Re: Re: Timing
- To: mathgroup at smc.vnet.net
- Subject: [mg88008] Re: [mg87971] Re: Timing
- From: "W_Craig Carter" <ccarter at mit.edu>
- Date: Mon, 21 Apr 2008 14:39:28 -0400 (EDT)
- References: <480A0EE6.3090109@csl.pl> <200804211038.GAA28042@smc.vnet.net>
Hello Artur,
I am not sure I understand your question:
Are you asking for
Member[expr,Integers] && Element[n,Integers]
for 1 <= n <= 17 ?
If so, this may be a hint, although it does not exactly check the above:
Map[Element[Rationalize[#, .00000001], Integers] & ,
Table[expr /. n -> i, {i, 1, 30}]]
and it is reasonably fast. This approximate method gets worse for
large n; so I don't trust the result except for n=1,3,5
Note, that
Map[Element[Rationalize[Expand[#], .00000001], Integers] & ,
Table[expr /. n -> i, {i, 1, 10}]]
Gives a different result and is much slower, thus emphasizing the
result for large n.
I am curious to see the better solutions....
On Mon, Apr 21, 2008 at 6:38 AM, Artur <grafix at csl.pl> wrote:
> 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
>
>
--
W. Craig Carter
- References:
- Re: Timing
- From: Artur <grafix@csl.pl>
- Re: Timing