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:
\(\ds C_n\) | \(=\) | \(\ds \dfrac {T_n} {n^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {n^3 \paren {n^3 + 1} } {2 n^2}\) | Sum of Terms of Magic Cube | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {n \paren {n^3 + 1} } 2\) |
$\blacksquare$
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.