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Consider the following function $f: \omega\to \{0,1\}$:

  • Set $f(n) = 1$ if for all $k\in \omega$ there are prime numbers $p,q > k$ such that $n = p-q$, and
  • set $f(n) = 0$ otherwise.

(Trivially, if $n$ is odd, we have $f(n) = 0$. Moreover, the question whether $f(2) = 1$ is the subject of the twin prime conjecture.)

Is $f$ computable?

Dominic van der Zypen
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I once heard Harvey Friedman suggest that the set of prime-differences, that is, the set of all natural numbers $n$ for which there are primes $p,q$ with $p-q=n$, as a possible candidate for all we knew for an intermediate Turing degree — a noncomputable set between $0$ and $0'$ — that was natural, not specifically constructed to have that feature.

Joel David Hamkins
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  • Am I right in assuming that the following implication holds? If the twin prime conjecture is undecidable, then $f:\omega\to{0,1}$ as described above is non-computable. – Dominic van der Zypen Sep 22 '15 at 10:51
  • I don't know how to prove that. – Joel David Hamkins Sep 22 '15 at 10:55
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    Is there a specific reason for the assumption that the set of prime differences has a reasonable chance to be noncomputable? -- I think the basic heuristics pretty clearly suggests it's just the set of even natural numbers. – Stefan Kohl Sep 22 '15 at 10:55
  • He didn't mention a reason, except that of course the question is not yet settled. – Joel David Hamkins Sep 22 '15 at 10:58
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    Dominic, I'm not sure that this is actually the answer to your question. – Joel David Hamkins Sep 22 '15 at 11:43
  • If we just require the existence of one difference $p-q=n$, we need only run up to $q=1117$ to verify that all even $n$ not exceeding $7\cdot 10^7$ admit such a representation. However, if we require (like in the question), arbitrarily large $q$ values for each single $n$, then no explicit $n$ is known to belong to $f^{-1}( { 1 } )$, but Zhang's recent result shows that at least one such $n$ exists below $7 \cdot 10^7$. – Jeppe Stig Nielsen Sep 22 '15 at 13:19
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    @JoelDavidHamkins - I accepted your answer because it seemed to me that the answer to my question depends on the decidability of the Twin Primes Conjecture, but now I'm confused whether that's the case... – Dominic van der Zypen Sep 22 '15 at 14:00
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    @DominicvanderZypen There's no direct relationship - the function could be computable even if the twin prime conjecture fails, or is undecidable. – Noah Schweber Sep 22 '15 at 14:32
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    @NoahSchweber, how can you assert such a definitive statement about what is possible? It seems to me that we don't really know anything about it with certainty. – Joel David Hamkins Sep 22 '15 at 15:33
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    @JoelDavidHamkins Sorry, that was my bad, you're absolutely right. I was unclear - I should have said "there is no obvious direct relationship." I was just getting at the confusion between "an atomic fact about the function is undecidable in a specific system" and "the function is not computable", which is a subtle issue (at least, it tripped me up when I was learning this). – Noah Schweber Sep 22 '15 at 15:37
  • @NoahSchweber Can you clarify the issue in your last comment for me? I would expect that if the twin prime conjecture is undecidable then we have no way of computing the value of $f(2)$ for if we had, then we could just do the computation and hence decide the twin prime conjecture. Where do I go wrong? – Vincent Jan 17 '20 at 14:19
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    @Vincent The mere knowledge that something is computable doesn't usually give us much concrete information. E.g. let $X={n:$ ($n=0$ and P$\not=$NP) or ($n=1$ and P$=$NP)$}$. Trivially $X$ is computable, but we don't know what algorithm computes $X$. – Noah Schweber Jan 17 '20 at 15:03
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    Basically, something can be "nonconstructively computable" - that is, we can have a definable set/function $X$ and a program $\Phi$ which computes $X$ but such that there is no proof in our theory that $\Phi$ (or any other specific program) computes $X$. This is a weird mix of classical and constructive flavors, but is a key feature of the subject. – Noah Schweber Jan 17 '20 at 15:05