Is the sequence $\{0,2,6,12,20,30,...,n(n+1)\}$ admissible for every natural $n$?

Question

Look here :

https://en.wikipedia.org/wiki/Prime_k-tuple

for the definition of an admissible sequence.

I wonder if the sequence of differences of primes can be $\{0,2,4,6,8,...,2n\}$ for every natural nuber $n$. A weaker version is, that $p+(j-1)j$ is prime for every $j$ with $1\le j\le n+1$.

For $n=7$, the smallest example for the weaker version is $11$, but $128981$ is the smallest example for the sequence $\{0,2,4,6,8,10,12,14\}$.

For $\{0,2,4,6,8,10,12,14,16,18\}$, the smallest example is $2426256797$.

The questions :

• Is the sequence $\{0,2,4,6,8,...2n\}$ of differences possible for every $n$ ?
• If not, is there at least an example for the weak version for every $n$ (Equivalent : $\{0,2,6,12,20,30,...n(n+1)\}$ is admissible) ?

Answer 1

$(0,2,4)$ is already inadmissible according to the definition: it contains all residues mod 3. So the answer to the first question is negative. The second sequence is clearly admissible (it contains at most $p-2$ different non-zero residues mod $p$).

Answer 2

The sequence is admissible for all $n$. Given $p$, we only need to consider $j\in[0,p-1]$ to get all residues mod $p$ that are covered by the sequence. Since $(p-j+1)(p-j)\equiv j(j-1)\bmod p$, most of the residues are doubly covered, and hence roughly half are uncovered.