Re: Timing
- To: mathgroup at smc.vnet.net
- Subject: [mg87971] Re: Timing
- From: Artur <grafix at csl.pl>
- Date: Mon, 21 Apr 2008 06:38:28 -0400 (EDT)
- References: <480A0EE6.3090109@csl.pl>
- Reply-to: grafix at csl.pl
Who know how change bellow procedure to received reasonable timing?
Part:
Expand[((3 + 2 Sqrt[2])^(2^(n - 1) - 1) - (3 - 2 Sqrt[2])^(2^(n - 1) -
1))/(4 Sqrt[2])]/(2^n - 1)
is every time integer
> Timing[Do[ If[IntegerQ[Expand[((3 + 2 Sqrt[2])^(2^(n - 1) - 1) - (3 -
> 2 Sqrt[2])^(2^(n - 1) - 1))/(4 Sqrt[2])]/(2^n - 1)],
> Print[n]], {n, 1, 17}]]
>
I will be greatfull for any help! (Mayby some N[] or Floor[N[]] or
Int[N[]] will be quickest
Best wishes
Artur
- Follow-Ups:
- Re: Re: Timing
- From: Carl Woll <carlw@wolfram.com>
- Re: Re: Timing
- From: "W_Craig Carter" <ccarter@mit.edu>
- Re: Re: Timing
- From: "W_Craig Carter" <ccarter@mit.edu>
- Re: Re: Timing