The RationalSolution(eq, var, term) command returns the rational solution of the linear difference equation eq. If such a solution does not exist, the function returns NULL.

The hypergeometric term in the linear difference equation is specified by a name, for example, t. The meaning of the term is defined by the parameter term. It can be specified directly in the form of an equation, for example, $t=n!$, or specified as a list consisting of the name of term variable and the consecutive term ratio, for example, $\left[t\,n+1\right]$.

If the third parameter is omitted, then the input equation can contain a hypergeometric term directly (not a name). In this case, the procedure extracts the term from the equation, transforms the equation to the form with a name representing a hypergeometric term, and then solves the transformed equation.

The term "rational solution" means a solution $y\left(x\right)$ in $Q\left(x\right)\left(t\right)$. (See PolynomialSolution for the meaning of "polynomial solution".) Here we use the term "denominator" which is q in $Q\left(x\right)\left[t\right]$ to mean that $qy$ is in $Q\left(x\right)\left[t,t^{-1}\right]$ .

The search for a rational solution is based on finding a universal denominator which is u in $Q\left(x\right)\left[t\right]$ such that $uy$ is in $Q\left(x\right)\left[t,t^{-1}\right]$ for any rational solution y. By replacing y with $\frac{Y}{u}$ in the given equation, we reduce the problem to searching for a polynomial solution.

The solution is the function, corresponding to var. The solution involves arbitrary constants of the form, for example, _c1 and _c2.

$\mathrm{with}\left(\mathrm{LREtools}\left[\mathrm{HypergeometricTerm}\right]\right)\:$

$\mathrm{eq}≔y\left(n+1\right)\left(1+\left(n+1\right)t\right)-y\left(n\right)\left(1+t\right)$

$\mathrm{eq}≔y\left(n+1\right)\left(1+\left(n+1\right)t\right)-y\left(n\right)\left(1+t\right)$

$\mathrm{RationalSolution}\left(\mathrm{eq}\,y\left(n\right)\,t=n!\right)$

$\frac{{\mathrm{\_C}}_1}{1+t},\left[t\,n+1\right]$

$\mathrm{eq}≔\mathrm{numer}\left(\frac{y\left(n+2\right)\left(n+2+\left(n+2\right)\left(n+1\right)t\right)}{1+\left(n+2\right)\left(n+1\right)t}-\frac{y\left(n\right)\left(n+t\right)}{1+t}\right)$

$\mathrm{eq}≔y\left(n+2\right)n^2{t}^2-y\left(n\right)n^{3}t-y\left(n\right)n^2{t}^2+y\left(n+2\right)n^{2}t+3y\left(n+2\right)n{t}^2-3y\left(n\right)n^{2}t-3y\left(n\right)n{t}^2+4y\left(n+2\right)nt+2y\left(n+2\right)t^2-2y\left(n\right)nt-2y\left(n\right)t^2+y\left(n+2\right)n+4y\left(n+2\right)t-y\left(n\right)n-y\left(n\right)t+2y\left(n+2\right)$

$\mathrm{RationalSolution}\left(\mathrm{eq}\,y\left(n\right)\,t=n!\right)$

$\frac{{\mathrm{\_C}}_1\left(1+t\right)}{n+t},\left[t\,n+1\right]$

$\mathrm{eq}≔y\left(n+1\right)\left(2t+n+1\right)-y\left(n\right)\left(t+n\right)$

$\mathrm{eq}≔y\left(n+1\right)\left(2t+n+1\right)-y\left(n\right)\left(n+t\right)$

$\mathrm{RationalSolution}\left(\mathrm{eq}\,y\left(n\right)\,t=2^n\right)$

$\frac{{\mathrm{\_C}}_1}{n+t},\left[t\,2\right]$

$\mathrm{eq}≔y\left(n+1\right)\left(2^{n+1}+n+1\right)-y\left(n\right)\left(2^n+n\right)$

$\mathrm{eq}≔y\left(n+1\right)\left(2^{n+1}+n+1\right)-y\left(n\right)\left(2^n+n\right)$

$\mathrm{RationalSolution}\left(\mathrm{eq}\,y\left(n\right)\right)$

$\frac{{\mathrm{\_C}}_1}{t+n},\left[t\,2\right]$

Abramov, S.A., and Bronstein, M. "Hypergeometric dispersion and the orbit problem." Proc. ISSAC 2000.

Bronstein, M. "On solutions of Linear Ordinary Difference Equations in their Coefficients Field." INRIA Research Report. No. 3797. November 1999.