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Based on the comments on OEIS-A002387:

$a_{n}$ = 1, 2, 4, 11, 31, 83, 227, 616,...

it is likely, that the sequence $a_{n}$ coincides with $[ e^{n-\gamma} ]$ , where $\gamma$ is the Euler-Mascheroni constant and $[\cdot]$ is the rounding function (remark made by Dean Hickerson).

My Question: Is there a formal proof, that OEIS-A002387 is $[ e^{n-\gamma} ]$ ?

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  • $\begingroup$ Equivalently, there exist no integers $k \ge 1$ and $n$ with $\log(k+1/2) + \gamma < n < \log(k+1/2) + \gamma + 1/(24k^2)$. $\endgroup$
    – Gerald Edgar
    Commented Jun 23, 2013 at 12:00
  • $\begingroup$ As $a_{n}$ essentially grows as a geometric progression of common ratio $e$, maybe there is a way to derive the analogue of Binet's formula for the Fibonacci sequence. $\endgroup$
    – Sylvain JULIEN
    Commented Jun 23, 2013 at 12:09
  • $\begingroup$ @Gerald I suppose there might exists integers in your equality, yet the value of $H_k$ to be the floor of the lower bound and the conjecture will be still true. $\endgroup$
    – joro
    Commented Jun 23, 2013 at 12:45
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    $\begingroup$ I doubt a proof is known or can be obtained using any known techniques. Basically, it comes down to whether a bizarre numerical coincidence occurs with Euler's contant $\gamma$. The approximation properties of $\gamma$ seem intractable, which makes the problem feel hopeless, but I guess you never know where a nice argument might be hiding (or I might be overlooking some aspect of this). $\endgroup$
    – Henry Cohn
    Commented Jun 23, 2013 at 14:28
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    $\begingroup$ As a matter of fact, $H(k)-L(k)=1/24k^2+o(1/k^2)$. So for instance if we knew that $$\liminf_{k\to\infty}k^2 \operatorname{frac}H(k) > 1/24$$ we could decuce that the equality $a(n)=\lfloor e^{n-\gamma} +0.5\rfloor$ holds for $n$ large. $\endgroup$
    – Pietro Majer
    Commented Jun 23, 2013 at 23:28

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