A pronic number is a number that can be expressed as the product of two consecutive positive integers. For instance, $42 = 6 \cdot 7$ is a pronic number. I've become interested in solving for the sequence of all pronic numbers $x_{n}$ such that $2x_{n}$ is also a pronic number.
This is essentially then solving the equation $\dfrac{m(m+1)}{n(n+1)} = \dfrac{1}{2}$ for positive integer solutions, I think. My first thought was to take a purely computational approach, and generate a list of all pronic numbers to some limit and them check as to whether each pronic number $\times 2$ was in my list, but this is far too costly and it seems to me there should me some mathematics to simplify my problem.
I took a look at the generating function for pronic numbers, and found it to be $\dfrac{2x}{(1-x)^{3}}$. Is there some way I can exploit the power series fractional representation of my problem or something else in order to more concisely express integer solutions to my problem?
Matthew Conroy's link probably gives you everything you need to know in practice, but here's a bit of theory, coming from the following general phenomenon: setting two quadratics equal to each other often reduces to a "Pell equation".
Here we have $n(n+1)=2m(m+1)$; completing the square gives $(2n+1)^2 - 2(2m+1)^2 = -1$. In other words, we want the solutions of the equation $x^2-2y^2=-1$ where both $x$ and $y$ are odd.
The equation $x^2-dy^2=C$ is called a Pell equation (although it should have been named after much earlier mathematicians), and a method for solving them using continued fractions is well known and worth learning. The case $C=\pm1$ is particularly nice. Moreover, if there's one solution then there are infinitely many, even starting with the congruence restriction $x\equiv y\equiv1\pmod2$.