Why this behaviour of primes?

Question

I calculated $$(p_k \cdot p_{k+1})\mod p_{k+2}$$ for $k=1,2,...25$ and obtained a following numbers $$1,1,2,12,7,12,1,2,16,11,40,12,24,7,13,16,48,40,72,48,40,60,15,48,12$$

We can see that there are only $8$ odd numbers and that some numbers repeat.

Is this expected behavior? Does ratio $$\dfrac {N_o(n)}{N_e(n)}$$ gets smaller and smaller as we calculate this sequence for larger and larger $n$ ($N_e(n)$ is a number of even and $N_o(n)$ a number of odd numbers in the first $n$ members of this sequence)?

It would be nice to see how this sequence would look like for, for example, first 100 numbers, then 1000, then 10000, so an answer with computational data is also welcomed.

Answer 1

We would expect $p_k$, $p_{k+1}$, and $p_{k+2}$ to be relatively close.

Letting $p_k = p_{k+2} - a$ and $p_{k+1} = p_{k+2} - b$, we see that $a$ and $b$ are even, and that $p_k p_{k+1} \equiv ab \pmod {p_{k+2}}$, so most of the time the remainder should be even.

Answer 2

See OEIS sequence A182126. With $a_n = p_n p_{n+1} \mod p_{n+2}$, $a_n = (p_{n+2}-p_n)(p_{n+2}-p_{n+1})$ when $(p_{n+2}-p_n)(p_{n+2}-p_{n+1}) < p_{n+2}$. Cramér's conjecture implies this is true for sufficiently large $n$ (and it seems to be the case for $n > 61$). This implies in particular that $a_n$ is even for such $n$.