Re: Re: Timing

*To*: mathgroup at smc.vnet.net*Subject*: [mg88010] Re: [mg87971] Re: Timing*From*: Artur <grafix at csl.pl>*Date*: Mon, 21 Apr 2008 14:39:51 -0400 (EDT)*References*: <480A0EE6.3090109@csl.pl> <200804211038.GAA28042@smc.vnet.net> <8ef293d70804210554i3d187e9at22ab6f9e8c743125@mail.gmail.com>*Reply-to*: grafix at csl.pl

Dear Craig, My problem is find such numbers n that expression ((3 + 2 Sqrt[2])^(2^(n - 1) - 1) - (3 - 2 Sqrt[2])^(2^(n - 1) - 1))/(4 Sqrt[2]) is divisable by (2^n - 1) or another words (2^n - 1) is divisor of ((3 + 2 Sqrt[2])^(2^(n - 1) - 1) - (3 - 2 Sqrt[2])^(2^(n - 1) - 1))/(4 Sqrt[2]) My procedure 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}] is extremaly slow (probably Expand function do that as aslowly but without Expand we don't have integers. I don't have idea how improove them. Best wishes Artur W_Craig Carter pisze: > 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 >> >> >> > > > >

**References**:**Re: Timing***From:*Artur <grafix@csl.pl>