# A triangular array of numbers.

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

$$11$$

$$1\frac{3}{2}1$$

$$1\frac{6}{4}\frac{6}{4}1$$

$$1\frac{10}{7}\frac{10}{6}\frac{10}{7}1$$

$$1\frac{15}{11}\frac{15}{9}\frac{15}{9}\frac{15}{11}1$$

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

$$f(r,c)=\frac{r(r+1)}{(r-c)(r-c+1)+c(c+1)}$$

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?

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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 $1 \leq f(r,c) < 2$.
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