Variable coefficient difference equation I

Question

Consider the difference equation \begin{align} n \, \phi_{n+1} &= (2 \, n^{2} + 2 \, n -1) \, \phi_{n} + (n+1) \, \phi_{n-1}. \end{align} It is seen that if $\phi_{n} = \Gamma(n+2) \, \theta_{n}$ then the difference equation becomes \begin{align} (n+1)(n+3) \, \theta_{n+2} = [2(n+1)(n+2) -1] \, \theta_{n+1} + \theta_{n}. \end{align} Are there further reductions or shifts that can be made to reduce this equation further?

Answer 1

If we set: $$ f(x) = \sum_{n\geq 0}\frac{\phi_n}{n!}x^n $$ we have: $$ F(x)=\sum_{n\geq 1}\frac{\phi_{n-1}}{n!}x^n,\qquad f'(x) = \sum_{n\geq 0}\frac{\phi_{n+1}}{n!}x^{n},\qquad f''(x) = \sum_{n\geq 0}\frac{\phi_{n+2}}{n!}x^{n} $$ $$ x\, f(x) = \sum_{n\geq 1}\frac{n \phi_{n-1}}{n!}x^n,\qquad x\,f'(x)=\sum_{n\geq 0}\frac{n\phi_n}{n!}x^n,\qquad x\,f''(x)=\sum_{n\geq 0}\frac{n\phi_{n+1}}{n!}x^n $$ and: $$ x^2\,f''(x) = \sum_{n\geq 2}\frac{(n^2-n)\phi_n}{n!}x^n $$ hence our recurrence translates into: $$ x\, f'' = 2x^2\,f''+4x\,f'+f+xf+F $$ and we may recover $\phi_n$ by solving the differential equation:

$$ g+(x+1)g'+4x\,g''+(2x^2-x)g''' = 0.$$