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.


The pattern can be explained by the fact that the sums of the prime factors of the numbers in your sequence are respectively $10, 9, 8, 7, 6, 5, 4$. The sequence just represents a countdown to the same next prime.


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