We consider the sequence $n\longmapsto {n\choose k}+1$
for $k\geq 1$ a fixed integer. For $k\geq 3$ odd,
this sequence seems to contain surprisingly few prime numbers
while there are many primes (perhaps roughly the expected
amount) among the first terms of this sequence for even $k\geq 2$.


Is there an explanation for this observation (or is it an artefact)?


The following table gives the number of values $n\leq 10^5$
such ${n\choose k}+1$ is prime for $k=1,..,15$:
$$\begin{array}{cc}
k&\sharp(\mathcal P\cap \{{n\choose k}+1\ \vert\ n=k,\dots,10^5\})\\
1&9592\\2&9863\\3&3\\4&6765\\5&5\\6&6203\\
7&7\\8&3092\\9&8\\10&4364\\11&21\\12&2817\\13&19\\
14&2968\\15&16
\end{array}$$
(with $\mathcal P=\{2,3,5,7,11,\ldots\}$ denoting the set of primes).


The small subsets $\mathcal S\_k=\{n\_1,\ldots,n\_{i\_k}\}\_k$ of $\{1,\ldots,10^5\}$ corresponding to prime-numbers
${n\choose k}+1$ for $n\in \mathcal S\_k$ and $k\geq 3$ odd are given by
$$\{3,4,5\}\_3,\ \{5,6,9,11,14\}\_5,\ \{7,9,11,14,20,29,104\}\_7,$$
$$\{9,10,14,35,39,71,119,839\}\_9,$$
$$\{11,12,13,17,19,29,32,34,44,65,69,76,83,109,153,197,279,791,1385,6929,13859\}\_{11}$$
$$\{13,17,20,27,34,44,51,55,69,87,104,119,142,209,251,263,359,857,923\}\_{13},$$
$$\{15,16,17,19,38,83,89,90,131,714,1091,1286,2001,2309,4003,6551\}\_{15}.$$


For $k=3$, I checked that there are no additional primes (of the form ${n\choose 3}+1$) for $n$ up to $10^6$.


**Added after accepting the answer of user334725:** The analogous phenomenon exists for ${n\choose k}-1$ and $k$ even.
(This question is by the way an illustration of the fact that experimental maths cannot be done simultaneously with clear thinking: Playing with the computer shuts generally my brain down!)



There seem to be only few primes of the form ${n\choose k}+1$ if $k\geq 3$ is odd
When $k$ is odd, writing $f\_k(n)={n\choose k}+1$ you have $f\_k(-1)=0$ as a polynomial evaluation, and removing the denominator gives you the extra $k!$ in integers.


I.e., $5!f\_5(n)=n(n-1)(n-2)(n-3)(n-4)+120$, so $$5!f\_5(-1)=(-1)(-2)(-3)(-4)(-5)+120=-120+120=0.$$ As $-1$ is a root of $f\_k(n)$, then $n+1$ divides it (again consider denominators in the application).


Note also that all members of the "small sets" have $n+1$ dividing $k!$, showing that such sets are finite.


