Let B(i) be the square root of the reciprocal triangular numbers, that is to say, B(i) = sqrt(2 / (i(i+1))).

Let A(i) = -sqrt(2i / (i+1)) = -sqrt(i^2 * (2 / (i(i+1)))). That is to say, A(i) is the negative square root of the square numbers divided by the triangular numbers.

Define the following set of points in n-space:

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`(B(1), B(2), B(3), ... , B(n-2), B(n-1), B(n))`

(A(1), B(2), B(3), ... , B(n-2), B(n-1), B(n))

( 0, A(2), B(3), ... , B(n-2), B(n-1), B(n))

( 0, 0, A(3), ... , B(n-2), B(n-1), B(n))

...

( 0, 0, 0, ... , 0, A(n-1), B(n))

( 0, 0, 0, ... , 0, 0, A(n))

The convex hull of these points form a regular n-simplex centered on the origin with edge length 2. The apex lies along the last coordinate axis (so in a sense the simplex is "upright") and the base has a plane of symmetry, with these two properties recursively true for the base (n-1)-simplex. So this is no accidental coincidence... the triangular numbers give rise to a "nice" orientation of the n-simplices.

The Greeks were on to something, I tell ya!