calculator sequence inspired by my calendar

[franktaw at netscape.net]

Not the most useful, but it suffices for small values:

If n is a 2k+1 digit number for which f(n) > 0, f(n) >= 10^k.
If n is a 2k digit number for which f(n) > 0, f(n) >= 9*10^(k-1).

(Here f(n) is the reversal of n minus n.)

So to determine if 15 is a possible difference, it suffices to check 
through n=999.
---

And yes, the sequence is certainly infinite. If n is a k digit integer 
with f(n) defined and positive, then f(10^(k+1)+10n+1) = 10 f(n).

Franklin T. Adams-Watters

-----Original Message-----
From: jos.koot at telefonica.net

  Hi,
  Indeed, a practical test on all numbers below 1000000 has shown me 
that 15 does not occur as such a difference for these numbers. Of 
course there is a systematic way (a procedure, not an algorithm) of 
forming a sequence of all such differences. Things very much depend on 
the order in which you want form this sequence, I think. Furthermore I 
think the sequence has unlimited length, but that makes it all the more 
interesting. I guess you want a decision algorithm that given any 
number can decide whether or not it is the difference between a number 
and its upside down (more precisely rotated over 180 degrees) This is 
interesting, but I am not sure it is within my capacities. Nevertheless 
I intend to look into this problem, but this has to wait a week or so 
(Chrismas time you know)
 Greetings, Jos Koot

 ----- Original Message -----
  From: Tanya Khovanova

I am looking for the numbers that can be represented as the difference 
of a number and its ipside down version on a calculator. I think I can 
prove that 15 is not there, but I do not have a systematic way to build 
this sequence.


---------- Original Message ----------------------------------
From: "jos koot" <jos.koot at telefonica.net>
Date: Thu, 21 Dec 2006 00:27:15 +0100

>Here are some solutions:
>Greetings, Jos Koot
>mabel 12, zane 21
>mabel 15, zane 51
>mabel 16, zane 91
>mabel 18, zane 81



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