[seqfan] Re: morphism in A284940 compared with A080580

[Joe Slater]

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
itself.

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 k+4.

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 bunch
> 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 nth
>> 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 in
>> 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 places
>> 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., if
>> 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
>>
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>>
>
>
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