The complete count of the special order-three magic squares

The special classes of order-three magic squares, counted completely.

Let $P_{3}(t)$ be the number of panmagic (pandiagonal) squares of order three and line sum $t$ — those $3\times3$ arrays of nonnegative integers whose rows, columns and all six broken diagonals sum to $t$ — and let $S_{3}(t)$ be the number of symmetric magic squares of order three and line sum $t$. The theorem determines both, for every $t$:

What the answers say. Where MacMahon's count for the plain magic squares of order three is quadratic ($M_{3}(3e)=2e^{2}+2e+1$), the two special classes are degenerate in opposite ways. The six broken diagonals are so restrictive that they collapse the whole two-parameter family to the single constant square: for line sum $3e$ that is the array all of whose entries equal $e$. Symmetry, by contrast, removes only three of the eight line conditions, and what survives is a genuine one-parameter family

giving the linear count $2e+1$. In both cases the divisibility obstruction for order-three magic squares — the centre equals one third of the line sum, by center_of_order_three — forces the vanishing off multiples of three.

Proof. The two evaluations at $t=3e$ are supplied by pan_three_card and by symm_three_bij composed with the cardinality of the parameter interval $\{0,\dots,2e\}$; the two vanishing cases are supplied by pan_three_otherwise and symm_three_otherwise, both of which reduce to the centre identity. Splitting on whether $3\mid t$ and substituting $t=3e$ finishes the count.

Context. This is the fourth and last instalment of the order-three programme: after counting all squares (Mission I), all semi-magic ones (Mission II) and classifying the normal ones (Mission III), the special classes complete the picture. The comparison between the three counts — quadratic, linear and constant — is the point of the order-three study, and it is what is lost at order four, where no closed form is known.