[seqfan] Re: morphism in A284940 compared with A080580
Andrew Weimholt
andrew.weimholt at gmail.com
Tue Apr 25 21:06:59 CEST 2017
Correction
b=A284939.
Let a=A080580 and A be the set of terms of a
Let b=A284939
b(n)=0 => n in A => a(n) = a(n-1)+2 => b(a(n-1)+2),b(a(n-1)+3) = 0,1
b(n)=1 => n not in A => a(n) = a(n-1)+4 => b(a(n-1)+2), b(a(n-1)+3),
b(a(n-1)+4), b(a(n-1)+5) = 1,1,0,1
One should be able to prove this rigorously using induction, since the
initial terms agree.
Andrew
> On Tue, Apr 25, 2017 at 3:13 AM, Neil Sloane <njasloane at gmail.com> wrote:
>
>> In other words, RJM is saying that it appears that
>>
>> "the positions of 0's in the fixed point of the morphism 0 -> 01, 1 ->
>> 1101"
>> are given by
>> "a(1)=1; for n>1, a(n)=a(n-1)+2 if n is already in the sequence,
>> a(n)=a(n-1)+4 otherwise".
>>
>> That is a pretty interesting conjecture (A284940 =? A080580)! It
>> would be worth checking it for a few million terms, or more.
>>
>> Clark Kimberling (if you are on this list), I know you
>> have recently been studying many similar "Positions of 0's in fixed
>> point of ..."
>> sequences. Have you observed any other apparent coincidences of this
>> type?
>> Best regards
>> Neil
>>
>> Neil J. A. Sloane, President, OEIS Foundation.
>> 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>> Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
>> Phone: 732 828 6098 <(732)%20828-6098>; home page: http://NeilSloane.com
>> Email: njasloane at gmail.com
>>
>>
>>
>> On Tue, Apr 25, 2017 at 3:28 AM, Richard J. Mathar
>> <mathar at mpia-hd.mpg.de> wrote:
>> > The first 1000 terms (at least) of A284940 equal the first
>> > 1000 terms of A080580. Can this be demonstrated for all general values?
>> >
>> > Richard
>> >
>> > --
>> > Seqfan Mailing list - http://list.seqfan.eu/
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>
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