A binomial identity from counting permutations with cycles of length two.

Question

Let $n\ge 1$ be an integer. Consider the following sequence:

where $\left\{l_j\right\}_{j=1}^n$ are indices.By analyzing all possible decompositions of the sequence above into distinct pairs we have discovered the following identity: \begin{equation} n!\sum\limits_{j=0}^{\lfloor \frac{n}{2} \rfloor}\frac{(2j-1)!!}{2^j j!} \binom{n}{2 j} = (2n-1)!! \end{equation} Note that the $j$th term in the sum is just equal to the number of decompositions into pairs where there are exactly $j$ pairs where the first and the last element of thew pair are both zeros.

Now, my question would be is there some alternative way of proving this identity, maybe using analytical methods?

Answer 1

Suppose we seek to prove that

$$n! \sum_{q=0}^{\lfloor n/2 \rfloor} \frac{(2q-1)!!}{2^q q!} {n\choose 2q} = (2n-1)!!,$$

which is

$$n! \sum_{q=0}^{\lfloor n/2 \rfloor} \frac{1}{2^q q!} \frac{(2q)!}{2^q q!} {n\choose 2q} = \frac{(2n)!}{2^n n!}$$

or

$$\sum_{q=0}^{\lfloor n/2 \rfloor} \frac{1}{4^q} {2q\choose q} {n\choose 2q} = \frac{1}{2^n} {2n\choose n}.$$

Now we have

$${2q\choose q} {n\choose 2q} = \frac{n!}{q!\times q!\times (n-2q)!} = {n\choose q} {n-q\choose q}.$$

The LHS becomes

$$\sum_{q=0}^{\lfloor n/2 \rfloor} \frac{1}{4^q} {n\choose q} {n-q\choose n-2q}$$

and we introduce

$${n-q\choose n-2q} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-2q+1}} (1+z)^{n-q} \; dz$$

This controls the range and vanishes when $2q\gt n$ so we may extend $q$ to $n$, getting for the sum

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n} \sum_{q=0}^n \frac{1}{4^q} {n\choose q} \frac{z^{2q}}{(1+z)^q} \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n} \left(1+\frac{z^2}{4(1+z)}\right)^n \; dz \\ = \frac{1}{4^n} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} (z+2)^{2n}\; dz.$$

Extracting coefficients we find

$$\frac{1}{4^n} {2n\choose n} 2^n$$ which is

$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{2^n} {2n\choose n}}$$

as claimed.

Answer 2

The general approach to tackling such questions is the Sister Celine's algorithm.One can prove that whenever the term under the sum is a hypergeometric function then that term satisfies a certain linear recursion relation. To be specific let us denote: \begin{equation} F[n,k] := n! \frac{(2k-1)!! }{2^k k!} \binom{n}{2 k} \end{equation} then there exists a linear recursion relation of the form: \begin{equation} \sum\limits_{i=0}^I \sum\limits_{j=0}^J a_{i,j} F[n-i,k-j] = 0 \end{equation} for some values of $I,J$. In practice the search for the coefficients goes as follows. Firstly, assume certain integer values for $I,J$. Secondly divide the left hand side by $F[n,k]$ bring the whole expression to a common denominator and simplify to reduce it to a polynomial form. Thirdly, express the numerator as a polynomial in $k$ set it to zero and solve for the unknown coefficients $a_{i,j}$. If there is no non-zero solution then go back to the beginning and start with some bigger values of $I,J$. In our case it is $I=2$ and $J=1$ produce a non-zero solution.In other words we have: \begin{equation} -F[n-1,k](2n-1)+F[n-2,k] (n-1)^2 - F[n-2,k-1](n-1)^2 + F[n,k] =0 \end{equation} Now define our sum as $S[n] := \sum\limits_{k=0}^\infty F[n,k]$. Then from the recursion above we have: \begin{equation} -S[n-1] (2n-1) + S[n] =0 \end{equation} subject to $S[1]=1$. This clearly gives $S[n] = (2n-1)!!$ as it should be.