Abramov's algorithm

The main concept in Abramov's algorithm is a universal denominator. Let ${\textstyle \mathbb {K} }$ be a field of characteristic zero. The dispersion ${\textstyle \operatorname {dis} (p,q)}$ of two polynomials ${\textstyle p,q\in \mathbb {K} [n]}$ is defined as

$${\displaystyle \operatorname {dis} (p,q)=\max\{k\in \mathbb {N} \,:\,\deg(\gcd(p(n),q(n+k)))\geq 1\}\cup \{-1\},}$$

where ${\textstyle \mathbb {N} }$ denotes the set of non-negative integers. Therefore the dispersion is the maximum ${\textstyle k\in \mathbb {N} }$ such that the polynomial ${\textstyle p}$ and the ${\textstyle k}$-times shifted polynomial ${\displaystyle q}$ have a common factor. It is ${\textstyle -1}$ if such a ${\textstyle k}$ does not exist. The dispersion can be computed as the largest non-negative integer root of the resultant ${\textstyle \operatorname {res} _{n}(p(n),q(n+k))\in \mathbb {K} [k]}$.[3][4] Let ${\textstyle \sum _{k=0}^{r}p_{k}(n)\,y(n+k)=f(n)}$ be a recurrence equation of order ${\textstyle r}$ with polynomial coefficients ${\displaystyle p_{k}\in \mathbb {K} [n]}$, polynomial right-hand side ${\textstyle f\in \mathbb {K} [n]}$ and rational sequence solution ${\textstyle y(n)\in \mathbb {K} (n)}$. It is possible to write ${\textstyle y(n)=p(n)/q(n)}$ for two relatively prime polynomials ${\textstyle p,q\in \mathbb {K} [n]}$. Let ${\textstyle D=\operatorname {dis} (p_{r}(n-r),p_{0}(n))}$ and

$${\displaystyle u(n)=\gcd([p_{0}(n+D)]^{\underline {D+1}},[p_{r}(n-r)]^{\underline {D+1}})}$$

where ${\textstyle [p(n)]^{\underline {k}}=p(n)p(n-1)\cdots p(n-k+1)}$ denotes the falling factorial of a function. Then ${\textstyle q(n)}$ divides ${\textstyle u(n)}$. So the polynomial ${\textstyle u(n)}$ can be used as a denominator for all rational solutions ${\textstyle y(n)}$ and hence it is called a universal denominator.[5]

Let again ${\textstyle \sum _{k=0}^{r}p_{k}(n)\,y(n+k)=f(n)}$ be a recurrence equation with polynomial coefficients and ${\textstyle u(n)}$ a universal denominator. After substituting ${\textstyle y(n)=z(n)/u(n)}$ for an unknown polynomial ${\textstyle z(n)\in \mathbb {K} [n]}$ and setting ${\textstyle \ell (n)=\operatorname {lcm} (u(n),\dots ,u(n+r))}$ the recurrence equation is equivalent to

$${\displaystyle \sum _{k=0}^{r}p_{k}(n){\frac {z(n+k)}{u(n+k)}}\ell (n)=f(n)\ell (n).}$$

As the ${\textstyle u(n+k)}$ cancel this is a linear recurrence equation with polynomial coefficients which can be solved for an unknown polynomial solution ${\textstyle z(n)}$. There are algorithms to find polynomial solutions. The solutions for ${\textstyle z(n)}$ can then be used again to compute the rational solutions ${\textstyle y(n)=z(n)/u(n)}$. [2]

algorithm rational_solutions is
    input: Linear recurrence equation ${\textstyle \sum _{k=0}^{r}p_{k}(n)\,y(n+k)=f(n),p_{k},f\in \mathbb {K} [n],p_{0},p_{r}\neq 0}$.
    output: The general rational solution ${\textstyle y}$ if there are any solutions, otherwise false.

    ${\textstyle D=\operatorname {disp} (p_{r}(n-r),p_{0}(n))}$
    ${\textstyle u(n)=\gcd([p_{0}(n+D)]^{\underline {D+1}},[p_{r}(n-r)]^{\underline {D+1}})}$
    ${\textstyle \ell (n)=\operatorname {lcm} (u(n),\dots ,u(n+r))}$
    Solve ${\textstyle \sum _{k=0}^{r}p_{k}(n){\frac {z(n+k)}{u(n+k)}}\ell (n)=f(n)\ell (n)}$ for general polynomial solution ${\textstyle z(n)}$
    if solution ${\textstyle z(n)}$ exists then
        return general solution ${\textstyle y(n)=z(n)/u(n)}$
    else
        return false
    end if

The homogeneous recurrence equation of order ${\textstyle 1}$

$${\displaystyle (n-1)\,y(n)+(-n-1)\,y(n+1)=0}$$

over ${\textstyle \mathbb {Q} }$ has a rational solution. It can be computed by considering the dispersion

$${\displaystyle D=\operatorname {dis} (p_{1}(n-1),p_{0}(n))=\operatorname {disp} (-n,n-1)=1.}$$

This yields the following universal denominator:

$${\displaystyle u(n)=\gcd([p_{0}(n+1)]^{\underline {2}},[p_{r}(n-1)]^{\underline {2}})=(n-1)n}$$

and

$${\displaystyle \ell (n)=\operatorname {lcm} (u(n),u(n+1))=(n-1)n(n+1).}$$

Multiplying the original recurrence equation with ${\textstyle \ell (n)}$ and substituting ${\textstyle y(n)=z(n)/u(n)}$ leads to

$${\displaystyle (n-1)(n+1)\,z(n)+(-n-1)(n-1)\,z(n+1)=0.}$$

This equation has the polynomial solution ${\textstyle z(n)=c}$ for an arbitrary constant ${\textstyle c\in \mathbb {Q} }$. Using ${\textstyle y(n)=z(n)/u(n)}$ the general rational solution is

$${\displaystyle y(n)={\frac {c}{(n-1)n}}}$$

for arbitrary ${\textstyle c\in \mathbb {Q} }$.