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I have a problem to find a general form of the sequence \begin{align} - \frac{{n\left( {n - 1} \right)}}{{2\left( {2n - 1} \right)}},\frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{2 \cdot 4 \cdot \left( {2n - 1} \right)\left( {2n - 3} \right)}}, - \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)\left( {n - 5} \right)}}{{2 \cdot 4 \cdot 6 \cdot \left( {2n - 1} \right)\left( {2n - 3} \right)\left( {2n - 5} \right)}}, \cdots :=a_n(k),\qquad n\ge 2 \end{align}

and then to find the sum $\sum |a_n(k)|^2$, $1\le k\le n$.

I have tried as follows:

$a_n(k)=\frac{(-1)^kP(n,2k)}{(2k)!!A_n(k)}$, where $P(n,2k)=\frac{n!}{(n-2k)!}$ and $\,\,A_n(k):=(2n-1)(2n-3)(2n-5)\cdots (2n-2k+1),\qquad k\ge1$

Is this true. If yes, can we write $A_n(k)$ in a closed form?! After all of that what is the sum of $|a_n(k)|^2$, $1\le k\le n$.

Mohammad W. Alomari
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  • Do you want to treat the first term of the sequence as $k=0$ or as $k=1$? Are you familiar with falling factorials? Are you familiar with double factorials? E.g. $n(n-1)(n-2)(n-3)=n^{\underline{4}}$. You may be more familiar with the notation $P(n,4)$ or $\frac{n!}{(n-4)!}$. Also of use are the double factorials I mention, $8!!=8\cdot 6\cdot 4\cdot 2$ for example. – JMoravitz Nov 17 '16 at 21:03
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    Thanks. I need $k\ge1$. I tried as follows $a_n=\frac{P(n,2k)}{(2k)!! A_n(k)}$ I cannot treat $A_n(k)$; the right terms in the denominator. Is $A_n(k) = (2n-1)(2n-3)\cdots(2n-2k+1)$, then how one can write such $A_n(k)$ in a closed form. – Mohammad W. Alomari Nov 17 '16 at 21:07
  • Good, that should be added to the main post for the question so it shows that you've at least put some effort into thinking about the problem. Note that double factorials are used for odd numbers as well as for even numbers. Similar to how $P(n,4)=\frac{n!}{(n-4)!}$ can you think about a way to find the ratio of two double factorials to leave only the terms you want? – JMoravitz Nov 17 '16 at 21:09
  • Its ok I will add what I have tried. Thanks – Mohammad W. Alomari Nov 17 '16 at 21:13
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    As for writing $A_n(k)$, notice, what is $(2n-1)!!$? What is $(2n-1-2k)!!$? What is the ratio of these? As for the question of finding the sum, perhaps now that we see the sequence written in a closed form, we can find some more convenient way to organize things, perhaps by writing things with binomial coefficients that yields a nicer identity. I do not personally see a convenient method of attack right away, perhaps someone else will. – JMoravitz Nov 17 '16 at 21:25
  • For the sum, are we doing something along the lines of $a_n(1)+a_n(2)+\dots$ or $a_1(k)+a_2(k)+\dots$? – Simply Beautiful Art Nov 17 '16 at 21:26
  • @SimpleArt the "$1\leq k\leq n$" implies to me that it is the first one, i.e. limits of summation are $\sum\limits_{k=1}^n$ – JMoravitz Nov 17 '16 at 21:27
  • @JMoravitz Thanks for the clarification. Wasn't completely sure. – Simply Beautiful Art Nov 17 '16 at 21:28
  • Instead of the double factorial for even numbers, I'd recommend using $(2k)!!=2^kk!$ – Simply Beautiful Art Nov 17 '16 at 21:30
  • Thanks JMoravitz and SimpleArt. I have found that $A_n(k)=\frac{(2n-1)!!}{(2n-2k+3)!!}$, $1\le k \le n$. So how we can proceed now? – Mohammad W. Alomari Nov 17 '16 at 21:47
  • SimpleArt's suggestion seems to be (and I agree with him) try to rewrite all doublefactorials as single factorials and exponentials noting that evens can be converted noting $(2k)!!=2^kk!$ and odds can be converted noting $(2k+1)!!=\frac{(2k+1)!}{(2k)!!}=\frac{(2k+1)!}{2^kk!}$. In doing so, might recognize binomial coefficients or some other nicer pattern or cancellations. – JMoravitz Nov 17 '16 at 21:58
  • Ok, I will try that, Thanks all – Mohammad W. Alomari Nov 17 '16 at 22:00

1 Answers1

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Here is a more compact representation as sum formula, most of it was already stated in the comment section.

Since \begin{align*} a_n(k)=\frac{(-1)^kn(n-1)\cdots (n-2k+1)}{2\cdot4\cdots (2k)\cdot(2n-1)(2n-3)\cdots(2n-2k+1)}\qquad\qquad 1\leq k\leq n \end{align*}

We obtain \begin{align*} a_n(k)&=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{(2k)!!}\cdot\frac{(2n-2k-1)!!}{(2n-1)!!}\tag{1}\\ &=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{(2k)!!}\cdot\frac{(2n-2k)!}{(2n-2k)!!}\cdot\frac{(2n)!!}{(2n)!}\tag{2}\\ &=(-1)^k\frac{n!}{(n-2k)!}\cdot\frac{1}{2^kk!}\cdot\frac{(2n-2k)!}{2^{n-k}(n-k)!}\cdot\frac{2^nn!}{(2n)!}\tag{3}\\ &=(-1)^k\frac{n!n!}{(2n)!}\cdot\frac{1}{k!(n-k)!}\cdot\frac{(2n-2k)!}{(n-2k)!}\\ &=\frac{(-1)^k\binom{n}{k}\binom{2n-2k}{n}}{\binom{2n}{n}} \end{align*}

Comment:

  • In (1) we use double factorials $(2n)!!=(2n)(2n-2)\cdots4\cdot 2$

  • In (2) we use $(2n)!=(2n)!!(2n-1)!!$

  • In (3) we use $(2n)!!=2^nn!$

I don't think that the series \begin{align*} \sum_{k=1}^n\left|a_n(k)\right|^2=\binom{2n}{n}^{-2}\sum_{k=1}^{n}\binom{n}{k}^2\binom{2n-2k}{n}^2\qquad\qquad n\geq 1 \end{align*} has a nice closed formula. The first few terms are \begin{align*} 0,4,144,3636,82000,1764400,37164736,\ldots \end{align*} but they are not known to OEIS.

I've also tried some standard techniques and checked for instance section 2.9 in Riordan Array Proofs of Identities in Gould’s Book which contains binomial identities of the type we need, but without success.

Wolfram Alpha provides following representation via hypergeometric series \begin{align*} \sum_{k=1}^n\left|a_n(k)\right|^2 &={}_{4}F_{3}\left(\frac{1}{2}-\frac{n}{2},\frac{1}{2}-\frac{n}{2},-\frac{n}{2},-\frac{n}{2};1,\frac{1}{2}-n,\frac{1}{2}-n;1\right)-1 \end{align*}

Markus Scheuer
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