Re: How do little quickest ?
- To: mathgroup at smc.vnet.net
- Subject: [mg93487] Re: [mg93475] How do little quickest ?
- From: Artur <grafix at csl.pl>
- Date: Tue, 11 Nov 2008 07:45:42 -0500 (EST)
- References: <gf3mj0$en2$1@smc.vnet.net> <200811091026.FAA20573@smc.vnet.net> <200811100831.DAA26388@smc.vnet.net>
- Reply-to: grafix at csl.pl
Now the quickest procedure is:
aa = {}; Do[Print[x]; rmin = 10^10; k = 2^x; w = Floor[(k - 1)/2];
Do[m = FactorInteger[k n (k - n)]; rad = 1;
Do[rad = rad m[[s]][[1]], {s, 1, Length[m]}];
If[rad < rmin, rmin = rad], {n, 1, w, 2}];
AppendTo[aa, rmin], {x, 2, 30}]; aa
because all possible numbers was odd <{n, 1, w, 2} inspite
{n, 1, w}> and GCD checking was delated because GCD[2^x-(2n-1),(2n-1)]=1
What to do yet because time is still expotential and for x=27 my computer need whole day to count.
Best wishes
Artur
Mathematical problem is following:
I need help with finding rule on follwing problem
Sequences:
B={1, 3, 7, 15, 27, 63, 125, 243, 343, 999, 1805, 3721, 8181, 16335,
32761, 65533, 112847, 190269, 519375, 1046875, 1953125}
C={2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384,
32768, 65536, 131072, 262144, 524288, 1048576, 2097152}
A={1, 1, 1, 1, 5, 1, 3, 13, 169, 25, 243, 375, 11, 49, 7, 3, 18225,
71875, 4913, 1701, 144027}
C(n) = 2^n
A(n) = C(n) - B(n)
A(n) < B(n) < C(n)
is obtained by algorhitm such
that GCD[A(n),B(n),C(n)] = 1 and product of distinct prime divisors of
A(n)*B(n)*C(n) have minimal value.
I'm looking for formula for B(n).
Who can help?
Artur pisze:
> Dear Mathematica Gurus!
> Who know how do quickest following prcedure:
>
> aa = {}; Do[Print[x]; rmin = 10^10; k = 2^x; w = Floor[(k - 1)/2];
> Do[If[GCD[n, k, k - n] == 1, m = FactorInteger[k n (k - n)]; rad = 1;
> Do[rad = rad m[[s]][[1]], {s, 1, Length[m]}];
> If[rad < rmin, rmin = rad]], {n, 1, w}];
> AppendTo[aa, rmin], {x, 2, 30}]; aa
>
> Best wishes
> Artur
>
>
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- How do little quickest ?
- From: Artur <grafix@csl.pl>
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