Erase to increment

[franktaw at netscape.net]

There is also a tie to the Hoffstadter-Conway $10000 sequence.

Start the set S with 1,3; the sequence a then starts with 1,2.  For n 
 >= 3, Take S(n) = S(n-1) + a(n-1).  Then S is A088359 with an initial 
1, and a is A051135 except that the first term is 1 instead of 2.

(Incidently, the formula in A088359 does not belong there; it is the 
formula for A004001: the Hoffstadter-Conway sequence itself.)

Franklin T. Adams-Watters


-----Original Message-----
From: franktaw at netscape.net

A further note: taking S (as described below) to be the set of squares, 
we get a sequence 
which is very similar to A023117 - it first differs at n = 36. 
Coincidently (well, not entirely 
coincidence) the C. Kimberling paper "Interspersions" 
( http://faculty.evansville.edu/ck6/integer/intersp.html ) referenced 
there is closely related 
to all this. 
 
Given a dispersion, take the set S to be its first column. Then a(n) is 
the column of 
the dispersion that n occurs in; and the sequence X that generates the 
dispersion is the 
set S'. This works both ways; of course, if S is finite, the dispersion 
will have only finitely 
many rows. 
 
Incidently, the erasable sequence a associated with the Wythoff array 
is A035612. 
 
Franklin T. Adams-Watters 
 
-----Original Message----- 
From: franktaw at netscape.net 
 
I just submitted the following comment:  
  
%I A082850  
%S A082850 
1,1,2,1,1,2,3,1,1,2,1,1,2,3,4,1,1,2,1,1,2,3,1,1,2,1,1,2,3,4,5,1,1,2,1  
%N A082850 Let S(0) = {}, S(n) = {S(n-1), S(n-1), n}; sequence gives 
S(infinity).  
%C A082850 Sequence counts up to successive values of A001511; i.e., 
apply the  
morphism k -> 1,2,...,k to A001511.  
If all 1's are removed from the sequence, the resulting sequence b has 
b(n) = a(n)+1.  
A101925 is the positions of 1's in this sequence.  
%F A082850 a(2^m-1) = m. If n = 2^m-1 + k, with 0 < k < 2^m, then a(n) 
= a(k).  
%e A082850 S(1) = {1}, S(2) = {1,1,2}, S(3) = {1,1,2,1,1,2,3}, etc.  
%Y A082850 Cf. A001511, A101925.  
%O A082850 1  
%K A082850 ,nonn,  
%A A082850 Frank Adams-Watters (FrankTAW at Netscape.net), Aug 16 2006  
  
(I got to this from the relation described by the comment about 
A001511.)  
  
What I want to explore is the property I noticed, listed next: if all 
1's are removed from  
the sequences, the resulting sequence b has b(n) = a(n) + 1. (One can, 
of course,  
define a similar property for sequences of non-negative integers, 
removing 0's instead  
of 1's. Everything that follows can be recast into this context.)  
  
Obviously, any such sequence must have a(1) = 1. (I am assuming here 
that the  
sequence offset is 1.) Otherwise, you would get b(1) = a(1).  
  
When you start to determine the values of a sequence that has this 
property, it  
quickly becomes obvious that the values of a(n) are completely 
determined by  
the indices n such that a(n) = 1.  
  
In fact, let S be the set of such indices, and S' be the complement of 
S (with  
respect to the positive integers). Regard S' as a sequence with offset 
1. Define  
a sequence c by c(n) is the index of n in S', or 0 if n is not in S'. 
a(n) is then the  
number of iterations of c required to reach 0. This construction works 
as long as S'  
is not finite - if S' is finite, the sequence resulting from removing 
the 1's from a is  
finite.  
  
Using this fact, we can quickly find some such sequences in the OEIS: 
A001511 itself,  
with S = the odd integers; A059981, S = primes U {1}; A049076, S = 
non-primes.  
(Not to mention A000027, S = {1}.)  
  
Now, returning to A082850, the positions of the 1's are A101925. (This 
is not quite  
trivial, but follows easily from the fact noted in A101925 that its 
first differences are  
A001511.) So we are interested in the complement of A101925. This 
appears to  
be A045412. But the definition of A045412 is quite different.  
  
Can anybody prove that A045412 is, in fact, the complement of A101925?  
  
Franklin T. Adams-Watters