Re: 2^[1,3]

Bill Allombert on Wed, 29 Jan 2025 22:08:02 +0100

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Re: 2^[1,3]

To: pari-users@pari.math.u-bordeaux.fr
Subject: Re: 2^[1,3]
From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
Date: Wed, 29 Jan 2025 22:07:57 +0100
References: <d43e2330-fb29-4889-8a13-b90b99dd7f01@isolution.nl> <Z5owN4oJqB1X4pAA@seventeen> <f80e7470-a2b1-4770-86a4-e6618f0bf9c4@isolution.nl> <Z5pAbp1TOsX6J7jQ@seventeen>

On Wed, Jan 29, 2025 at 03:51:26PM +0100, Bill Allombert wrote:
> On Wed, Jan 29, 2025 at 03:24:02PM +0100, Ruud H.G. van Tol wrote:
> > 
> > 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

I geuss you can do vecsum([2^i|i<-[1,3]])

Cheers,
Bill.

Follow-Ups:
- Re: 2^[1,3]
  - From: "Ruud H.G. van Tol" <rvtol@isolution.nl>

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>
- Re: 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>

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