calculator sequence inspired by my calendar

[jos koot]

Hi,
This means that there is an algorithm that decides whether or not a number is the difference between itself and its pocket-calculator upside down.
Thanks, Jos Koot

PS:I did not consider 7 and 1 to mirror. The updside down of 7 looks like 1, but not reversely.
  ----- Original Message ----- 
  From: franktaw at netscape.net 
  To: seqfan at ext.jussieu.fr 
  Sent: Thursday, December 21, 2006 8:43 PM
  Subject: Re: calculator sequence inspired by my calendar


  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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