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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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Equivalently, there exist no integers $k \ge 1$ and $n$ with $\log(k+1/2) + \gamma < n < \log(k+1/2) + \gamma + 1/(24k^2)$. –  Gerald Edgar Jun 23 '13 at 12:00
    
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. –  Sylvain JULIEN Jun 23 '13 at 12:09
    
@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. –  joro Jun 23 '13 at 12:45
2  
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). –  Henry Cohn Jun 23 '13 at 14:28
    
Let me summarize. $k=a(n)$ means $$n\in I_k:=[H(k-1),H(k) )$$ $k=\lfloor e^{n-\gamma} +0.5\rfloor$ is equivalent to $$n\in J_k:=[L(k-1), L(k) ),$$ for $L(k):=\log(k+1/2)+\gamma$. Asymptotics for $H$ and $L$ imply that $I_k$ and $J_k$ are quite close intervals, which makes more and more likely that $I_k$ meets $\mathbb{N}$ iff $J_k$ does (the thesis being that this is true for all $k$). Moreover, it is encouraging that $H(k)$ is never an integer for $k > 1$ (a consequence of the Bertrand's postulate), that is, if an integer $n$ is in $I_k$, it is not on its boundary. –  Pietro Majer Jun 23 '13 at 21:12

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