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The recurrence $a_{n+1}=a_n(n-1/2)$ is related to $\Gamma(n+1/2)$ ( not difficult to prove) and it could be represented in a way like $\frac {(2n-1)!!} {2^n}$ Also I know that $(2n-1)!!$ is the number of permutations of 2n whose cycle type consists of n parts equal to 2; these are the involutions without fixed points (A).

Also, for each $n \in N$, let $f(n)$ is the number of subsets of set $[n]=\ {1,2,...,n}$. Then $f(n)=2^n$ (B)

I wonder about the understanding of the meaning ( sense) of the ration: A/B? What could be the meaning of $\frac {(2n-1)!!} {2^n}$ in terms of sets?

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Can you clarify your question? Your first question asks about A/B, which is then $\frac{(2n-1)!!}{4^n}$, and you second asks about A which you seem to have answered already. – Mitch Feb 23 '11 at 23:56
I changed the tags. – JDH Feb 24 '11 at 5:20
up vote 4 down vote accepted

This is a good question, although its statement may need to be cleaned up a little bit. It occasionally happens that looking for patterns between seemingly different types of problems and seemingly different branches of mathematics produces real results. The issue with this particular problem is not that there aren't any interpretations that this quantity can be given, but that there are too many. The solution is not to throw this problem away, but to refine it to the point that it admits a clear answer. This is a natural process in mathematical research.

I suspect that your looking for a combinatorial interpretation to the formula $\frac{\left(2n-1\right)!!}{2^n}$. Since $\mbox{gcd}\left(2^n,\left(2n-1\right)!!\right) = 1$ for all $n \geq 1$, this formula cannot be interpreted as enumerating the points in some specified finite set. Since $2^n < \left(2n-1\right)!!$ for all $n\geq 3$, this formula cannot be interpreted as a probability of some kind.

The formula $\frac{\left(2n-1\right)!!}{n!2^n}$ can be interpreted as giving the probability that if two people each flip two separate fair coins $n$ times, then each person gets heads the same number of times. I worked this out by unraveling $\frac{A}{B}$ using binomial identities until I got something that looked like the probability of some easily described random event.

Edit: In the light of this question and its answers, it seems that there is a connection between this simple game I've described and the numbers you are interested in. It's strange how these things happen.

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Thank you, it's very helpful. Actually, there is one problem that the real recurrence is $a(n+1)=a(n)(n-1/2)+o(1/n). Do you know a way to select an interval when your interpretation works for the slightly changed recurrence? – Mikhail G Feb 24 '11 at 19:41
I don't know of any ways to modify this game with respect to small changes in your recurrence in such a way that the new game has the same relationship to the new recurrence as the old game had to the original recurrence. I'm not even sure what I meant by "same relationship" in the previous sentence. This doesn't mean that it's not possible to do such a thing though. – Albert Steppi Feb 26 '11 at 23:14

@Albert: The recurrence $M(n+1)/M(n)=n-1/2+o(1/n)$ is related to Kendall-Mann property

Could you look at the answer from Moron please Recurrence representation(s): $a(n+1)=a(n)(n-1/2)+o(1/n)$ and $a(n+1)=a(n)(n-1/2+o(1/n))$ It seems to me that your game is a good interpretation, am I right?

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I'm glad I could be of help before, but I don't think I know enough to add anything more. It seems that $M\left(n\right)$ will tend to $C\frac{\left(2n-1\right)!!}{2^n}$, for some constant $C$, but I don't know how to connect the above game to the Mahonian distribution. It might have something to do with the Mahonian and Binomial distributions both being asymptotically normal, but I don't know much about this stuff, and you probably shouldn't listen to me. – Albert Steppi Feb 28 '11 at 16:32
Well, I just know that "Mixing of Diffusing Particles" tends to be Mahonian But I do not know any answer about the meaning for the large number n in the process. It looks like you jump from one level to another($M(n+1)/M(n)$) and this produce some results. Also, it has some relation to Markov chains. All in one, everything is unclear now to me. – Mikhail G Feb 28 '11 at 17:41
Having glanced through the paper you linked to, I now think that it's no accident that the coin tossing game I described is related to this Mahonian distribution business. When I come to understand this connection a little better, I'll add something along these lines to my answer. – Albert Steppi Feb 28 '11 at 19:08
Yes, you are right, there is no accident here, Moreover, please look at Lane Clark work about An Asymptotic Expansion for the Number of Permutations with a Certain Number of Inversions you'll see that there is deep connection. But by now I am trying to understand the meaning of the recurrence. – Mikhail G Feb 28 '11 at 19:18

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