Recurrence Relation $k_{n+2}=\frac{1-n}{n+2}k_n$

Question

There was an interesting question posted on here earlier today but it seems to have disappeared. With due to respect to the OP, I'll post the same question here from memory. If anyone finds the original please mark this as a duplicate and I'll remove it accordingly.

Given the recurrence relation $$k_{n+2}=\frac{1-n}{n+2}k_n$$ express $k_n$ in terms of $k_0$ and $k_1$.

Answer 1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{k_{n + 2} = {1 - n \over n + 2}\,k_{n}.\ \mbox{Express}\ k_{n}\ \mbox{in terms of}\ k_{0}\ \mbox{and}\ k_{1}:\ {\large ?}}$. Lets introduce the generating function

$$ {\cal K}\pars{z} \equiv \sum_{n\ =\ 0}^{\infty}k_{n}z^{n}\,,\qquad \verts{z} < 1 $$

such that \begin{align} 0&=\sum_{n\ =\ 0}^{\infty}\pars{n + 2}k_{n + 2}\,z^{n} +\sum_{n\ =\ 0}^{\infty}\pars{n - 1}k_{n}\,z^{n} =\sum_{n\ =\ 2}^{\infty}n\,k_{n}\,z^{n - 2} +\sum_{n\ =\ 0}^{\infty}n\,k_{n}\,z^{n} - \sum_{n\ =\ 0}^{\infty}k_{n}\,z^{n} \\[5mm]&=-\,{k_{1} \over z} +\pars{{1 \over z^{2}} + 1}\sum_{n\ =\ 0}^{\infty}n\,k_{n}\,z^{n} - \sum_{n\ =\ 0}^{\infty}k_{n}\,z^{n} \\[5mm]&=-\,{k_{1} \over z} +\pars{{1 \over z^{2}} + 1}\,z\,\partiald{}{z}\sum_{n\ =\ 0}^{\infty}k_{n}\,z^{n} -\sum_{n\ =\ 0}^{\infty}k_{n}\,z^{n} \\[5mm]&=-\,{k_{1} \over z} +\pars{{1 \over z^{2}} + 1}\,z\,\partiald{{\cal K}\pars{z}}{z} -{\cal K}\pars{z} \end{align}

$$ \partiald{{\cal K}\pars{z}}{z} -{z \over 1 + z^{2}}\,\,{\cal K}\pars{z}={k_{1} \over 1 + z^{2}} \ \imp\ \partiald{}{z}\bracks{{\cal K}\pars{z} \over \root{1 + z^{2}}} ={k_{1} \over \pars{1 + z^{2}}^{3/2}} $$

$$ {{\cal K}\pars{z} \over \root{1 + z^{2}}} - k_{0} =k_{1}\ \overbrace{\int_{0}^{z}{\dd t \over \pars{1 + t^{2}}^{3/2}}} ^{\dsc{z \over \root{1 + z^{2}}}}\ =\ {k_{1}\,z \over \root{1 + z^{2}}} $$

\begin{align} {\cal K}\pars{z}&=k_{1}\,z + k_{0}\root{1 + z^{2}} =k_{1}\,z + k_{0}\sum_{n\ =\ 0}^{\infty}{1/2 \choose n}z^{2n} =k_{1}\,z + k_{0} \sum_{\atop{n\ =\ 0 \atop n\ \mbox{even}}}^{\infty}{1/2 \choose n/2}z^{n} \end{align}

$$\color{#66f}{\large k_{n}} =\color{#66f}{\large\left\{\begin{array}{lcl} 0 & \mbox{if} & n\ \mbox{is odd}\ \wedge\ n \not= 1 \\[2mm] k_{1} & \mbox{if} & n = 1 \\[2mm] k_{0}{1/2 \choose n/2} & \mbox{if} & n\ \mbox{is even} \end{array}\right.} $$

Answer 2

Case 1, $n=2h, h\in \mathbb{N}$

$k_{2(h+1)}=(-1)\left(1-\dfrac{3}{2(h+1)}\right)k_{2h}=(-1)^{h+2}\displaystyle\prod_{i=0}^h\left(1-\dfrac{3}{2(i+1)}\right)k_0$

Case 2, $n=2h+1, h\in \mathbb{N}$

Similar analysis.

Answer 3

Putting $n=1$, we have

$$k_3=0$$

It is clear that, for $n=3, 5, 7, ...$, $$k_{n+2}=0$$ Hence $k_n=0$ for all odd $n\geq3$. $\qquad \blacksquare$

Now consider the case for even $n$, i.e. where $n=2m$.

The recurrence relation becomes $$\begin{align} k_{2m+2} &=\frac{1-2m}{2m+2}k_{2m}\\[10pt] &=-\frac{2m-1}{2m+2}k_{2m}\\[10pt] &=\left(\prod_{r=1}^{m}(-1)^r\frac{2r-1}{2r+2}\right)\color{red}{k_2}\\[10pt] &=(-1)^m\frac{1\cdot 3\cdot 5\cdots (2m-1)}{4\cdot6\cdot 8\cdots (2m+2)}\cdot \color{blue}{\frac{2m+1}{2m+1}}\cdot\color{red}{\frac{k_0}2} \\[10pt] &=(-1)^m\frac{1}{\color{blue}{2m+1}}\cdot \frac{1\cdot 3\cdot 5\cdots \color{blue}{(2m+1)}}{\color{red}2\cdot 4\cdot6\cdots (2m+2)}\color{red}{k_0}\\[10pt] &=(-1)^m\frac{1}{2m+1}\cdot \frac{1\cdot 3\cdot 5\cdots (2m+1)}{2^{m+1}(1\cdot 2\cdot 3\cdots (m+1))}\cdot\color{green}{\frac{2\cdot 4\cdot6\cdots (2m+2)}{2^{m+1}(1\cdot 2\cdot 3\cdots (m+1))}}k_0\\[10pt] &=(-1)^m\frac{1}{2m+1}\cdot \frac1{2^{2m+2}}\cdot\frac{(2m+2)!}{(m+1)!(m+1)!}k_0\\[10pt] &=(-1)^m\frac{1}{(2m+1)2^{2m+2}}\binom{2m+2}{m+1}k_0\\[10pt] k_{2m}&=(-1)^{m-1}\frac{1}{(2m-1)2^{2m}}\binom{2m}{m}k_0\qquad \blacksquare \end{align}$$

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EDITED: Alternatively, as inspired by the answer from Felix Marin below, the case for even $n$ can be derived as follows:

$$\begin{align} k_{2m+2} &=\frac{1-2m}{2m+2}k_{2m}\\[10pt] &=-\frac{2m-1}{2m+2}k_{2m}\\[10pt] &=\left(\prod_{r=1}^{m}(-1)^r\frac{2r-1}{2r+2}\right)\color{red}{k_2}\\[10pt] &=(-1)^m\cdot \frac{\color{green}1 \cdot 1\cdot 3\cdot 5\cdots (2m-1)}{\color{red}2\cdot 4\cdot6\cdot 8\cdots (2m+2)}\color{red}{k_0}\\\\ &=\frac{\frac12\left(-\frac 12\right)\left(-\frac32\right)\left(-\frac52\right)\cdots \left(-\frac{2m-1}2\right)}{1\cdot 2\cdot 3\cdots (m+1)}k_0\\\\ &=\binom{1/2}{m+1}k_0\\\\ k_{2m}&=\binom{1/2}{m}k_0\\\\ k_{n}&=\binom{1/2}{n/2}k_0\qquad\text{(for even $n$)}\qquad \blacksquare \end{align}$$

Answer 4

Clearly $k_3=0$, and by an easy induction $k_{2m+1}=0$ for $m\ge 1$. It’s not hard to work out and then prove by induction on $m$ that

$$\begin{align*} k_{2m}&=(-1)^{m-1}\cdot\frac{(2m-3)(2m-5)\ldots(3)(1)(1)}{(2m)(2m-2)\ldots(4)(2)}k_0\\\\ &=(-1)^{m-1}\cdot\frac{(2m-3)!!}{(2m)!!}k_0\\\\ &=(-1)^{m-1}\frac{(2m-3)!!}{2^mm!}k_0\\\\ &=(-1)^{m-1}\frac{(2m-2)!}{2^mm!\cdot2^{m-1}(m-1)!}k_0\\\\ &=(-1)^{m-1}\frac{(2m-2)!}{2^{2m-1}m(m-1)!^2}k_0\\\\ &=\frac{(-1)^{m-1}}{2^{2m-1}m}\binom{2m-2}{m-1}k_0\\\\ &=\frac{(-1)^{m-1}}{2^{2m-1}}C_{m-1}k_0\;, \end{align*}$$

where $C_{m-1}$ is the $(m-1)$-st Catalan number.