James Tanton asked on Twitter:

As there are “more” triang nmbrs than sq nmbrs http://www.jamestanton.com/?p=1009 let f(N) = nmbr triangs >= N^2 but < (N+1)^2. Curious:What graph like?

The th triangular number is about (more precisely, it’s .) So there are about triangular numbers less than . Therefore, “on average”, in each interval there are triangular numbers.

For example, in the interval [9, 16) there are two triangular numbers, namely 10 and 15; this is f(3). In the interval [16, 25) there is one triangular number, namely 21; this is f(4).

Let’s write down an explicit formula for f(n). Let g(x) be the number of triangular numbers less than x. To figure this out, I’ll introduce a function t(x), which takes as input x and outputs the index of x in the triangular-number sequence. For example, t(10) = 4, t(15) = 5. But we also want to be able to compute, say, t(12). But that’s fine! t(n) is just the inverse of the function which takes n to the nth triangular number, the function ; in particular, solving the quadratic,

So ; .

Next we write g(x) in terms of t(x). It’s tempting to say that g(x) = \lfloor t(x) \rfloor, but it’s not. t(10) = 4, for example, but we want g(10) = 3. We’ll say that — we’ll only need this formula to work when x is an integer. So, for example, , and the index of 9.875 in the triangular number sequence, whatever that means, is between 3 and 4. But .

Why the constant 1/8? Because

which makes the formula marginally easier to write.

Finally . Take the number of triangular numbers less than , and subtract the number less than , and you get the number in the interval in between. For example ; there are 6 triangular numbers less than 25, namely 1, 3, 6, 10, 15, and 21. And . Thus , indicating the triangular number 21. So at long last we have the formula

.

In particular the arguments of these two floor functions differ by , which is between 1 or 2, so f(n) is always either 1 or 2. The graph that Tanton asked about is below.

You can see some hints of periodicity in the function; for example, from a quick glance at the graph it might look like has period 12, each period containing five 2s and seven 1s. But this can’t hold, not unless . In fact can’t be periodic, because is irrational.