Re: symbol manipulation

["Ruud H.G. van Tol" <rvtol@isolution.nl>]

On 2022-11-20 15:09, Bill Allombert wrote:
> On Sun, Nov 20, 2022 at 01:45:26PM +0100, Ruud H.G. van Tol wrote:
> > I'm looking for a simpler way to investigate how a formula evolves, for
> > formulas with inner loops.
> >
> > My test-formula:
> > A071521(n)= my(t=1/3; sum(k=0, logint(n, 3), t*=3; logint(n\t, 2)+1)
> >
> > {
> > f(n)=my( t= 1/3
> > , add_(v1,v2)=  Str("((",v1,")+(",v2,"))")
> > , mul_(v1,v2)=  Str("((",v1,")*(",v2,"))")
> > , idiv_(v1,v2)= Str("((",v1,")\\(",v2,"))")
> > , logint_(v,b)= Str("logint(",v,",",b,")")
> > , sum_(f,t,s)=  Str("sum(k_=",f,",",t,",",s,")")
> > , a_(n)= sum_(
> >      k=0, logint_(n,3), add_(logint_(idiv_(n, (mul_(t,3))), 2), 1))
> > );
> > a_(n)
> > }
> >
> > my(f_); eval(Str("f_()=",f(2))); f_
> > %3 = ()->my(f_);sum(k_=0,logint(5,3),((logint(((5)\(((1/3)*(3)))),2))+(1)))
>
> This does not seem quite right, since this does not depend on k.

In this case the k is only used as an iteration count.
Do you mean that in such a case, sum() is not the best function to use?

Compare:
[sum(k=0, logint(n, 3), 1+logint(3*n\(t*=3),2)) |n<-[1..20]; t<-[1]]


> > The ultimately desired return value is "logint(x,2)".
> > How feasible/doable is that?
>
> It is unclear what purpose this would serve you.

"code lineage". For example to show per call what the inner loop looks 
like when it would be pre-resolved/inlined.


> One trick is to convert functional values to variables this way:
> ? eval(Str("'logint_",x,"_",2))
> %3 = logint_x_2
>
> that you can then add and multiply as needed.
Thanks.

-- Ruud