Re: symbol manipulation

[Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>]

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.

> The ultimately desired return value is "logint(x,2)".
> How feasible/doable is that?

It is unclear what purpose this would serve you.

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.

Cheers,
Bill