What other sequences solve this equation?

[franktaw at netscape.net]

 From the increasing and complementary properties, we must have that when a(i) = b(j) + 1, then the numbers a(1), ..., a(i), b(1), ..., b(j) contains each number 1, ..., i+j exactly once, with a(i) = i+j and b(i) = i+j-1.  Substituting the given equation a(b(n)) = b(a(n)) + 1, we have:
 
a(b(n)) = a(n) + b(n)
b(a(n)) = a(n) + b(n) - 1
 
Suppose b(1) = 1.  Substituting in the first equation above, we get a(1) = a(1) + b(1) = a(1) + 1.  So we must have a(1) = 1.
 
Suppose a(2) = 2.  Substitute in the second equation to get b(2) = a(2) + b(2) - 1 = b(2) + 1.  So we must likewise have b(1) = 2.
 
Now the first equation gives a(2) = a(1) + b(1) = 3.
 
 From this point we start having choices.  b(2) can be any value 4 or larger.  Fill in the smaller values 4 through b(2)-1 in a(3) through a(b(2)-2).  Now, iteratively, for each n from 2 on, we know what a(b(n)) and b(a(n)) must be.  This leaves a certain number of holes in the a and b sequences, which can be filled by assigning the intermediate values to the two sequences as we see fit.  (Sometimes there will  be no holes in one or both sequences, hence no real choice, but as we keep going, eventually there will be.)
 
For example, we might at one point have sequence a starting 1,3,4,6,8,9,11 and b starting 2,5,7,10.  We now have a(10) = 16 and b(6) = 15.  This leaves values 12,13,14 to be distributed between a(8), a(9), and b(5).  Since we must have a(8) < a(9), there are 3 ways to do this.
 
Franklin T. Adams-Watters
 
 
-----Original Message-----
From: Kimberling, Clark ck6 at evansville.edu

 
In this query, a(n) and b(n) denote increasing complementary sequences, such as (1,3,5,7,...) and (2,4,6,8...), as well as pairs of Beatty sequences.
 
The question is, how can we account for all solutions of the equation 
 
a(b(n)) - b(a(n)) = 1 ?
 
Among the solutions are a = (1,3,5,7,....) and the lower Wythoff sequence, a(n) = Floor[tau*n)], where tau is the golden mean.
 
Anyone want to look into this?
 
Thanks.
 
Clark Kimberling
 
 
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