[seqfan] Re: morphism in A284940 compared with A080580

jean-paul allouche jean-paul.allouche at imj-prg.fr
Tue Apr 25 16:00:34 CEST 2017

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?

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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