[seqfan] Re: morphism in A284940 compared with A080580
Joe Slater
seqfan at slatermold.com
Tue Apr 25 14:45:47 CEST 2017
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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