Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

So I was given the triangular array of numbers below (the first line consists of two "1")






and I was told to find a function of two variables $f(r,c)$ that takes as input the row number $r$ (starting with $1$) and the column number $c$ (starting with $0$) and outputs the correct number. So for example $f(3,1)=\frac{6}{4}$. The general statement is


My question is, how can I figure out if this triangle of numbers contains all rational numbers somewhere and if not, how do I figure out which fractions will never appear in this pattern?

share|cite|improve this question
None of the numbers will be less than $1$ and greater than $2$. Do you mean it will hit all rationals in $[1,2)$? – user17762 Dec 11 '11 at 8:00

Note that $$f(r,c) = \frac{r(r+1)}{r(r+1) + 2c(c-r)}$$ Since $c \in \{0,1,2,\ldots,r\}$, we have that $2c(c-r) \leq 0$. Hence, $f(r,c) \geq 1$. Further, along each row the maximum occurs as $c = \left \lfloor \frac{r}{2} \right \rfloor$ (AM-GM). Hence, the maximum in each row is at-most $2 \left( \frac{r+1}{r+2} \right) < 2$. Hence, we have $1 \leq f(r,c) < 2$.

share|cite|improve this answer
Sorry, you're correct. I meant all the rationals in the interval [1, 2). What branch of mathematics deals with this type of problem? – Hautdesert Dec 11 '11 at 8:35
I do have another question though, just for you. How did you notice that my original expression for $f(r,c)$ could be simplified? I know I thought about expanding it and perhaps finding another equivalent expression, but I figured it probably couldn't be aesthetically simplified. But I was wrong. How did you do it? – Hautdesert Dec 11 '11 at 8:43
@Hautdesert the denominator is a symmetric polynomial in $c$ and $r-c$, which means you can express it as a polynomial in terms of $c+r-c = r$ and $c(r-c)$. It turns out that when you do this you get this simpler expression. – mercio Dec 11 '11 at 9:58

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.