Re: log_int_rat

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

On Wed, Nov 29, 2023 at 07:44:11PM +0100, Ruud H.G. van Tol wrote:
> 
> ? a054414(n) = 1 + n + floor( n * log(2) / log(3/2) );
> 
> ? [ a054414(n) |n<-[0..20]]
> % [1, 3, 6, 9, 11, 14, 17, 19, 22, 25, 28, 30, 33, 36, 38, 41, 44, 47, 49,
> 52, 55]
> 
> Is there a cleaner way, similar to logint, to do that floor-expression?

Using logint(2^n,3/2) would be much slower even if it worked, because that would
force you to compute 2^n.

Really, the only defect of your function is that you do not take the accuracy
into account so

? a054414(2^1000)
  *** floor: precision too low in truncr (precision loss in truncation).

but this is easily fixed:

a054414(n) = localbitprec(logint(n+2,2));1 + n + floor( n * log(2) / log(3/2) );

? a054414(2^1000)
%8 = 29032646699514618648207963388262277036465203935797085941659905401800597040332207967488444244921882872990640968909754332584729580377856680178651318617855116668095773049095433042501905586769687829752260479426676858529835781762375806913040422676074996080564646620120844538289072080537948895898995900461439
  ***   last result computed in 0 ms.

If being slow is not an issue, I suppose you can do:

a054414(n)=my(a=2^n,b=1);while(a>=b,a*=2;b*=3;n++);n

and

table(n)=my(a=1/2,b=1);vector(n+1,i,a*=2;while(a>=b,a*=2;b*=3);valuation(a,2));

? table(20)
%12 = [1,3,6,9,11,14,17,19,22,25,28,30,33,36,38,41,44,47,49,52,55]

Cheers,
Bill.