How many elements in a 3rd order tensor with certain symmetries?

Question

I have a tensor $A_{ijk}$ were indices $i$, $j$, and $k$ run from $1$ to $N$.

Obviously there are $N^3$ elements, but how many unique elements are there if the following symmetries exist?

1. $A_{iij} = A_{iji} = A_{jii} $
2. $A_{ijk} = A_{ikj} = A_{jik} = A_{jki} = A_{kij} = A_{kji}$

These are all the symmetries that would exist with differentiation, since this tensor actually represents the third derivative of a function.

Answer 1

There are $N$ unique elements of the form $A_{iii}$.

For elements of the form $A_{iij}$, there are $N$ ways to choose $i$ and $N-1$ ways to choose $j$. Notice that order matters here.

For elements of the form $A_{ijk}$, there are $\binom{N}3$ ways to choose $i$, $j$ and $k$. Notice that order does not matter here, which is why we used the binomial coefficient here (and not in the previous case).

This gives us $N + N(N-1) + N(N-1)(N-2)/6 = \frac16(n^3+3n^2+2n)$.

Answer 2

This amounts to counting the number of integer solutions of

$$n_1 + n_2 + n_3 +\cdots + n_N= 3$$

The number of solutions is:

$$\binom{N+2}{3} = \frac{1}{6}N(N+1)(N+2)$$