Why if $a_n=-n\sum_{m=1}^{n}\frac{(-1)^{m}}{m}\binom{m+n-1}{2m-1}b_m$ then $b_m = \sum_{n=1}^{m}(-1)^{n+1}\binom{2m}{m-n}a_n?$

Question

I tried different equality of binomial coefficients, but with no success. Also I tried with inverse binomial coefficients, but I couldn't prove the equality. If someone can help, I will appreciate it.

Answer 1

First, we need to fix the indexing, so that both sequences start with index $0$, not $1$. After doing that, we can rephrase your question as

Why, if $\displaystyle a_m=\sum_{n=0}^{m}(-1)^{n}\frac{m+1}{n+1}\binom{m+n+1}{2n+1}b_n$, then $\displaystyle b_m = \sum_{n=0}^{m}(-1)^{n}\binom{2m+2}{m-n}a_n$ ?

Basically, we have matrix multiplication here. Let $\vec{a}=[a_m]_{m\ge 0}$ and $\vec{b}=[b_m]_{m\ge 0}$ be infinite column vectors, and let $$ X=\left[(-1)^{n}\frac{m+1}{n+1}\binom{m+n+1}{2n+1}\right]_{m,n\ge 0}, \qquad Y=\left[(-1)^{n}\binom{2n+2}{n-m}\right]_{m,n\ge 0}. $$ Note that both $X$ and $Y$ are lower triangular matrices (as both formulas yield $0$ when $m<n$ with all $1$'s on the diagonal. Your question is, therefore:

Why, if $\vec{a}=X\vec{b}$, then $\vec{b}=Y\vec{a}$ ?

The answer, obviously, is because $XY=I$, or $Y=X^{-1}$. To show this, consider the generating functions of the columns of $X$ and $Y$. The generating function of the $n$th column of $X$ (where $n\ge 0$) is $$ \begin{split} \sum_{m=0}^{\infty}(-1)^{n}\frac{m+1}{n+1}\binom{m+n+1}{2n+1}z^m&=\sum_{m=n}^{\infty}(-1)^{n}\frac{m+1}{n+1}\binom{m+n+1}{2n+1}z^m\\ &=\sum_{m=0}^{\infty}(-1)^{n}\frac{m+n+1}{n+1}\binom{m+2n+1}{2n+1}z^{m+n}\\ &=\frac{(-1)^n}{n+1}\frac{d}{dz}\left(\sum_{m=0}^{\infty}\binom{m+2n+1}{2n+1}z^{m+n+1}\right)\\ &=\frac{(-1)^n}{n+1}\frac{d}{dz}\left(z^{n+1}\sum_{m=0}^{\infty}\binom{m+2n+1}{2n+1}z^m\right)\\ &=\frac{(-1)^n}{n+1}\frac{d}{dz}\left(z^{n+1}\frac{1}{(1-z)^{2n+2}}\right)\\ &=\frac{(-1)^n}{n+1}\frac{d}{dz}\left(\left(\frac{z}{(1-z)^2}\right)^{n+1}\right)\\ &=\frac{(-1)^n}{n+1}(n+1)\left(\frac{z}{(1-z)^2}\right)^{n+1}\left(\frac{1}{z}+\frac{2}{1-z}\right)\\ &=\frac{1+z}{(1-z)^3}\left(-\frac{z}{(1-z)^2}\right)^n. \end{split} $$ When, for any $n\ge 0$, we can write the generating function of the $n$th column as $gf^n=g(z)(f(z))^n$ for some fixed $g$ and $f$, we call such a matrix the Riordan array $(g,f)$. Thus, $$ X=\left(\frac{1+z}{(1-z)^3},\frac{-z}{(1-z)^2}\right). $$ Similarly, the $n$th column of $Y$, for any $n\ge 0$, has the generating function $$ \begin{split} \sum_{m=0}^{\infty}(-1)^{n}\binom{2m+2}{m-n}z^m&=\sum_{m=n}^{\infty}\binom{2m+2}{m-n}(-z)^m\\ &=(-1)^{n}\sum_{m=0}^{\infty}\binom{2m+2n+2}{m}z^{m+n}\\ &=(-z)^n\sum_{m=0}^{\infty}\binom{2m+2n+2}{m}z^m\\ &=(-z)^nB(z)C(z)^{2n+2}=\\ &=B(z)C(z)^2(-zC(z)^2)^n=\\ &=(BC^2)(-zC^2)^n, \end{split} $$ and thus $Y$ is also a Riordan array, $$ Y=(BC^2,-zC^2), $$ where $B=B(z)=\dfrac{1}{\sqrt{1-4z}}$ and $C=C(z)=\dfrac{1-\sqrt{1-4z}}{2z}$. Note that $C^2=1+zC^2$ and $B=\dfrac{1}{1-2zC}$.

Why do we need that? Here are some basic facts about Riordan arrays.

1. Given a sequence $\vec{c}=(c_n)_{n\ge 0}$ with generating function $h(z)$ and a Riordan array $X=(g,f)$, the generating function of $X\vec{c}$ is $(g,f)*h=g\cdot(h\circ f)=g(z)h(f(z))$.

2. From (1), it follows that the product of Riordan arrays is also a Riordan array, namely, $(g,f)(G,F)=(g\cdot(G\circ f), F\circ f)$. The identity matrix is also a Riordan array, $I=(1,z)$.

3. From (2), if $X=(g,f)$ has $g(0)\ne 0$, $f(0)=0$, $f'(0)\ne 0$ (so $g$ has a reciprocal and $f$ has an inverse), then $X$ is lower triangular and invertible, and so is $X^{-1}$. Moreover, $$ X^{-1}=(g,f)^{-1}=\left(\frac{1}{g\circ \bar{f}}, \bar{f}\right), $$ where $\bar{f}$ denotes the compositional inverse of $f$.

Now we have all we need. Taking the composition of the second components of $X$ and $Y$, we get $$ \frac{-z}{(1-z)^2}\circ(-zC^2)=\frac{zC^2}{(1+zC^2)^2}=\frac{zC^2}{C^2}=z, $$ so they are compositional inverses of each other. Thus, for $g=g(z)=\dfrac{1+z}{(1-z)^3}$, $f=f(z)=\dfrac{-z}{(1-z)^2}$, we have $\bar{f}=-zC^2$, so $$ \frac{1}{g\circ \bar{f}}=\frac{(1-z)^3}{1+z}\circ(-zC^2)=\frac{(1+zC^2)^3}{1-zC^2}=\frac{C^3}{C-2zC^2}=\frac{C^2}{1-2zC}=BC^2, $$ the first component of $Y$. Thus, $Y=X^{-1}$, and we are done.

Answer 2

Starting with the question posed by @AlexanderBurstein we ask, why, if

$$a_n = \sum_{m=0}^{n}(-1)^{m}\frac{n+1}{m+1} {n+m+1\choose 2m+1} b_m$$ then

$$b_n = \sum_{m=0}^{n}(-1)^{m} {2n+2\choose n-m} a_m?$$

We substitute into the formula for $b_n$:

$$b_n = \sum_{q=0}^{n}(-1)^{q} {2n+2\choose n-q} a_q \\ = \sum_{q=0}^{n}(-1)^{q} {2n+2\choose n-q} \sum_{m=0}^{q}(-1)^{m}\frac{q+1}{m+1} {q+m+1\choose 2m+1} b_m \\ = \sum_{m=0}^n (-1)^m b_m \sum_{q=m}^n (-1)^q {2n+2\choose n-q} \frac{q+1}{m+1} {q+m+1\choose 2m+1}.$$

This means we require

$$(-1)^m [[n=m]] = \sum_{q=m}^n (-1)^q {2n+2\choose n-q} \frac{q+1}{m+1} {q+m+1\choose 2m+1}$$

which is

$$[[n=m]] = \sum_{q=0}^{n-m} (-1)^q {2n+2\choose n-m-q} \frac{q+m+1}{m+1} {q+2m+1\choose 2m+1}.$$

We get two components, the first is

$$\sum_{q=0}^{n-m} (-1)^q {2n+2\choose n-m-q} {q+2m+1\choose 2m+1} \\ = [z^{n-m}] (1+z)^{2n+2} \sum_{q\ge 0} (-1)^q z^q {q+2m+1\choose 2m+1} \\ = [z^{n-m}] (1+z)^{2n+2} \frac{1}{(1+z)^{2m+2}} = [z^{n-m}] (1+z)^{2n-2m} \\ = {2n-2m\choose n-m}.$$

The second component is (this is zero when $n=m$ and we will assume $n\gt m$ in the following):

$$\frac{1}{m+1} \sum_{q=0}^{n-m} (-1)^q {2n+2\choose n-m-q} q {q+2m+1\choose q} \\ = \frac{2m+2}{m+1} \sum_{q=1}^{n-m} (-1)^q {2n+2\choose n-m-q} {q+2m+1\choose q-1} \\ = - 2\sum_{q=0}^{n-m-1} (-1)^q {2n+2\choose n-m-1-q} {q+2m+2\choose q} \\ = -2 [z^{n-m-1}] (1+z)^{2n+2} \sum_{q\ge 0} (-1)^q z^q {q+2m+2\choose 2m+2} \\ = - 2 [z^{n-m-1}] (1+z)^{2n+2} \frac{1}{(1+z)^{2m+3}} = -2 [z^{n-m-1}] (1+z)^{2n-2m-1} \\ = -2 {2n-2m-1\choose n-m-1}.$$

Collecting everything,

$${2n-2m\choose n-m} - 2 [[n \ne m]] {2n-2m-1\choose n-m-1}.$$

This evaluates to one when $n=m$ and when $n\gt m$ we get

$${2n-2m\choose n-m} - 2 {2n-2m-1\choose n-m-1} = {2n-2m\choose n-m} - 2 \frac{n-m}{2n-2m} {2n-2m\choose n-m} = 0.$$

We may conclude.