I noticed $49=1\cdot7^2$ and $50=2\cdot5^2$, and I was curious if there exists longer runs of this form. Specifically, consecutive positive integers such that the first is a square, the second is twice a square, the third is $3$ times a square, etc..
Messing around trying to find a run of $3$, I figure the first number $x$ must be odd because the second number is even, and $x$ must be one more than a multiple of $3$, because the third number is a multiple of $3$. This means the first number is one more than a multiple of $6$, and therefore square of a number one more or one less than a multiple of $6$.
Using similar logic, I figured that the second number is twice the square of a number one more or one less than a multiple of $6$, and the third number is $3$ times the square of an odd number, but that was all I was able to figure.
Any help on getting further would be great.
Edit:
Trivially, $1=1\cdot1^2, 2=2\cdot1^2, 3=3\cdot1^2, $ etc.
besides $49$ and $50$, I also found $1681=1\cdot41^2$ and $1682=2\cdot29^2$
Best Answer
To find all runs of $2$ terms $\{a^2,2b^2\}$, solve the Pell(ish) equation $a^2 - 2b^2 = -1$.
These numbers (solutions) are already well documented in A002315.
For instance, they satisfy $a_n = 6a_{n-1} - a_{n-2}$ and $a_n=\frac{(1+\sqrt2)^{2n+1} + (1-\sqrt2)^{2n+1}}{2}$.
They are also called NSW Numbers, starting as $a_n=1,7, 41, 239, 1393,\dots$
On the other hand, there are no non-trivial runs of $3$ (or more) terms.
This was proven in the following MSE question (equivalent problem): Does this system of simultaneous Pell-like equations have any non-trivial positive integer solutions?.