[seqfan] Self expulsion array (variation on A007063).

[David Sycamore]

Let A = A007063 and B = A006852.
(A and B are mutually inverse). 

The first row of the expulsion array (whose diagonal terms give A) is the natural numbers. If we change the first row to another infinite sequence, then generate an array using the same (Kimberling “shuffle”) system as for the original array, then of course we end up with a different sequence of diagonal terms. 

Question: What would  the first row sequence have to look like for the diagonal sequence to be identical to that row? In other words: Is there a self expulsion variant of the original (Kimberling) idea ?

There could be several different answers to this. The sequence below is a draft attempt to find a systematic way to do it (though it’s a bit long). If this result is interesting for the oeis then it would need some improvement to sharpen definitions and perhaps find a way to code it, etc, which is why I take a deep breath and post it here. 

For integer n>=1 define the set:
[n]={x; A^r(x)=n}U{y; B^r(y)=n}; 
(r =0, 1,2,3......... ; A^0(n)=B^0(n)=n). This gives n and all the “upstream” and “downstream” numbers linked to  it in A007063.

If a number m is in [n],  then [m]=[n], so we can most conveniently name the set by its  smallest member k, and also for convenience we can list the contents in their natural order. The sequence of such numbers k is:

C= 1,2,4,6,8,11,13,14,19,21,26,27....

(How best to Name this?)

The collection of distinct sets [k], where k is in C, is a partition of the natural numbers, assuming that every natural number is a term in A (Has that been proved yet? If not there is a least n with no finite path to the diagonal, and at that point the sequence below would be indeterminate). Putting that minor point aside for the moment, we continue with examples of [k] for the list given above (ordered, and limited to elements <100):

[1]={1}, 
[2]={2,3,5,9,10,12,17,20,23,24,25,36, 42,43,56,61,79,84...}
[4]={4}, 
[6]={6,7,15,16,22,28,59,66,81...}, 
[8]={8},
[11]={11,18,29,32,35,70,76..}
[13]={13,31,39,44,54,60,90....}
[14]={14}
[19]={19,33,34,48..}
[21]={21,53,67,69..}
[26]={26,30,37,38,41,47,55,58,65,68, 71,95,99,..}, 
[27]={27,62,88...}.

This is just enough to produce the first 25 terms of the self expulsion sequence, and corresponds to the point at which 2 arrives in the diagonal of the original array (A(25)=2). 

With this data we can modify the first row of the array, as follows: 

If a number n is in [k], replace it with k.  This gives sequence D:

1,2,2,4,2,6,6,8,2,2,11,2,13,14,6,6,2,11,
19,2,21,6,2,2,2,26,27,6,11,26,13,11,19...

With this sequence replacing the first row (all the natural numbers) we can “shuffle” the rows as per the original, and recover the same terms in the same order in the diagonal. See photo below, showing derivation of first 25 terms. Sorry, visual quality is poor because the work is in pencil (essential for making errors!). I could send zoomed in versions of fewer terms if anyone wants them. 

Well this is as far as I’ve got with it. 

Whereas in the A007063 array, a number from the first row can appear only once in any subsequent row, in this sequence multiple occurrences appear, except for those k for which [k]={k}; namely k= 4, 8, 14 (are there more?).

If sequences C and D are to be submitted to oeis, the long descriptions given above would need to be refined into exact definitions and Names. At the moment I can’t see the best way to do that. It would also be very helpful to have a code and thereby establish more terms by confirming that the self expulsion continues for as long as....?  

I have no idea how to code this but maybe there is someone out there with the skills and interest to do it? 

Any comments or offers for cooperation with any of the above would be most welcome.

Best regards,

David.


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