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God plays dice

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 $latex k$th triangular number is about $latex k^2/2$ (more precisely, it’s $latex (k^2+k)/2$.) So there are about $latex \sqrt{2} n$ triangular numbers less than $latex n^2$. Therefore, “on average”, in each interval $latex [N^2, (N+1)^2)$ there are $latex \sqrt{2}$ 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…

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