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Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.

However, for entries $\{a,b,n\}$ a 90 degree rotation can interchange $b$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $\binom{n-1}{2}$ unique entry pairs.

That gives a total of $\binom{n+2}{3} + \binom{n}{3} - \binom{n-1}{2}$ which is $\binom{n+2}{3} + \binom{n-1}{3}$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.

Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.

However, for entries $\{a,b,n\}$ a 90 degree rotation can interchange $b$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $\binom{n-1}{2}$ unique entry pairs.

That gives a total of $\binom{n+2}{3} + \binom{n}{3} - \binom{n-1}{2}$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.

Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.

However, for entries $\{a,b,n\}$ a 90 degree rotation can interchange $b$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $\binom{n-1}{2}$ unique entry pairs.

That gives a total of $\binom{n+2}{3} + \binom{n}{3} - \binom{n-1}{2}$ which is $\binom{n+2}{3} + \binom{n-1}{3}$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.

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Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.

However, for entries where $\{a,1,n}\}$$\{a,b,n\}$ a 90 degree rotation can interchange $1$$b$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $(n-2)$$\binom{n-1}{2}$ unique entry pairs (not times 3 since we already set entries at $(a,1,n)$, $(n,a,1)$ and $(1,n,a)$ equal.

That gives a total of $\binom{n+2}{3} + \binom{n}{3} - (n-2)$$\binom{n+2}{3} + \binom{n}{3} - \binom{n-1}{2}$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.

Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.

However, for entries where $\{a,1,n}\}$ a 90 degree rotation can interchange $1$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $(n-2)$ unique entry pairs (not times 3 since we already set entries at $(a,1,n)$, $(n,a,1)$ and $(1,n,a)$ equal.

That gives a total of $\binom{n+2}{3} + \binom{n}{3} - (n-2)$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.

Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.

However, for entries $\{a,b,n\}$ a 90 degree rotation can interchange $b$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $\binom{n-1}{2}$ unique entry pairs.

That gives a total of $\binom{n+2}{3} + \binom{n}{3} - \binom{n-1}{2}$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.

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Since rotations give only even permutations, the 3 distinct index triples $\{a,b,c\}$ with $a \neq b \neq c \neq a$ give twice as much distinct entries as you gave: entries at $(a,b,c)$ and $(b,a,c)$ may be distinct in this case. This adds $n$ choose $3$ distinct entries.

However, for entries where $\{a,1,n}\}$ a 90 degree rotation can interchange $1$ and $n$ because $N$ is odd and $n=(N+1)/2=1-n$ modulo $N$, which is an odd permutation identifying $(n-2)$ unique entry pairs (not times 3 since we already set entries at $(a,1,n)$, $(n,a,1)$ and $(1,n,a)$ equal.

That gives a total of $\binom{n+2}{3} + \binom{n}{3} - (n-2)$ distinct entries/dimensions in your tensor space. Check for instance for $n=3, N=5$: giving $10$ distinct entries.