Let $n$ and $k$ be integers such that $n\ge3$ and $k\ge 2$ and $g(n)$ is the prime gap where $n$ lies
$$k\le g(n)+2\implies \gcd\left(\binom{n+j}{k} , j\in \{ 1,k-1 \} \right)\gt1$$
$\binom{n+j}{k}$ is a binomial coefficient and more precisely: $p(n)$ and $q(n)$ being two consecutive primes such that $p(n)\lt n \le q(n)$ let $g(n)=q(n)-p(n)$
The above proposition seems true, but I am having hard time finding a proof. Any help is welcome.Thank you!
This is the begining ($1\le n,k \le 20$) of the table $d_{n,k}=\gcd\left(\binom{n+j}{k} , j\in \{ 1,k-1 \}\right)$ with the entries where $k=g(n)+2$ in red color. ($g(1)=g(2)=4$, because $p(0)=p(1)=-2$)
There would be an analogous proposition for the Greatest Common Divisors in rows of the Pascal Triangle
$$k\le \lfloor\frac{n+1}{2}\rfloor- g(n+1)+1\implies \gcd\left(\binom{n} {k+j} ,j\in \{ 0,k \} \right)\gt1$$
Here is the table $D_{n,k}=\gcd\left(\binom{n}{k+j} , j\in \{ 0,k \}\right)$,
$1\le n\le 30,0\le k\le 15 $, with the entries where $k=\lfloor\frac{n+1}{2}\rfloor- g(n+1)+1$ in red color
The latter proposition should be more easy to prove, even though it does not involve only prime gaps but also the floor function, since it is easy to see that any entry $\binom{n}{m}$ in Pascal Triangle is a multiple of $P=p(n)$ (or $P=n$ if $n$ is prime) provided that $n-P\lt m \lt P$