Re: A306044(n)

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

On 2022-11-16 16:56, Bill Allombert wrote:
> On Wed, Nov 16, 2022 at 04:21:10PM +0100, Ruud H.G. van Tol wrote:
> > I added PARI-code to https://oeis.org/A306044
> > "Powers of 2, 3 and 5."
> > but I'm not happy with it.
> >
> > {
> > a(n)=
> > my(f= [2, 3, 5]);  /* ordered co-primes (?) */
> > for(i= 1, #f
> > , my(p= (n-1)\f[i], d= -1, j= 0, m= 0);
> >    while( j < n
> >    , d++;
> >      m= f[i] ^ (p+d);
> >      j= 1; for(k=1, #f, j+= logint(m, f[k])); /* j= index */
> >      if( (j > n) && 0 == d, j=0; p--; d=-1) /* retry */
> >    );
> >    if(j==n, return(m))  /* found */
> > )
> > }
> >
> > The initial value of 'p' is a (simple) guess.
>
> Seems to me you want to invert
> b(n)=logint(n,2)+logint(n,3)+logint(n,5)+1;
>
> an approximation is thus
> h(n)=exp((n-1)/(1/log(2)+1/log(3)+1/log(5)))
>
> This function is a good approximation of a(n):
> g(n)=my(f=[2,3,5],e,a=round(exp((n+1)/(1/log(2)+1/log(3)+1/log(5))),&e));vecmax([p^logint(a,p)|p<-f]);

Thanks!

I used it for a better guess:
p= logint(exp((n+1)/(1/log(2)+1/log(3)+1/log(5))),f[i])

That already leads to much less steps per call.

And I'm now investigating how to remove the backtracking.
(IOW: check if it indeed doesn't overshoot anymore)

-- Ruud