p\ge 2p$ or $p\le k.$ (We have
classified by the number of fixed points).
We get for the second piece
$$\sum_{j=1}^k {n+k-j\choose 2k}
\sum_{p=0}^{j} (-1)^{j-p} {k-p\choose j-p}
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
(-1)^p \sum_{j=p}^k (-1)^j
{n+k-j\choose 2k} {k-p\choose j-p}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
\sum_{j=0}^{k-p} (-1)^j
{n+k-j-p\choose 2k} {k-p\choose j}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
[z^{2k}] (1+z)^{n+k-p} \sum_{j=0}^{k-p} (-1)^j
(1+z)^{-j} {k-p\choose j}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
[z^{2k}] (1+z)^{n+k-p}
\left(1-\frac{1}{1+z}\right)^{k-p}
\\ = \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
[z^{2k}] (1+z)^{n} z^{k-p}
= \sum_{p=0}^k
\left\{\! \left\{ k+p\atop p\right\} \! \right\}
{n\choose k+p}.$$
With this piece we get exactly the same reasoning as with the first
one, namely it evaluates to ${n\brace n-k}$. We write it as $\left\{\!
\left\{ k+p\atop p\right\} \! \right\} {n\choose n-k-p}$ in choosing
the number of singletons, of which there are $n-k-p.$ The remaining
$k+p$ elements are distributed into $p$ disjoint sets of at least two
elements for a total of $n-k$ sets. We once more have the condition
that $k+p\ge 2p$ or $p\le k.$ (We have classified by the number of
singleton sets.)
This concludes the argument.
