Terms and Terms Differences -> 1,2,3,4,...

[Leroy Quet]

Here are some sequences (not in the EIS) which are interesting (to me) 
because they involve recursive sequences and their compliments (sequences 
of positive integers not in original sequence).

First, the least simply expressed, perhaps, sequence:

Let a(1) = 1;

Let a(m+1) =
a(m) +(number of elements of {a(1),a(2),..a(m)} which are <= m)

(figured by hand)

1, 2, 4, 6, 9, 12, 16, 20, 29, 34, 39,..

So, every positive integer is either of the form
a(m) +m-1
or
a(m+1)-a(m)+m,
but not of both forms.

a(m) +m-1: 1, 3, 6, 9, 13, 17, 22, 27, 32,..

a(m+1) -a(m) +m: 2, 4, 5, 7, 8, 10, 11, 12, 14,..

-

b(1) = 1;

for m >= 2,
(b(m+1)-b(m)) is lowest positive integer not among
{b(1),b(2),..,b(m)} or among {b(2)-b(1),b(3)-b(2),...,b(m)-b(m-1)}.

b(m):        1, 3, 7, 12, 18, 26, 35, 45,..
 
b(m+1)-b(m): 2, 4, 5, 6, 8, 9, 10,..

This sequence IS in the EIS.

http://www.research.att.com/projects/OEIS?Anum=A005228

But if we instead take the absolute value of the difference, and require 
the c-sequence to descend whenever possible, each term remaining unique 
and positive, we get:
 
c(1) = 1;

for m >= 2,
|c(m+1)-c(m)| is lowest positive integer not among
{c(1),c(2),..,c(m)}
or among {|c(2)-c(1)|,|c(3)-b(2)|,..,|c(m)-c(m-1)|}.

c(m):         1, 3, 7, 12, 18, 10, 19, 30, 17,..

|c(m+1)-c(m)|: 2, 4, 5, 6*, 8, 9, 11, 13,..

*(note that subtracting 6 from 12 would have gotten another 6, which is 
forbidden.)

But I wonder if this c-sequence is infinite. It might, since we must 
subtract whenever we    can, have 2 subtractions in a row, and then the 
sequence could get "caught", where neither adding nor subtracting the 
lowest unused integer gets an unused integer.

We could, in this case, just require that there be at most one 
subtraction in a row.
(But I do not know, with the few terms I have calculated by hand, if 
there is ever {with the original rules} more than one subtraction in a 
row anyway.)

thanks,
Leroy Quet