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I give here a heuristics that suggests that the quantity $\displaystyle{G_{k}:=\liminf_{n\to\infty}p_{n+k}-p_{n}}$ should be approximately equal to $k(1+H_{k})$, where $H_{k}$ is the $k$-th harmonic number. It seems that whenever $k$ doesn't divide $G_{k}$ then we have $$\dfrac{G_{k}}{k}-(1+H_{k})=-\dfrac{k+\delta_{k}}{M_{k}}$$ where $M_{k}$ is the least common multiple of the first $k$ positive integers, and $\delta_{k}$ is the number of integers $m\lt k$ such that $M_{m}=M_{k}$.

My question is: has this already been conjectured? If so, could I get some references? Of course, I also welcome any insight.

Thanks in advance.

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The Hardy-Littlewood prime tuples conjecture implies that $G_k$ equals the smallest diameter of an admissible $(k+1)$-tuple. In particular, it implies that $G_{10}=36$ (cf. here).

Your conjecture implies for $k>3$ that $$ G_k>kH_k+k-\frac{2k^2}{M_k}>kH_k+k-3.$$ In particular, it implies that $G_{10}>36$.

So your conjecture contradicts a widely believed old conjecture.

P.S. I have not checked if the OP's first display yields an integral value for $G_k$. If not, then it is false for the trivial reason that $G_k$ is an integer.

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As proved in the polymath 8b paper found here: http://arxiv.org/pdf/1407.4897v4.pdf we have the bounds:

$$(\frac{1}{2}+o(1))k\log(k) \leq H(k) \leq (1+o(1))k\log(k)$$

where $H(k)$ is the diameter of the smallest admissible $k$-tuple. The Hardy-Littlewood prime-tuples conjecture implies $G_{k}=H(k+1)$. See Theorem 3.3 in the paper for an even better upper bound.

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  • $\begingroup$ Unfortunately, this does not falsify the OP's conjecture. See also my response. $\endgroup$ – GH from MO Apr 28 '15 at 2:15
  • $\begingroup$ These bounds are only conditional, if I understood well. $\endgroup$ – Sylvain JULIEN Apr 28 '15 at 2:34
  • $\begingroup$ @Sylvain: Right. I've corrected my post. $\endgroup$ – Pace Nielsen Apr 28 '15 at 3:01
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    $\begingroup$ @GH: It doesen't falsify the OP's second conjecture, but it is a step towards verifying his initial guess that $G_k$ should be approximately $k*H_k$. $\endgroup$ – Pace Nielsen Apr 28 '15 at 3:06
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    $\begingroup$ @PaceNielsen: The thread I meant is mathoverflow.net/questions/185354/… $\endgroup$ – GH from MO Apr 28 '15 at 19:23
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Sorry for answering my own question, but apparently Zhi-Wei Sun conjectured that for all $k\gt 4$ one has $0\lt \dfrac{G_{k}}{k}-H_{k}\lt \dfrac{2+\gamma}{\log k}$ (where $\gamma$ is Euler's constant) which contradicts my own conjecture.

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