OEIS Server: Installed new b-file as b056744.txt.  Old b-file is now b056744_1.txt.

Sean A. Irvine: I haven't looked too closely at this sequence, but doesn't this amount to finding the appropriate de Bruijn sequence: https://en.wikipedia.org/wiki/De_Bruijn_sequence  (There are already algorithms for doing this)

Davis Smith: I just added a comment regarding my proof and the implications of it.

Jon E. Schoenfield: (I started reading through your proof today, and then decided I needed to try again when I had more time to think through everything.  I can be slow sometimes.)  ?:-/

Davis Smith: If the proof needs improvement, please let me know, I am a philosophy professor specializing in mathematical logic, so there may be some wording or phrasing which I am unfamiliar with.

Davis Smith: It centers around a Reductio Ad Absurdum argument, starting with the assumption that a(n) is not congruent to 2^floor(log_2(n)) (mod 2^(floor(log_2(n))+1)). After that there are four different possibilities, two which obviously lead to a contradiction, the other two take some extra reasoning to lead to a contradiction.

Jon E. Schoenfield: *If* we had a proof that, for n > 1, the binary expansion of a(n) begins with that of k1 = 2^floor(log_2(n-1)) + 1, then there's a very quick and easy proof that a(n)'s binary expansion always ends with that of k2 = 2^floor(log_2(n)): suppose it doesn't end with k2. Then the binary expansion of a(n) starts with k1, has k2 somewhere later, and has some string of additional bits following it. Cut the bit string into two pieces, immediately after the last bit in k2; then swap the positions of the left and right pieces, and shorten the result by having the two pieces overlap by at least one bit (which is always doable). This yields a shorter string of bits that still contains all the required substrings, proving that the original string wasn't really a(n) -- a contradiction.  (But I have no idea how to prove that a(n) starts with k1 for all n > 1. Maybe that would require a much longer proof than your proof that a(n) ends with k2, I don't know.)  ?:-/

Jon E. Schoenfield: I just now saw your 21:44 and 21:49 posts.  Thanks!  No, I don't see anything needing improvement in your proof.  (I may be needing improvement in my brain!)

Davis Smith: Jon, I was thinking about doing a similar proof to the one which I posted for the first element of Order(a(n),n) being 2^floor(log_2(n)) + 1 for n >= 2^floor(log_2(n)) + 1 and it being 2^floor(log_2(n - 1)) + 1 for n = 2^floor(log_2(n)).  It would likely follow a very similar logical structure: 1) Assume for Reductio that the first element of Order(a(n),n) is not 2^floor(log_2(n-1)) + 1. 2) Find the different possibilities for that situation. 3) Show that each of these lead to some manner of contradiction (either with the initial assumption or some derived fact).

Davis Smith: I can say, thinking about the problem a little bit, where b = 2^floor(log_2(n)) + 1 and x is another required substring, Overlap(b,x) = 1 and Overlap(x,b) = m such that m is the position of the least significant 'on' bit in x (2^m is the greatest power of 2 evenly dividing x).

Davis Smith: O, another option would be to 'kill two birds with one stone' and prove that both the first element of Order(a(n),n) = b1 = 2^floor(log_2(n)) + 1 for n >= 2^floor(log_2(n)) + 1 and 2^floor(log_2(n - 1)) + 1 for n = 2^floor(log_2(n)) and the last element of Order(a(n),n) = b2 = 2^floor(log_2(n)). For this, assume that the first and last elements are neither b1 nor b2. One of these will lead to a contradiction. This gives you that either the first element is b1 and the last is b2 or the first element is b2 and the last is b1. However, the binary expansion of the one starting with b2 will be at least one longer than the binary expansion of the one starting with b1. Thus proving that Order(a(n),n) starts with b1 and ends with b2.

Jon E. Schoenfield: I've uploaded an updated version of the file containing a table of the binary and decimal values (and numbers of 0- and 1-bits), using the values from the b-file.  I've also moved most of the content of my recent Comments section addition to a file in the Links.  Additionally, I've shortened the table in the Example section (since we have a longer table in the Links).  Also, I've taken some of the content I had put in the Example section earlier and moved it to the Comments section instead, because it seems to fit better there.

Jon E. Schoenfield: @Davis, rather than having it in my Comments entry, do you think it'd be better if there were a comment from you in the Comments section regarding your proof?