Re: factor(x+1)

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

On 2021-12-29 14:38, Bill Allombert wrote:
> On Wed, Dec 29, 2021 at 02:12:14PM +0100, Ruud H.G. van Tol wrote:

> > Example:
> > 103
> > -> 2^3*13
> > -> 2^3*(2*7-1)
> > -> 2^3*(2*(2^3-1)-1)
>
> It seems to me you should only keep the exponents and work backward:
> Try this:
>
> tofp1(x)=my(v=valuation(x,2));x>>=v;if(x==1,return([v]),concat(v,tofp1(x+1)))
> fromfp1(v)=my(n=1);v=Vecrev(v);for(i=1,#v,n=2^v[i]*n-1);n+1
>
> ? tofp1(103)
> %18 = [0,3,1,3]
> ? 2^0*(2^3*(2^1*(2^3-1)-1)-1)
> %20 = 103
> ? fromfp1([0,3,1,3])
> %21 = 103

Thanks, that really helps!

I forgot to mention
that I'd like to stop at some minimal factor,
being 3 in my example. (so no 3 -> 2^2-1 step)

Am now considering using basically your approach,
with each element a factor-Vec.

With only 2 and 3 around,
I could also use 2 elements per step,
but "flattening" always bites some day.

-- Ruud