[seqfan] Re: Rule needed
Maximilian Hasler
maximilian.hasler at gmail.com
Sun Nov 9 23:46:08 CET 2008
> B={1, 3, 7, 15, 27, 63, 125, 243, 343, 999, 1805...
> C={2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048,...
> A={1, 1, 1, 1, 5, 1, 3, 13, 169, 25, 243, 375, 11, ....
> A(n) < B(n) < C(n)
A(1)=B(1) so you must mean "An <= Bn"
gcd( An, Bn, Cn) = 1 with Cn=2^n
implies that An or Bn is odd, but then automatically
gcd( An, Bn, Cn) = 1 since Cn=2^n
so this is of no use. Do you rather mean gcd(An,Bn)=1 ?
But in fact, you always have
gcd(An,Bn) = gcd(An,2^n-An) = gcd( An,2^n) = 1
(for odd An) ! So this condition is also useless.
Note that in the product of distinct prime divisors = square free
kernel = A007947
you can omit Cn=2^n which just contibutes a factor 2 each time
and you can consider A007947(An) * A007947(Bn) since their gcd is 1.
Thus the problem can be simplified as follows:
A(n) = 2k-1, 0 < k <= 2^(n-2), such that
A007947(2k-1)*A007947(2^n-2k+1) is minimized.
That said, this seems quite nontrivial. I have no idea if an
elementary formula could be found.
In PARI/gp: sfk(n) = prod(i=1,#n=factor(n)[,1]~,n[i]);
AJ(n)={ local( /*best:*/ b=1, /*minimal sfk:*/ m=2^n-b);
forstep( a=3,2^(n-1),2, sfk(a)*sfk(2^n-a)<m | next; m =
sfk(b=a)*sfk(2^n-a)); b }
for(n=1,99,print1(AJ(n)","))
1,1,1,1,5,1,3,13,169,25,243,375,11,49,7,3,18225,71875,4913,1701,144027,1825,3483,2197,
***break
Maximilian
On Sun, Nov 9, 2008 at 12:17 PM, Artur <grafix at csl.pl> wrote:
> Dear Seqfans,
> 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?
> Best wishes
> Artur
>
>
>
>
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