[seqfan] Re: morphism in A284940 compared with A080580
njasloane at gmail.com
Thu Apr 27 18:49:55 CEST 2017
Joe, Many thanks for that proof! I have added a (slightly
edited) version to A080580, and I've recycled A284940, and moved the b-file
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; home page: http://NeilSloane.com
Email: njasloane at gmail.com
On Wed, Apr 26, 2017 at 8:08 AM, Joe Slater <seqfan at slatermold.com> wrote:
> This may be a more direct explanation:
> The series A284939 consists of a concatenation of the groups 01 and 1101,
> each group representing a single element from the previous iteration of the
> transformation taken in order. Because the series has reached its fixed
> point we don't need to compare our groups of elements to the elements of
> the previous iteration: we can actually relate the present iteration to
> I.e., this table relates elements of A284939 to its groups. You would end
> up with the same series by concatenating either the elements or the groups:
> 0 -> 01
> 1 -> 1101
> 1 -> 1101
> 1 -> 1101
> 0 -> 01
> 1 -> 1101
> ... ...
> Consider any arbitrary element A284939(n) and its successor A284939(n+1) .
> They relate to the groups n and n+1, and since neither of the possible
> groups have sequential zeroes there are only three possibilities for the
> two elements (0,1; 1,0; and 1,1) and therefore three possibilities for the
> groups: 01 1101, 1101 01, or 1101 1101. We can see that the only time there
> is a gap of three 1's between the first and second zeros will be when the
> two elements A284939(n) and A284939(n+1) are (1,0). Therefore, if the nth
> zero of the series (corresponding to the element A284939(n)) is at position
> k, it will be followed by a zero at position k+2 if the element
> A284939(n+1)=0, but otherwise it will be followed by a zero at position
> Now, let's relate this to series A080580:
> We know series A284939 starts with a zero in position 1 (i.e.,
> A284939(1)=0), so we can make a series A284940 listing the position of the
> zeroes with A284940(1)=1. From our earlier discussion we know that the nth
> zero of A284939 relates to the nth element of A284939. Therefore, if
> A284940(n)=k, the following element 284940(n+1) will be +4 for if
> A284939(n+1)=1 and +2 for if A284939(n+1)=0. But we don't actually need to
> refer to A284939 at all! When we come to A284940(n+1) we can just see
> whether (n+1) appears in our series. If the value (n+1) already appears in
> A284940 then we know that A284939(n+1)=0, so A284940(n+1)=A284940(n)+2. If
> the value (n+1) does not appear in A284940 then we know that
> A284939(n+1)=1, so A284940(n+1)=A284940(n)+4.
> And this rule is exactly the rule of A080580, which means that A080580 and
> A284940 are identical, and A080580 manages to predict the positions of the
> zeroes in A284939 without ever referring to the series itself.
> Joe Slater
> On Wed, Apr 26, 2017 at 12:00 AM, jean-paul allouche <
> jean-paul.allouche at imj-prg.fr> wrote:
> > Dear J oe
> > I am n ot sure I understand y our reas oning: if A284939(n) = 0,
> > A284939(n+1) = 1, s o this bl ock 01 (beginning at p ositi on n)
> > will be transf ormed int o 011101, but there can be (and there is) a
> > of intermediate 0's and 1's inbetween (e.g., if
> > yvu l o ok at the bl ock 01 beginning at index 5, its image under the m
> > orphism d oes occur at p ositi on 15, s o that y ou missed
> > tw o intermediate 0's. May be I missed s omething?
> > best
> > jpa
> > Le 25/04/17 à 14:45, Joe Slater a écrit :
> >> The correspondence between A080580 and the zeroes of A284939 is not
> >> merely coincidental. In effect, A080580 calculates the position of the
> >> zero in A284939 based on whether A284939(n) is a zero.
> >> A080580(n) = 1, by definition, which corresponds to the first zero in
> >> A284939. For subsequent elements of A080580, the test "is n is already
> >> the sequence A080580" is equivalent to "is A284939(n)=0", because by
> >> definition if n were in the sequence it would only be because the nth
> >> element of A284939 was a zero.
> >> As we move through the elements of A284939 there are two possibilities.
> >> If A284939(n)=0 then it will be transformed to 01, and A284939(n+1) will
> >> be transformed to 1101. Consequently, the next zero will appear four
> >> after A284939(n).
> >> Alternatively, if A284939(n)=1 then it will be transformed to 1101.
> >> Consequently, the next zero will appear two places after A284939(n).
> >> These two rules are equivalent to the rules in A080580 which stipulate
> >> that A080580(n) = A080580(n-1)+2 if n is already in the sequence [i.e.,
> >> A28439(n)=0] but A080580(n) = A080580(n-1)+4 otherwise.
> >> Therefore, the enumeration of the zeroes of A284939 in A284940 is
> >> equivalent to A080580.
> >> Joe Slater
> >> --
> >> Seqfan Mailing list - http://list.seqfan.eu/
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> Seqfan Mailing list - http://list.seqfan.eu/
More information about the SeqFan