Find the positive $n$ such $2^{n-2i}|\sum_{k=i}^{\lfloor\frac{n-1}{2}\rfloor}\binom{k}{i}\binom{n}{2k+1}$ Question:

if such $\forall i\in\{1,2,\cdots,\lfloor\dfrac{n-1}{2}\rfloor\}$,have
$$2^{n-2i}|\sum_{k=i}^{\lfloor\frac{n-1}{2}\rfloor}\binom{k}{i}\binom{n}{2k+1}$$ Find all the positive intger $n$

I want to find this sum $\sum_{k=i}^{\lfloor\frac{n-1}{2}\rfloor}\binom{k}{i}\binom{n}{2k+1}$?this step is right?
this problem is post my frend xi yong wang
 A: We have the sum
$$S_{n,q} = \sum_{k=q}^{\lfloor (n-1)/2\rfloor}
{k\choose q} {n\choose 2k+1}.$$
Introducing 
$${n\choose 2k+1} = {n\choose n-2k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k}} (1+z)^n
\; dz$$
this controls the range and we may extend $k$ to infinity, getting for
the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n}} (1+z)^n
\sum_{k\ge q} {k\choose q} z^{2k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2q}} (1+z)^n
\sum_{k\ge 0} {k+q\choose q} z^{2k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2q}} (1+z)^n
\frac{1}{(1-z^2)^{q+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2q}} (1+z)^{n-q-1}
\frac{1}{(1-z)^{q+1}}
\; dz.$$
This is
$$\sum_{p=0}^{n-2q-1} {n-q-1\choose p} {n-2q-1-p+q\choose q}
\\ = \sum_{p=0}^{n-2q-1} {n-q-1\choose n-q-p-1} {n-q-1-p\choose q}.$$
Observe that
$${n-q-1\choose n-q-p-1} {n-q-1-p\choose q}
= \frac{(n-q-1)!}{p! q! (n-2q-1-p)!}
\\ = {n-q-1\choose q} {n-2q-1\choose p}$$
so that the sum becomes
$${n-q-1\choose q} \sum_{p=0}^{n-2q-1} {n-2q-1\choose p}
= 2^{n-2q-1} {n-q-1\choose q}.$$
where  we have  $q\le  \lfloor (n-1)/2\rfloor.$  If this  is  to be  a
multiple of $2^{n-2q}$ then the  binomial coefficient must be even. We
apply                                                          Lucas'
Theorem  which  says
that a binomial coefficient ${n\choose p}$  is odd iff all the bits of
$p$ are less  than or equal to the corresponding  bits of $n.$ Suppose
first that  $n$ is not  a power of  two and take  $q = n  - 2^{\lfloor
\log_2 n\rfloor}.$ We  have $q\ge 1$ as required. To  verify the upper
range  start from  $\lceil  (n-1)/2\rceil \le  n/2$  which follows  by
cases. This is  $\lceil (n-1)/2\rceil \le 2^{-1 + \log_2  n}$ which is
less  than  $2^{\lfloor  \log_2  n\rfloor}$ so  that  $n-1  -  \lfloor
(n-1)/2\rfloor  \lt 2^{\lfloor  \log_2 n\rfloor}$  or $n  - 2^{\lfloor
\log_2 n\rfloor}  \lt \lfloor (n-1)/2\rfloor  + 1$ or $n  - 2^{\lfloor
\log_2  n\rfloor} \le  \lfloor (n-1)/2\rfloor.$  This proves  that the
choice of $q$  is within the range.  We thus  have $n-q-1 = 2^{\lfloor
\log_2 n\rfloor} - 1$, a string of  one bits, which are larger than or
equal to the  bits of $q$ by  inspection.  Hence we have  found an odd
binomial coefficient  and the claim  does not hold  when $n$ is  not a
power of two.  The  remaining case is for $n$ a power  of two. We thus
have $n-1$ a  string of one bits. Hence the  bits corresponding to $q$
in $n-1-q$ are the bits of $q$, but flipped. Now since all $q$ have at
least one bit that is set we  get a correspondong zero bit in $n-1-q.$
Hence  the binomial  coefficient is  even.   This works  for all  $q.$
Therefore the  answer to the query  is that $S_{n,q}$ is  divisible by
$2^{n-2q}$ with $q$ in the proposed range iff $n$ is a power of two.
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