Magic Constant of Magic Cube

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Theorem

The magic constant of a magic cube of order $n$ is given by:

$C_n = \dfrac {n \paren {n^3 + 1} } 2$


Proof

Let $M_n$ denote a magic cube of order $n$.

By Sum of Terms of Magic Cube, the total of all the entries in $M_n$ is given by:

$T_n = \dfrac {n^3 \paren {n^3 + 1}} 2$

There are $n^2$ rows in $M_n$, each one with the same magic constant.

Thus the magic constant $C_n$ of the magic cube $M_n$ is given by:

\(\displaystyle C_n\) \(=\) \(\displaystyle \dfrac {T_n} {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {n^3 \paren {n^3 + 1} } {2 n^2}\) Sum of Terms of Magic Cube
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {n \paren {n^3 + 1} } 2\)

$\blacksquare$


Sequence

The sequence of the magic constants of magic cubes of order $n$ begins:

$1, (9,) \, 42, 130, 315, 651, 1204, 2052, 3285, 5005, 7326, 10 \, 374, 14 \, 287, 19 \, 215, 25 \, 320, 32 \, 776, \ldots$

However, note that while $9 = \dfrac {2 \paren {2^3 + 1} } 2$, a magic cube of order $2$ does not actually exist.