Meanwhile, I found one answer myself.
https://fricas.github.io/api/RecurrenceOperator
ROZE==>RecurrenceOperator(Integer, Expression Integer)
getEq(res)$ROZE
I do not like this abbreviated name. That goes against the usual rule of
having fully expanded names. We have ==> to abbreviate.
Other question...
aList(n)==[n*factorial(n+2*k-1)/(factorial(n + k)*factorial(k))*(-1)^k_
for k in 0..14];
g := guessPRec(aList 1)
[[f(n): (n+2)*f(n+1)+(4*n+2)*f(n)=0, f(0)=1]]
rec := getEq(first g)$ROZE
(n+2)*f(n+1)+(4*n+2)*f(n)=0
Such a recurrence cries for turning it into a program that can compute
the elements of the sequence.
Of course, it doesn't work to simply define
fexpr := subst(f(n+1)-rec/(n+2),n=n-1)
f(0) == 1
f(n) == fexpr
because then f(1) would return exactly the expression involving n.
also
f(k) == subst(fexpr, n=k)
doesn't work, since it does not trigger recursion.
(1) Is it actually possible to turn an expression into a program?
(2) How do I transform rec to an expression of the form
f(n) = ... f(n-1) + ... f(n-2)
but NOT doing by hand.
(3) Can I extract the initial values from the result of guessPRec?
Thank you
Ralf
On 10/6/20 11:30 AM, Ralf Hemmecke wrote:
> Suppose I have
>
> aList(n)==[n*factorial(n+2*k-1)/(factorial(n + k)*factorial(k))*(-1)^k_
> for k in 0..14]
> res := guessAlg(aList 1)
>
> The result shows as:
>
> n 2 2 3 4
> [[[x ]f(x): x f(x) + f(x) - 1 = 0, f(x) = 1 - x + 2 x - 5 x + O(x )]]
>
> with type List(Expression(Integer)).
>
> I can do
>
> (7) -> k := first rest tower res.1
>
> n 2 2 3 4
> [[x ]f(x): x f(x) + f(x) - 1 = 0, f(x) = 1 - x + 2 x - 5 x + O(x )]
> Type: Kernel(Expression(Integer))
> (8) -> operator k
>
> (8) rootOfADE
> Type: BasicOperator
> (9) -> argument k
>
> (9) [n, %infoser0()]
> Type: List(Expression(Integer))
>
> But if I want the equation? How do I get it from res?
>
> In general it would also for the other guess functions to be able to do
> something with the result other than displaying it.
>
> Ralf
>
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