The diffeqtorec(deq, y(z), u(n)) command converts a linear differential equation, deq, into a recurrence.

Let $f$ be a power series solution of the differential equation.

If u(n) is the nth Taylor coefficient of $f$ around zero, the diffeqtorec function returns a linear recurrence for the numbers u(n), with rational coefficients in n.

The syntax is the same as that of dsolve.  Combined with algeqtodiffeq, this function produces a linear recurrence for the Taylor coefficients of an algebraic function.

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

$\mathrm{diffeqtorec}\left(y\left(z\right)=a\mathrm{diff}\left(y\left(z\right)\,z\right)\,y\left(z\right)\,v\left(n\right)\right)$

$v\left(n\right)+\left(-an-a\right)v\left(n+1\right)$

$\mathrm{deq}≔\mathrm{algeqtodiffeq}\left(y=1+z\left(y^2+y^3\right)\,y\left(z\right)\,\varnothing \right)\:$

$\mathrm{diffeqtorec}\left(\mathrm{deq}\,y\left(z\right)\,u\left(m\right)\right)$

$\left\{\left(-2m^2-m\right)u\left(m\right)+\left(-18m^2-30m-9\right)u\left(m+1\right)+\left(46m^2+227m+279\right)u\left(m+2\right)+\left(-4m^2-26m-42\right)u\left(m+3\right)\,u\left(0\right)=1\,u\left(1\right)=2\,u\left(2\right)=10\right\}$