Rigorous or not? I want to prove
$$T(n,k)=\frac{n}{\left\lfloor\frac{n+k+1}{2}\right\rfloor}(T(n-1,k)+T(n-1,k-1))=\binom{n}{k}\binom{n-k}{\left\lfloor\frac{n-k}{2}\right\rfloor}$$
First we know only that
$$T(n,k)=0, n<k$$
$$T(n,k)=0, n<0, k<0$$
$$T(0,0)=1$$
so obviously
$$T(n,0)=\frac{n}{\left\lfloor\frac{n+1}{2}\right\rfloor}T(n-1,0)=\binom{n}{\left\lfloor\frac{n}{2}\right\rfloor}$$
next we may notice (for $n\geqslant k$)
$$T(n,k)=\frac{n}{\left\lfloor\frac{n+k+1}{2}\right\rfloor}(\frac{\left\lfloor\frac{n-k+1}{2}\right\rfloor}{n}T(n,k)+\frac{k}{n}T(n,k))$$
then
$$T(n,k)=\frac{n}{k}T(n-1,k-1)=\binom{n}{k}T(n-k,0)=\binom{n}{k}\binom{n-k}{\left\lfloor\frac{n-k}{2}\right\rfloor}$$
which agrees with
$$T(n,k)=\frac{n}{\left\lfloor\frac{n-k+1}{2}\right\rfloor}T(n-1,k)$$
Can we say that all steps looks clear? Is there another way to prove it?

Although there is already a good answer, let me present a different approach.  
We can better restate the problem as
$$
\left\{ \matrix{
  T(n,k) = 0\quad \left| {\;n < 0\; \vee \;k < 0} \right. \hfill \cr 
  T(n,k) = \binom{n}{k} \binom{n-k}{\left\lfloor {{{n - k} \over 2}} \right\rfloor} =  \hfill \cr 
   =  \left[ {0 = n} \right]\left[ {0 = k} \right] + \left[ {1 \le n} \right]{n \over {\left\lfloor {{{n + k + 1} \over 2}} \right\rfloor }}
 \left( {T(n - 1,k) + T(n - 1,k - 1)} \right) \hfill \cr}  \right.
$$
where $[condition]$ denotes the Iverson bracket
Let's recall some useful properties of the floor and ceil functions. We have that, for integer $m$
$$
\eqalign{
  & m = \left\lfloor {{m \over 2}} \right\rfloor  + \left\lceil {{m \over 2}} \right\rceil   \cr 
  & \left\lceil {{m \over 2}} \right\rceil  = \left\lfloor {{{m + 1} \over 2}} \right\rfloor  \cr} 
$$
Calling $B(n,k)$ the binomial expression, we can write it as
$$
B(n,k) = \binom{n}{k} \binom{n-k}{\left\lfloor {{{n - k} \over 2}} \right\rfloor}
 = \left[ {0 \le k \le n} \right]{{n!} \over {k!}}{1 \over {\left\lfloor {{{n - k} \over 2}} \right\rfloor !\left\lceil {{{n - k} \over 2}} \right\rceil !}}
$$
Then we have
$$
\eqalign{
  & {{B(n - 1,k)} \over {B(n,k)}} =   \cr 
  &  = \left[ {0 \le k \le n - 1} \right]{{\left( {n - 1} \right)!k!\left\lfloor {{{n - k} \over 2}} \right\rfloor !\left\lceil {{{n - k} \over 2}} \right\rceil !}
 \over {k!\left\lfloor {{{n - 1 - k} \over 2}} \right\rfloor !\left\lceil {{{n - 1 - k} \over 2}} \right\rceil !n!}} =   \cr 
  &  = \left[ {0 \le k \le n - 1} \right]{1 \over n}{{\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor !} \over {\left\lfloor {{{n - 1 - k} \over 2}} \right\rfloor !}} =   \cr 
  &  = \left[ {0 \le k \le n - 1} \right]{1 \over n}\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor  \cr} 
$$
and
$$
\eqalign{
  & {{B(n - 1,k - 1)} \over {B(n,k)}} =   \cr 
  &  = \left[ {1 \le k \le n} \right]{{\left( {n - 1} \right)!k!\left\lfloor {{{n - k} \over 2}} \right\rfloor !\left\lceil {{{n - k} \over 2}} \right\rceil !}
 \over {\left( {k - 1} \right)!\left\lfloor {{{n - k} \over 2}} \right\rfloor !\left\lceil {{{n - k} \over 2}} \right\rceil !n!}} =   \cr 
  &  = \left[ {1 \le k \le n} \right]{k \over n} \cr} 
$$
So we conclude that
$$
\eqalign{
  & {{B(n - 1,k) + B(n - 1,k - 1)} \over {B(n,k)}} =   \cr 
  &  = \left[ {0 \le k \le n - 1} \right]{1 \over n}\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor  + \left[ {1 \le k \le n} \right]{k \over n} =   \cr 
  &  = \left[ {0 \le k} \right]\left[ {1 \le n} \right]\left[ {k\left( { \le n - 1} \right) \le n} \right]{1 \over n}\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor
  + \left[ {0 \le \left( {1 \le } \right)k} \right]\left[ {1 \le n} \right]\left[ {k \le n} \right]{k \over n} =   \cr 
  &  = \left[ {0 \le k} \right]\left[ {1 \le n} \right]\left[ {k \le n} \right]{1 \over n}\left( {\left\lfloor {{{n + 1 - k} \over 2}} \right\rfloor  + k} \right) =   \cr 
  &  = \left[ {0 \le k} \right]\left[ {1 \le n} \right]\left[ {k \le n} \right]\;{{\left\lfloor {{{n + 1 + k} \over 2}} \right\rfloor } \over n} \cr} 
$$
and the thesis is demonstrated.
