Re: How do little quickest ?

```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;
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;
>  AppendTo[aa, rmin], {x, 2, 30}]; aa
>
> Best wishes
> Artur
>
>
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```

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