Inverse tetrahedral numbers

The previous post looked at the tetrahedral numbers:

1, 4, 10, 20, 35, …

We could invert the process of creating tetrahedral numbers and ask for what n is a given number the nth tetrahedral number. So the inverse of 1 is 1, the inverse of 4 is 2, the inverse of 10 is etc.

We can find the nth tetrahedral number by evaluating a cubic polynomial in n

and so we could invert this for a given x by solving p(n) = x for n. The result is complicated, but it can be written in closed form:

$n = \sqrt[3]{3x+\sqrt{9{x^2}-\frac{1}{27}}} +\sqrt[3]{3x-\sqrt{9{x^2}-\frac{1}{27}}} -1$

We could do the same for generalized tetrahedral numbers. The nth tetrahedral number of dimension d is given by a polynomial in n of degree d, and so in principle we could look for the roots of these polynomials, though we shouldn’t expect closed forms expressions for the roots in general.

Now if x is a tetrahedral number, i.e. x is an integer such that p(n) = x for some positive integer n, then this value of n is unique as far as positive integers go. The sequence of tetrahedral numbers is increasing, so it has a unique inverse, if it has an inverse at all.

Now suppose x is not a tetrahedral number, and maybe not even an integer. We can still solve the equation p(n) = x. And if x > 0 then there is a real root given by the equation above. What are the other roots like?

We can find out by computing the discriminant of the equation p(n) – x = 0, which works out to be

$\Delta = \frac{1}{364} - \frac{3}{4}x^2$

When the discriminant is positive, there are three real roots. When the discriminant is negative, there is one real root and two complex roots.

So, for example, the tetrahedral inverse of 10 is 3, but you could say that -3 ± √11 i are also inverses.

For another example, the real tetrahedral inverse of 7 is 2.57189… and there is also a conjugate pair of complex inverses.