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 found next formula of counting all rational numbers (or integer pairs, from which rational numbers can be builded):

$N(i,j) = ((i - j - 1)(i - j) - 2j - 1)(1 + j / |j|)/2 + ((i + j - 1)(i + j) + 2j)(1 - j/|j|)/2$

or this formula, which excluded repeating number:

$N(i,j) = [(i + j - 1)(i + j - 2) + 2j + 2δ_{0,i}](1 + j/|j|)/2 + [(i - j - 1)(i - j - 2) - 2j + 2δ_{0,i} - 1](1 - j/|j|)/2 + 1 - δ_{0,i} $

where $δ_{0,i}$ is Kronecker delta.

The first values of last functions are displayed in this table.

enter image description here

But what is the inverse function of it?

share|cite|improve this question
up vote 2 down vote accepted

You write $N(1,1)=2$, but also $N(1,1)=4$, so there is something wrong with your calculations. I'll just ignore the first two lines of the table.

Note we have
$i+|j|=2$ for $3\le N(i,j)\le4$
$i+|j|=3$ for $5\le N(i,j)\le8$
$i+|j|=4$ for $9\le N(i,j)\le14$
$i+|j|=5$ for $15\le N(i,j)\le22$
and so on. I'll rewrite these as
$i+|j|=2$ for $1\le 4N(i,j)-11\le5$
$i+|j|=3$ for $9\le 4N(i,j)-11\le21$
$i+|j|=4$ for $25\le 4N(i,j)-11\le45$
$i+|j|=5$ for $49\le 4N(i,j)-11\le77$
and then as
$i+|j|=2$ for $1\le \sqrt{4N(i,j)-11}\lt3$
$i+|j|=3$ for $3\le \sqrt{4N(i,j)-11}\lt5$
$i+|j|=4$ for $5\le \sqrt{4N(i,j)-11}\lt7$
$i+|j|=5$ for $7\le \sqrt{4N(i,j)-11}\lt9$
and finally as
$i+|j|=2$ for $2\le (\sqrt{4N(i,j)-11}+3)/2\lt3$
$i+|j|=3$ for $3\le (\sqrt{4N(i,j)-11}+3)/2\lt4$
$i+|j|=4$ for $4\le (\sqrt{4N(i,j)-11}+3)/2\lt5$
$i+|j|=5$ for $5\le (\sqrt{4N(i,j)-11}+3)/2\lt6$
So we see that $$i+|j|=\left[{\sqrt{4N(i,j)-11}+3\over2}\right]$$ where I use $[x]$ to indicate the greatest integer not exceeding $x$. So, for example, if $N(i,j)=12345$ then we compute $4N(i,j)-11=49369$, $\sqrt{49369}=222.19\dots$, add $3$ to get $225.19\dots$, divide by $2$ to get $112.59\dots$, and round down to get $i+|j|=112$.

Now let's do some computations on the last 10 rows of your table:

$$\matrix{i&j&N&4N-11&A&B&C\cr5&-1&23&81&0&0&5\cr5&1&24&85&4&0&5\cr4&-2&25&89&8&1&4\cr\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\cr1&5&32&117&36&4&1\cr}$$ where I am using $$Q=\sqrt{4N-11},\quad z=2\left[{Q+1\over2}\right]-1=9,\quad A=4N-11-z^2,\quad B=\left[{A\over8}\right],\quad C=\left[{Q+1\over2}\right]-B$$ The first and last columns are identical, so we have a formula for $i$. If you unravel everything, it comes down to $$i={[M]^2+[M]\over2}-\left[{N-3\over2}\right]$$ where $M=(\sqrt{4N-11}+1)/2$ (although it's probably a good idea to check over my algebra). For $N=12345$, this works out to $i=45$.

Now that we have formulas for $i$ and for $i+|j|$, we subtract to find $|j|$. Then it's just a question of deciding whether $j$ is positive or negative, and it's positive when $N$ is even, negative when $N$ is odd, so the formula is $j=(-1)^N|j|$. In our example, this gives $j=-67$.

share|cite|improve this answer

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.