Reference request - identity in central factorial numbers Knuth (in arXiv:math/9207222 [math.CA], page 10)  gives an odd polynomial identity like
$$n^{2m-1} = \sum_{k=1}^{m} (2k-1)! T(2m,2k) \binom{n+k-1}{2k-1},$$
where $T(m,k)$ is cental factorial numbers. However, the following identity holds as well
$$(2k-1)! T(2m,2k) = \frac{1}{k} \sum_{j=0}^{k} (-1)^j \binom{2k}{j} (k-j)^{2m} \quad \quad (1.0)$$
Or by symmetry of binomial coefficients
$$(2k-1)! T(2m,2k) = \frac{1}{k} \sum_{j=0}^{k} (-1)^{k-j} \binom{2k}{k-j} j^{2m} \quad \quad (1.1)$$
The question: Is there any reference to identities (1.0), (1.1) ?
Alternative question: Prove the identities (1.0), (1.1)
PS It might seem that (1.0) is got from stirling numbers of second kind
PSS Bounty started
 A: Some references: We find in


*

*Combinatorial Identities by J. Riordan (1963), ch. 6.5 the formula (24):
\begin{align*}
 k!T(n,k)=\sum_{j=0}\binom{k}{j}(-1)^j\left(\frac{1}{2}k-j\right)^n\tag{24}
 \end{align*}

*The divided central differences of zero by  L. Carlitz and J. Riordan (1961) the formula (10a):
\begin{align*}
 K_{rs} = \frac{1}{(2s)!}\sum_{t=0}^{2s}(-1)^t\binom{2s}{t}(s-t)^{2r+2}\tag{10a}
 \end{align*}

*Interpolation by J.F. Steffenson (1927) section 58:
The development of $x^r$ in central factorials
\begin{align*}
 x^r=\sum_0^r x^{[\nu]}\frac{\delta^{\nu}0^r}{\nu!}
 \end{align*}
leads  to central differences of nothing, that is
\begin{align*}
 \delta^m0^r=\sum_0^m(-1)^{\nu}\binom{m}{\nu}\left(\frac{m}{2}-\nu\right)^r
 \end{align*}
Comment: The meaning of the left hand side $\delta^m0^r$ is given in the derivation below.

Here I'd like to show the derivation of (24) above following J. Riordan. It is based upon three ingredients: operators, a recurrence relation and Newton's formula.
Operators: We recall  the shift operator $E^a$ and the difference operator $\Delta$:
\begin{align*}
 E^af(x)&=f(x+a)\\
 \Delta f(x)&=f(x+1)-f(x)
 \end{align*}
 and introduce the central difference operator $\delta$:
 \begin{align*}
 \delta f(x)=f\left(x+\frac{1}{2}\right)-f\left(x-\frac{1}{2}\right)
 \end{align*}

We can write the  $\delta$ operator using shift and difference operator as:
  \begin{align*}
 \delta f(x)&= \left(E^{\frac{1}{2}} - E^{-\frac{1}{2}}\right)f(x)\tag{1}\\
 &=\Delta E^{\frac{1}{2}}f(x)=E^{\frac{1}{2}}\Delta f(x)\tag{2}\\
 \end{align*}
  We obtain from (1) by successive application of $\delta$
\begin{align*}
 \delta^kf(x)&=\left(E^{\frac{1}{2}}-E^{-\frac{1}{2}}\right)^kf(x)\\
 &=\sum_{j=0}^k\binom{k}{j}(-1)^jE^{-\frac{j}{2}}E^{\frac{k-j}{2}}f(x)\\
 &=\sum_{j=0}^k\binom{k}{j}(-1)^jf\left(x-j+\frac{k}{2}\right)\tag{3}
 \end{align*}

Note that (3) has already the shape of (24). So, this step looks promising and it's interesting to see how J. Riordan continues. The next step is to introduce the main actor.
Central factorials: We denote with $x^{[n]}$ the central factorial defined as

\begin{align*}
 x^{[n]}&=x\left(x+\frac{n}{2}-1\right)^{\underline{n-1}}\\
 &=x\left(x+\frac{n}{2}-1\right)\left(x+\frac{n}{2}-2\right)\cdots\left(x+\frac{n}{2}-n+1\right)
 \end{align*}
   where we use Don Knuths notation for falling factorials $x^{\underline{n}}=x(x-1)\cdots(x-n+1)$.
The central factorials obey an important recurrence relation which is the key for all what follows. We have according to (2):
  \begin{align*}
 \delta x^{[n]}&=\Delta E^{-\frac{1}{2}}x^{[n]}\\
 &=\Delta \left(x-\frac{1}{2}\right)^{[n]}\\
 &=\Delta\left(x-\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{3}{2}\right)^{\underline{n-1}}\\
 &=\left(x+\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{1}{2}\right)^{\underline{n-1}}-\left(x-\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{3}{2}\right)^{\underline{n-1}}\\
 &=\left(x+\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{3}{2}\right)^{\underline{n-2}}\\
 &\qquad-\left(x-\frac{1}{2}\right)\left(x+\frac{n}{2}-\frac{3}{2}\right)^{\underline{n-2}}\left(x+\frac{n}{2}-\frac{3}{2}-n+2\right)\\
 &=nx^{[n-1]}\tag{4}
 \end{align*}

The recurrence relation reminds us on $\frac{d}{dx}x^n=nx^{n-1}$ and we will use it accordingly. 
Newton's formula:
We consider a series expansion of $f(x)$ in terms  of $x^{[n]}$ and apply the central difference operator $\delta$:
\begin{align*}
 f(x)&=\sum_{n\geq 0}a_nx^{[n]}\\
 \delta^j f(x)&=\sum_{n\geq 0}a_n\delta^j x^{[n]}=\sum_{n\geq 0}a_n n^{\underline{j}}x^{[n-j]}\tag{5}\\
 \delta^j f(0)&=\sum_{n\geq 0}a_n n^{\underline{j}}\delta_{n,j}=a_jj!\tag{6}
 \end{align*}
Comment:


*

*In (5) we apply the central difference operator $\delta$ $j$ times and use the recurrence relation (4).

*In (6) we evaluate $f(x)$ at $x=0$ using the Kronecker delta symbol.

We obtain from (6) the following representation of Newton's formula:
  \begin{align*}
 f(x)&=\sum_{n\geq 0}\frac{x^{[j]}}{j!}\delta^jf(0)\tag{7}
 \end{align*}
The final step is setting $f(x)=x^n$ in (7), denoting the coefficients $a_k$ with $T(n,k)$ and obtain with (6)
  \begin{align*}
 x^n&=\sum_{k=0}^nT(n,k)x^{[k]}\\
 \delta^k0^n&=T(n,k)k!\tag{8}\\
 \end{align*}
  and using (3) we obtain from (8):
  \begin{align*}
 k!T(n,k)=\sum_{j=0}^k\binom{k}{j}(-1)^jf\left(\frac{1}{2}k-j\right)^n
 \end{align*}
  which is formula (24) and the claim follows.

