I sent Knuth a email about this problem, and OH MY he wrote me back.
It seems Knuth made a small mistake. The hypergeometric expression calculated "directly" has a 0/0 when n=1. Then Knuth tried to calculate "separately" as when n is odd number and when n is even number, and then it seems the two answers can be combined as one. And that's how the ceiling came out: because odd number and even number are considered separately.
But the "combining" phase has a small mistake. It turns out the two cannot be combined, their lower parameters are different.
We can learn a lot in the way Knuth tried to solve the problem. So let's check it out closely :-)
First, we don't have to "transform it to $\sum_{k>0}$$ k\choose n-k$ to have 'infinite' sum". Hypergeometric function has a infinite form. It doesn't mean we have to transform the "real series" into infinite form to calculate its hypergeometric expression. Like page 206 says "A negative integer as upper parameter causes the infinite series to become finite".
So we can calculate directly like:
$$\sum_{k≥0} {n-k \choose k} z^k $$
$$\frac{t_{k+1}}{t_k} = \frac{(n-k-1)!}{(k+1)!(n-2k-2)!} \frac{k!(n-2k)!}{(n-k)!} z$$
$$=\frac{(k-\frac{n}{2})(k+\frac{1-n}{2})}{(k-n)(k+1)}(-4z)$$
So we can get the hypergeometric function:
$\displaystyle F{-\frac{n}{2},-\frac{n-1}{2}\choose -n}$ with constant -4z
But this formula has a problem: when n =1, it becomes
$$\sum \frac{(-1/2)^\bar k \cdot 0^\bar k}{(-1)^\bar k}(-4z)^k$$
Which expands like :$1 + \frac{0}{-1}(-4z) + \frac{0}{0}(-4z)^2...$
And 0/0 is prblem.
So Knuth tried to calculate it separately as n is odd number and even number:
$$\sum_{k≥0} {n-k \choose k} z^k = \sum_{k≥0} {2n-k \choose k} z^k + \sum_{k≥0} {2n+1-k \choose k}z^k$$
The even number term is:
$$\sum_{k≥0} {2n-k \choose k} z^k = \sum_{k≥0} {n+k \choose n-k}z^{n-k}$$
If you wonder why it's still k≥0 after we replaced k by n-k: $\sum_{k≥0} {2n-k \choose k} z^k$ has finite terms, which ends with the term $z^n$, there's no $z^{n+1}$ term. And $n+k \choose n-k$is 0 when k>n, so we don't have to say k<=n.
Same old way, we can get:
$$\frac{t_{k+1}}{t_k} = \frac{(n+k+1)!}{(n-k-1)!(2k+2)!} \frac{(n-k)!(2k)!}{(n+k)!} \frac{1}{z}$$
We can get the hypergeometric function:
$z^n\displaystyle F{1+n,-n \choose \frac{1}{2}}$ with constant -1/4z.
Now we calculate the odd number term$\sum_{k≥0} {2n+1-k \choose k}z^k$ in the same way.
Seems Knuth made a mistake. He got
$z^n\displaystyle F{2+n,-n \choose \frac{1}{2}}$ with constant -1/4z, then it seems these two F can be written as one. So came the ceiling function.
But I think the $\displaystyle F{2+n,-n \choose \frac{1}{2}}$ is wrong, it should be $\displaystyle F{2+n,-n \choose \frac{3}{2}}$.
At least that's what I'd got based on Knuth's hints ;-)
