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I know that the recurrence $\displaystyle a(n+1)=a(n)(n-1/2)$ can be represented like $\displaystyle \frac{(2n-1)!!}{2^n}$

Actually the initial recurrence is slightly different: $$\displaystyle a(n+1)=a(n)(n-1/2)+o(1/n)$$

I am looking for the new representation of it in terms of $\displaystyle \frac{(2n-1)!!}{2^n}$.

Probably the question has to be refined. The recurrence is very close to the property of Kendall-Mann numbers http://oeis.org/A181609 that is why I am trying to find an interval of $\displaystyle n$'s when the $\displaystyle \frac{(2n-1)!!}{2^n}$ or similar representation works fine.

I understand that $\displaystyle o(1/n)$ is quite big error for the representation.

I am also interested in the case the recurrence is

$$\displaystyle a(n+1)=a(n)(n-1/2+o(1/n))$$

which is closer to the recurrence satisfied by the Kendall-Mann numbers.

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    Clearly it makes a big difference whether you mean $a(n)(n-1/2+o(1/n))$ or $a(n)(n-1/2)+o(1/n)$. In the OEIS entry for the Kendall-Mann numbers you linked to, it is the former; do you actually mean the latter?2011-02-24
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    Fire and forget? mjq asked a valid question...2011-02-26
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    Sorry Moron and mjqxxxx, I was too invloved with "Bath towel on the rope" question. The actual recurrence to study futher: $M(n+1)/M(n)=n-1/2+o(1/n)$ from OEIS.2011-02-27
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    I'm not sure I understand the connection with the Kendall-Mann numbers. The K-M numbers at http://oeis.org/A000140 have a combinatorial interpretation but no closed form function or even a recurrence. The numbers defined at http://oeis.org/A181609, despite the title, just seem to be equal to K-M for the first few then start to diverge. How is A181609 associated with K-M?2011-03-04

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