Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have an ordinal list that I am trying to represent mathematically. The list is as follows:

10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, 2000, 3000, 4000, 5000, 6000, 7000, 8000, 9000, 10000, 20000, 30000, 40000, 50000, 60000, 70000, 80000, 90000, 100000.

So basically, 100, 1000 and 10000 are multiplied by 10. I am visualising a chunk of data and I need an easy way to communicate the distribution without resorting to log. I'd like to formalise this, however.

cheers all,


share|cite|improve this question
I removed the tag (ordinals) since this has nothing to do with ordinals (which is a well defined concept in set theory); I feel that there should be another tag brought in here, possibly to replace (problem-solving) altogether - however I'm not sure which tag fits. – Asaf Karagila Apr 13 '11 at 14:16
You could consider generating functions. They would help you get a formula for the $n$th item in the list, and give a mathematical structure that contains the list. – Matt Groff Apr 13 '11 at 14:32

See OEIS sequence A037124: Numbers that contain only one nonzero digit.

share|cite|improve this answer
which has the formula a(n) = [(n mod 9)+1] * 10^floor(n/9) and slotishtype wants the first 37 terms. – Henry Apr 13 '11 at 15:17
@slotishtype: For your list (containing 37 terms) use the above formula for $n=9,10,\ldots,45$. – Shai Covo Apr 13 '11 at 15:48

If the list stops at $100,000$, you have represented it by listing it. You could say something like $k10^n$ where $k \in \{1,2,3,4,5,6,7,8,9\}$ and $1 \le n \le 4$, but that is pretty complicated and leaves off the last term.

share|cite|improve this answer
Thanks Ross...I was just looking for some short hand notation but I am probably over complicating the issue... – slotishtype Apr 13 '11 at 14:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.