= Solution
For $0\le k\le n$, the adjacent <binomial coefficients> obey
$$
\binom n{k+1}=\binom nk\frac{n-k}{k+1},
\qquad
g(k)=\binom nk\frac{n-2k-1}{k+1}.
$$
Since $\binom nk>0$ and $k+1>0$, the sign is determined by the numerator:
$$
\boxed{g(k)>0\iff0\le k<\frac{n-1}{2}.}
$$
The printed <codomain> $\mathbb N_+$ is inconsistent with the displayed map: $g(n)=-1$, and when $n$ is odd, $g((n-1)/2)=0$. With the usual extension $\binom nk=0$ for $k>n$, $g$ is a <function> $\mathbb N_0\to\mathbb Z$, and $g(k)=0$ for $k>n$; the positive range above remains unchanged. Without that extension, the natural <domain> is $\{0,\ldots,n\}$.<Discrete mathematics, continuous mathematics, and probability examination (2015) by University of Oxford>[Printed page 3, Question 2(b): the map is printed with codomain N+]{ref}
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