I was looking at some cases of triangular numbers that are twice the amount of other triangular numbers, i.e. $\frac{n(n+1)}{2}=m(m+1)$ for some integers n and m. The first few cases of this are as follows in the form (n, m): (2, 3), (14, 20), (84, 119), (492, 696), (2870, 4059), and (16730, 23660), and there appear to be no other cases in between (though there are more after). Looking at this, a pattern appears to emerge; the frequency closely resembles the function log base $(3+\sqrt8)$ of n. If we instead switch to looking for cases where a triangular number is thrice another triangular number, the frequency appears to resemble log base $(2+\sqrt3)$ of n. Triangular quadruples and any $k^2$-tuples have no frequency as the only times when they are $k^2$ times another triangular number are when the formula $\frac{n(n+1)}{2}$ is equal to 0 (i.e. n=-1 or n=1). Quintuples seem to have a frequency that is two thirds the frequency of log base $\frac{3+\sqrt5}{2}$ of n. I haven't been able to find any patterns in k-tuples for k>5 but I haven't looked for a substantial amount of time. I was wondering if anyone has a proof for or a suggestion of a method of proving the convergence of these frequencies or a formula of k that finds the frequency of any triangular number being a k-tuple of another triangular number? I have not been able to find one yet.
Asked
Active
Viewed 25 times
0
-
Multiplying both sides by $8$, this becomes $$(2n+1)^2-1=2\left((2m+1)^2-1\right)$$ or $$(2n+1)^2-2(2m+1)^2=-1,$$ which is a negative Pell equation, $u^2-2v^2=-1,$ and we know the solutions of this are of the form $u+v\sqrt2=(1+\sqrt2)^{2k-1}$ for positive integers $k.$ – Thomas Andrews Jan 22 '24 at 16:38
-
In particular, $u$ is the nearest integer to $\frac12(1+\sqrt2)^{2k-1},$ and $v$ is the nearest integer to $\frac1{2\sqrt2}(1+\sqrt2)^{2k-1}.$ These turn out to always be odd. – Thomas Andrews Jan 22 '24 at 16:40
-
If course, $(1+\sqrt2)^2=3+2\sqrt2=3+\sqrt8.$ – Thomas Andrews Jan 22 '24 at 16:44