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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


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