# Formula:KLS:14.02:17

$\displaystyle {\displaystyle q^{-n}(1-q^n)(1-\alpha\beta q^{n+1})y(x) {}=B(x)y(x+1)-\left[B(x)+D(x)\right]y(x)+D(x)y(x-1) }$

## Substitution(s)

$\displaystyle {\displaystyle D(x)=\frac{q(1-q^x)(1-\delta q^x)(\beta-\gamma q^x)(\alpha-\gamma\delta q^x)} {(1-\gamma\delta q^{2x})(1-\gamma\delta q^{2x+1})}}$ &

$\displaystyle {\displaystyle B(x)=\frac{(1-\alpha q^{x+1})(1-\beta\delta q^{x+1})(1-\gamma q^{x+1})(1-\gamma\delta q^{x+1})} {(1-\gamma\delta q^{2x+1})(1-\gamma\delta q^{2x+2})}}$ &
$\displaystyle {\displaystyle y(x)=\qRacah{n}@{\mu(x)}{\alpha}{\beta}{\gamma}{\delta}{q}}$ &
$\displaystyle {\displaystyle \mu(x)=q^{-x}+\gamma\delta q^{x+1} =\lambda(x)=q^{-x}+cq^{x-N} =q^{-x}+q^{x+\gamma+\delta+1} =2a\cos@@{\theta}}$ &
$\displaystyle {\displaystyle \lambda(x)=x(x+\gamma+\delta+1)}$ &
$\displaystyle {\displaystyle \mu(x):=q^{-x}+\gamma\delta q^{x+1}}$ &
$\displaystyle {\displaystyle \mu(x)=q^{-x}+\gamma\delta q^{x+1} =\lambda(x)=q^{-x}+cq^{x-N} =q^{-x}+q^{x+\gamma+\delta+1} =2a\cos@@{\theta}}$ &

$\displaystyle {\displaystyle \lambda(x)=x(x+\gamma+\delta+1)}$

## Proof

We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.

## Symbols List

& : logical and
$\displaystyle {\displaystyle R_{n}}$  : $\displaystyle {\displaystyle q}$ -Racah polynomial : http://dlmf.nist.gov/18.28#E19
$\displaystyle {\displaystyle \mathrm{cos}}$  : cosine function : http://dlmf.nist.gov/4.14#E2