Define the function $f:\mathbb N^+\to\mathbb N^+$ as: $f(n)$ is the least semi-prime $pq$ such that $n+p+q$ is a prime. (One must prove that this really is a function, but it should be and my tests doesn't contradicts that).
There is a sequence of numbers $$21\to 14\to 15\to 10\to 9\to 6\to 4$$
such that if $f(n)\neq 4$ is a number in this sequence, then $f(n+1)$ equals to the successor of $f(n)$ in the sequence.
This is tested for all values of $n<100,000$.
I would like a proof of that $f$ really is a function, that is, for all $n\in\mathbb N^+$ it exists a semi-prime $pq$ with the property that $n+p+q\in\mathbb P$.
Also, if anyone can find a counter-example from the peculiar pattern, I would like to see that. And off course, if anyone can explain the pattern I would be very glad.
Is $f$ really a function?
Suppose that $n$ is even. Then since $n + p+ q$ is prime, it is odd, and thus $p+q$ is also odd. Thus necessarily either $p=2$ or $q = 2$. Supposing that $q = 2$, then $p$ and $p + n + 2$ are primes. Thus the existence of such a $p$ amounts to answer this question, already asked on this site: Can every even integer be expressed as the difference of two primes? As you will see from the answers, it seems to be an open question.