The Modular[ModularOrePoly](Ore, p) calling sequence computes the normal form of the Ore polynomial Ore modulo the prime p

The Modular[Content](Ore, p) calling sequence computes the content of the Ore polynomial Ore modulo the prime p. If the third (optional) argument is present, it is assigned the primitive part of Ore.

The Modular[Primitive](Ore, p) calling sequence computes the primitive part of Ore modulo the prime p. If the third (optional) argument is present, it is assigned the content of Ore.

The Modular[MonicAssociate](Ore, p) calling sequence computes the left monic associate of the Ore polynomial Ore modulo the prime p.

$\mathrm{with}\left(\mathrm{OreTools}\right)\:$

$\mathrm{Ore}≔'\mathrm{OrePoly}'\left(-\frac{n}{n-1}\,-\frac{-5n+n^2+3}{n-1}\,n-3\,\frac{38}{n-3}\right)$

$\mathrm{Ore}≔\mathrm{OrePoly}\left(-\frac{n}{n-1}\,-\frac{n^2-5n+3}{n-1}\,n-3\,\frac{38}{n-3}\right)$

$\mathrm{Modular}\left[\mathrm{ModularOrePoly}\right]\left(\mathrm{Ore}\,19\right)$

$\mathrm{OrePoly}\left(\frac{18n}{n+18}\,\frac{18\left(n^2+14n+3\right)}{n+18}\,n+16\right)$

$\mathrm{Modular}\left[\mathrm{ModularOrePoly}\right]\left(\mathrm{Ore}\,11\right)$

$\mathrm{OrePoly}\left(\frac{10n}{n+10}\,\frac{10\left(n^2+6n+3\right)}{n+10}\,n+8\,\frac{5}{n+8}\right)$

$\mathrm{Ore}≔'\mathrm{OrePoly}'\left(-n\,3n-n^2-1\,{\left(n-1\right)}^2\right)$

$\mathrm{Ore}≔\mathrm{OrePoly}\left(-n\,-n^2+3n-1\,{\left(n-1\right)}^2\right)$

$\mathrm{Modular}\left[\mathrm{Primitive}\right]\left(\mathrm{Ore}\,7\,'c'\right)$

$\mathrm{OrePoly}\left(6n\,6n^2+3n+6\,n^2+5n+1\right)$

$c$

$1$

$\mathrm{Modular}\left[\mathrm{Content}\right]\left('\mathrm{OrePoly}'\left(n^2-1\,n+10\right)\,11\,'\mathrm{pp}'\right)$

$n+10$

$\mathrm{pp}$

$\mathrm{OrePoly}\left(n+1\,1\right)$

$\mathrm{Ore}≔'\mathrm{OrePoly}'\left(n^2\,3n-n^2-1\,{\left(n-17\right)}^2\right)$

$\mathrm{Ore}≔\mathrm{OrePoly}\left(n^2\,-n^2+3n-1\,{\left(n-17\right)}^2\right)$

$\mathrm{Modular}\left[\mathrm{MonicAssociate}\right]\left(\mathrm{Ore}\,17\right)$

$\mathrm{OrePoly}\left(1\,\frac{16n^2+3n+16}{n^2}\,1\right)$