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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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    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

1 Answers 1

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For the recurrence $\displaystyle a_{n+1} = (n - 1/2) a_n + o(1/n)$

If you set $\displaystyle a_n = \frac{(2n-1)!!}{2^n} - e_n$,

you get $\displaystyle e_{n+1} - e_n = o\left(\frac{1}{n}\right)$ and thus $e_n = o(\log n)$.

Using Stirling's approximation, I believe we have that

$\displaystyle \frac{(2n-1)!!}{2^n} \approx \frac{Kn!}{\sqrt{n}}$

Thus we get

$\lim_{n \to \infty} \frac{2^n a_n}{(2n-1)!!} = 1 $

Now if the recurrence was $\displaystyle a_{n+1} = a_{n} (n - 1/2 + o(1/n))$

Set $\displaystyle a_{n+1} = e_n\frac{(2n-1)!!}{2^n}$

We get

$\displaystyle \frac{e_{n+1}}{e_n} = 1 + o(1/n^2)$

And thus

$\displaystyle e_n = e_1 \prod_{k=1}^{n-1} (1 + o(1/k^2))$

Now using $1 + x \le e^x$ we get that

$\displaystyle e_n = \mathcal{O}(1)$

and we can easily show that $\displaystyle e_n$ is convergent.

Thus

$a_n \sim \frac{C(2n-1)!!}{2^n}$

for some constant $\displaystyle C = \lim_{n \to \infty} e_n$ .

We could chose your $\displaystyle f(n) = o(1/n)$ as we like, to make this constant different from $\displaystyle 1$.

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    @Moron: Thank you very much. Your answer is fine, but I am trying to find a clear picture from fifferent sides. I hope the papers I referenced will light something new in a time. Will work on that.2011-02-28