Bdtyktl

Tetrahedral numbers (https://oeis.org/A000292) are:

1, 4, 10, 20, 35, 56, 84, etc

WHERE Tetrahedral(n) = (n * (n+1) * (n+2)) / 6

But what is the test for whether a given number is a tetrahedral number?
I also need to know, if it is a tetrahedral number, then which term in the tetrahedral sequence is it? I am looking for a formula or something that would enable me to get these results:

input   result ("is tetrahedral")

1       1

2       false

3       false

4       2

5       false

6       false

7       false

8       false

9       false

10      3

11      false

12      false

I can't find any formula on the Internet for this. Any pointers in the right direction would be great - thx

Notice that $n(n+1)(n+2)/6approx n^3/6$ for large values of $n$. Thus, if you have a suspected tetrahedral number $k$, and you compute $n_0=lfloorsqrt[3]{6k}rfloor$ you will have a pretty good guess as to which index you need to check to confirm or refute your suspicions.

This of course won't work for small $k$, and your initial guess $n_0$ need not be correct, but you have at least vastly reduced the number of operations you need to finish your problem. The full algorithm now would look something like this:

1) Take the input number $k$ and compute $n_0=lfloorsqrt[3]{6k}rfloor$.

2) If $text{Tetra}(n_0)>k$, begin testing all naturals $n<n_0$ (in decreasing order) until $text{Tetra}(n)leq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.

3) If $text{Tetra}(n_0)<k$, begin testing all naturals $n>n_0$ until $text{Tetra}(n)geq k$. If you achieve equality, then $k$ is tetrahedral with index $n$. If you don't have equality, then $k$ is not tetrahedral.

4) If $text{Tetra}(n_0)=k$, then you are done, and the index is $n_0$.

Another way to look at that: $frac{n(n+1)(n+2)}{6}=frac{(n+1)^3-(n+1)}{6}$.

So then, if we have some $m$ we want to test? Multiply it by $6$. Add its cube root, rounded up - $k^3-k > (k-1)^3$ for all $k > 1$. If we get a perfect cube ($6m+lceilsqrt[3]{6m}rceil=k^3$ for some integer $k$), it's a tetrahedral number (Specifically, for $n=k-1$). If not, it isn't.

For a given $k$, you want to know if it exists an integer $n$ such that
$$frac{n(n+1)(n+2)}{6}=k$$ A simple way is to look at the cubic equation
$$n^3+3n^2+2n-6k=0$$ which has only one real root. Using the hyperbolic solution for this case, the solution is given by
$$n=-1+frac{2 }{sqrt{3}}cosh left(frac{1}{3} cosh ^{-1}left(9 sqrt{3},
kright)right)$$ If this $n$ is an integer (or very very close to because of possible rounding errors), then you are done.