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Ruud H.G. van Tol on Wed, 29 Jan 2025 15:35:45 +0100
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- To: pari-users@pari.math.u-bordeaux.fr
- Subject: Re: 2^[1,3]
- From: "Ruud H.G. van Tol" <rvtol@isolution.nl>
- Date: Wed, 29 Jan 2025 15:24:02 +0100
- Delivery-date: Wed, 29 Jan 2025 15:35:45 +0100
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On 2025-01-29 14:42, Bill Allombert wrote:
> On Wed, Jan 29, 2025 at 02:28:46PM +0100, Ruud H.G. van Tol wrote:
>> ? vecsum(2^[1,3])
>> % 10.000000000000000000000000000000000000
>
> Use fromdigits:
> ? fromdigits([1,3],2)
> % = 10
Doesn't work:
? fromdigits([1,3],2)
% = 5
(because 1*2^1 + 3*2^0 = 2+3 = 5)
The (distinct) digits are exponents, are not positional.
2^1 + 2^3 = 10.
Example quest:
>> How to best convert something like (int) 6131786 to (int) 2^1 + 2^3
+ 2^6 +
>> 2^7 + 2^8?
>
> You can use the binary() function.
I don't see yet how.
-- Ruud
- Follow-Ups:
- Re: 2^[1,3]
- From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
- References:
- 2^[1,3]
- From: "Ruud H.G. van Tol" <rvtol@isolution.nl>
- Re: 2^[1,3]
- From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>