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

[Leroy Quet]

I wrote, in part,
>...
>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)-c(2)|,..,|c(m)-c(m-1)|}.

['b' replaced with 'c' in previous line]

>
>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.

No, the c-sequence has 27 terms, where
c(27) = 54. (If I did not err.)
|c(28)-c(27)| must be 43, so c(28) must be either 11 or 97.
But 11 is earlier in the difference-sequence and 97 is earlier in the 
c-sequence.



>
>We could, in this case, just require that there be at most one 
>subtraction in a row.

If this sequence is to be submitted to the EIS, the rules should be 
modified so there is at most one descent in a row so as to ensure the 
sequence is infinite.

  
>(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.)

I notice that the first case of 2 descents in a row leads to the sequence 
ending.


thanks,
Leroy Quet