[seqfan] Re: A178307

[Jeremy Gardiner]

Yes, I get:

3,3,2,3,3,2,3,4,2,3,3,4,2,3,3,2,3,2,4,4,2,3,3,5,2,3,4,2,4,2,2,3,4,2,3,2,3,3,
3,2,2,4,4,4,8,2,2,2,3,3,2,4,5,3,2,3,3,4,2,4,2,2,4,3,3,2,2,2,3,2,4,4,2,2,3,4,
2,2,4,3,3,4,5,3,5,2,2,2,6,4,4,2,4,2,2,8,2,2,2,2,3,2,3,3,3,3,5,2,2,4,5,3,2,3,
2,5,3,5,2,3,4,2,4,2,2,4,3,2,2,2,3,4,3,5,3,2,3,2,2,2,2,4,2,3,3,3,2,7,4,4,6,2,
2,2,2,3,2,2,4,2,2,3,2,3,4,3,3,3,3,3,4,2,5,3,2,2,5,2,5,2,3,2,2,4,6,4,2,4,4,2,
2,4,2,3,2,5,2,8,2,2,2,3,2,3,2,2,5,3,3,2,5,3,2,3,3,3,2,2,3,3,5,2,2,2,2,4,2,5,
2,2,3,3,2,3,3,2,2,3,5,3,2,4,5,3,2,3,3

Your sequence b(n) is A075930 (Positions of check bits in code in A075928),
see comment in that sequence.

I get the following sequence for the corresponding values of b(n)^m:

343,1331,169,2744,6859,441,10648,390625,676,21952,29791,1500625,1369,54872,6
8921,1764,85184,2209,5764801,6250000,2704,166375,175616,714924299,3721,23832
8


On 24/5/10 16:56, "Vladimir Shevelev" <shevelev at bgu.ac.il> wrote:

Dear  Jeremy,

I ask you also to extend a "dual" sequence:

%I A178307
%S A178307 3,3,2,3,3,2,3,4,2,3
%N A178307 Remove from A000069 powers of 2. Let b(n) be n-th term of the
remain sequence. Then a(n) is the least number m such that (b(n))^m is in
A001969
%F A178307 If k=b(n)=2^m*b(s), where b(s) is odd, then a(n)=a(s).
%Y A178307 A000069 A001969 A178253
%K A178307 nonn
%O A178307 1,1

Regards,
Vladimir