[seqfan] Re: 8/5 Sequence
Maximilian Hasler
maximilian.hasler at gmail.com
Tue Jun 12 19:19:51 CEST 2012
AFAICS Yasutoshi thought that A082010 should be the orbit of 1 (with a
prefixed 0),
but of course it is the 8/5 *map* itself, not the orbit of an element.
(PARI)
A082010(x)=if(bittest(x,0),8*x\5+1,x\2)
vector(99,i,t=if(i>1,A082010(t),1))
%2 = [1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1,...]
vector(99,i,t=if(i>1,A082010(t),3))
%3 = [3, 5, 9, 15, 25, 41, 66, 33, 53, 85, 137, 220, 110, 55, 89, ...]
Maximilian
On Tue, Jun 12, 2012 at 12:50 PM, Neil Sloane <njasloane at gmail.com> wrote:
> Yasutoshi, Thank you for posting that message.
> You say there is something wrong with A082010.
> The sequence looks correct to me, and so does
> the related sequence A152199.
> What exactly needs to be changed?
> Neil
>
> On Mon, Jun 11, 2012 at 5:01 AM, kohmoto <zbi74583.boat at orange.zero.jp>wrote:
>
>> Hi, Neil and Seqfan
>>
>>
>> I think A082010 is not correct. it and the sequence of definition are
>> different.
>> If you started from 0 then it must be the following.
>> 0,1,2,1,2,....
>>
>> It is not so interesting.
>> The first interesting case is the following.
>>
>>
>> 3,5,9,15,25,41,66,33,53,85,137,220,110,55,89,143,229,367,588,294,147,236,118,59,95,153,245,393
>>
>> http://list.seqfan.eu/pipermail/seqfan/2009-September/009818.html
>>
>> Ask Don Reble.
>> He has 2000000 terms of the sequence.
>>
>> Once Franklin wrote
>> >This can be simplified: if x is odd, then next x is floor(8/5*x)+1.
>> >No need for separate m and n; let r be the ratio (m/n), and if x is
>> >odd, the next x is floor(r*x)+1. And now we don't even need r to be
>> >rational.
>>
>>
>> You are right.
>> 8/5 sequence is one of the K-Sequence.
>>
>> a(n)=[A*a(n-1)+B]/p^m
>> Where [N] is integer part of N, p^m is the highest p power dividing
>> [A*a(n-1)+B]
>>
>> http://boat.zero.ad.jp/~zbi74583/ess0.htm
>>
>> 3 examples :
>> See A028948, A029580, A036982
>>
>> >When r is the golden ratio, (sqrt(5)+1)/2, starting with 3, we get
>> >A001595, a sequence of all odd numbers, so there is unlimited growth.
>> >This may shed some light on what is happening with 8/5.
>>
>> I think your opinion is interesting.
>> My observation is the following.
>>
>> p=2, a(0)=3, A=1.6, 1<=B<2
>> In this area all sequences are unlimited.
>>
>> You say the case of A=(1+root(5))/2 is also unlimited.
>> How did you find this fact?
>> I think it is your discovery.
>>
>> Once I conjectured as follows.
>>
>> "K-Sequence which represents Fibonacci Sequence exists"
>>
>> I knew that A001959 is one of the K-Sequence.
>> So my conjecture is almost right.
>>
>>
>>
>> Yasutoshi
>>
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
>
> --
> Dear Friends, I have now retired from AT&T. New coordinates:
>
> Neil J. A. Sloane, President, OEIS Foundation
> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA
> Phone: 732 828 6098; home page: http://NeilSloane.com
> Email: njasloane at gmail.com
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
More information about the SeqFan
mailing list