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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.

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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.

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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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