Abramov's algorithm

In mathematics, particularly in computer algebra, Abramov's algorithm computes all rational solutions of a linear recurrence equation with polynomial coefficients. The algorithm was published by Sergei A. Abramov in 1989.

Universal denominator

The main concept in Abramov's algorithm is a universal denominator. Let K {\textstyle \mathbb {K} } be a field of characteristic zero. The dispersion dis ⁡ ( p , q ) {\textstyle \operatorname {dis} (p,q)} of two polynomials p , q ∈ K [ n ] {\textstyle p,q\in \mathbb {K} [n]} is defined asdis ⁡ ( p , q ) = max { k ∈ N : deg ⁡ ( gcd ( p ( n ) , q ( n + k ) ) ) ≥ 1 } ∪ { − 1 } , {\displaystyle \operatorname {dis} (p,q)=\max\{k\in \mathbb {N} \,:\,\deg(\gcd(p(n),q(n+k)))\geq 1\}\cup \{-1\},}where N {\textstyle \mathbb {N} } denotes the set of non-negative integers. Therefore the dispersion is the maximum k ∈ N {\textstyle k\in \mathbb {N} } such that the polynomial p {\textstyle p} and the k {\textstyle k}-times shifted polynomial q {\displaystyle q} have a common factor. It is − 1 {\textstyle -1} if such a k {\textstyle k} does not exist. The dispersion can be computed as the largest non-negative integer root of the resultant res n ⁡ ( p ( n ) , q ( n + k ) ) ∈ K [ k ] {\textstyle \operatorname {res} _{n}(p(n),q(n+k))\in \mathbb {K} [k]}. Let ∑ k = 0 r p k ( n ) y ( n + k ) = f ( n ) {\textstyle \sum _{k=0}^{r}p_{k}(n)\,y(n+k)=f(n)} be a recurrence equation of order r {\textstyle r} with polynomial coefficients p k ∈ K [ n ] {\displaystyle p_{k}\in \mathbb {K} [n]}, polynomial right-hand side f ∈ K [ n ] {\textstyle f\in \mathbb {K} [n]} and rational sequence solution y ( n ) ∈ K ( n ) {\textstyle y(n)\in \mathbb {K} (n)}. It is possible to write y ( n ) = p ( n ) / q ( n ) {\textstyle y(n)=p(n)/q(n)} for two relatively prime polynomials p , q ∈ K [ n ] {\textstyle p,q\in \mathbb {K} [n]}. Let D = dis ⁡ ( p r ( n − r ) , p 0 ( n ) ) {\textstyle D=\operatorname {dis} (p_{r}(n-r),p_{0}(n))} andu ( n ) = gcd ( [ p 0 ( n + D ) ] D + 1 _ , [ p r ( n − r ) ] D + 1 _ ) {\displaystyle u(n)=\gcd([p_{0}(n+D)]^{\underline {D+1}},[p_{r}(n-r)]^{\underline {D+1}})}where [ p ( n ) ] k _ = p ( n ) p ( n − 1 ) ⋯ p ( n − k + 1 ) {\textstyle [p(n)]^{\underline {k}}=p(n)p(n-1)\cdots p(n-k+1)} denotes the falling factorial of a function. Then q ( n ) {\textstyle q(n)} divides u ( n ) {\textstyle u(n)}. So the polynomial u ( n ) {\textstyle u(n)} can be used as a denominator for all rational solutions y ( n ) {\textstyle y(n)} and hence it is called a universal denominator.

Algorithm

Let again ∑ k = 0 r p k ( n ) y ( n + k ) = f ( n ) {\textstyle \sum _{k=0}^{r}p_{k}(n)\,y(n+k)=f(n)} be a recurrence equation with polynomial coefficients and u ( n ) {\textstyle u(n)} a universal denominator. After substituting y ( n ) = z ( n ) / u ( n ) {\textstyle y(n)=z(n)/u(n)} for an unknown polynomial z ( n ) ∈ K [ n ] {\textstyle z(n)\in \mathbb {K} [n]} and setting ℓ ( n ) = lcm ⁡ ( u ( n ) , … , u ( n + r ) ) {\textstyle \ell (n)=\operatorname {lcm} (u(n),\dots ,u(n+r))} the recurrence equation is equivalent to∑ k = 0 r p k ( n ) z ( n + k ) u ( n + k ) ℓ ( n ) = f ( n ) ℓ ( n ) . {\displaystyle \sum _{k=0}^{r}p_{k}(n){\frac {z(n+k)}{u(n+k)}}\ell (n)=f(n)\ell (n).}As the u ( n + k ) {\textstyle u(n+k)} cancel this is a linear recurrence equation with polynomial coefficients which can be solved for an unknown polynomial solution z ( n ) {\textstyle z(n)}. There are algorithms to find polynomial solutions. The solutions for z ( n ) {\textstyle z(n)} can then be used again to compute the rational solutions y ( n ) = z ( n ) / u ( n ) {\textstyle y(n)=z(n)/u(n)}.

Example

The homogeneous recurrence equation of order 1 {\textstyle 1}( n − 1 ) y ( n ) + ( − n − 1 ) y ( n + 1 ) = 0 {\displaystyle (n-1)\,y(n)+(-n-1)\,y(n+1)=0}over Q {\textstyle \mathbb {Q} } has a rational solution. It can be computed by considering the dispersionD = dis ⁡ ( p 1 ( n − 1 ) , p 0 ( n ) ) = disp ⁡ ( − n , n − 1 ) = 1. {\displaystyle D=\operatorname {dis} (p_{1}(n-1),p_{0}(n))=\operatorname {disp} (-n,n-1)=1.}This yields the following universal denominator:u ( n ) = gcd ( [ p 0 ( n + 1 ) ] 2 _ , [ p r ( n − 1 ) ] 2 _ ) = ( n − 1 ) n {\displaystyle u(n)=\gcd([p_{0}(n+1)]^{\underline {2}},[p_{r}(n-1)]^{\underline {2}})=(n-1)n}andℓ ( n ) = lcm ⁡ ( u ( n ) , u ( n + 1 ) ) = ( n − 1 ) n ( n + 1 ) . {\displaystyle \ell (n)=\operatorname {lcm} (u(n),u(n+1))=(n-1)n(n+1).}Multiplying the original recurrence equation with ℓ ( n ) {\textstyle \ell (n)} and substituting y ( n ) = z ( n ) / u ( n ) {\textstyle y(n)=z(n)/u(n)} leads to( n − 1 ) ( n + 1 ) z ( n ) + ( − n − 1 ) ( n − 1 ) z ( n + 1 ) = 0. {\displaystyle (n-1)(n+1)\,z(n)+(-n-1)(n-1)\,z(n+1)=0.}This equation has the polynomial solution z ( n ) = c {\textstyle z(n)=c} for an arbitrary constant c ∈ Q {\textstyle c\in \mathbb {Q} }. Using y ( n ) = z ( n ) / u ( n ) {\textstyle y(n)=z(n)/u(n)} the general rational solution isy ( n ) = c ( n − 1 ) n {\displaystyle y(n)={\frac {c}{(n-1)n}}}for arbitrary c ∈ Q {\textstyle c\in \mathbb {Q} }.