The Magic Encyclopedia ™

Panmagic squares of squared order
(by Aale de Winkel)

In principle the prime order construction for panmagic squares also work for nonprime orders save for the fact that the LS-generating formulae does not produce panmagic latin squares as it does with prime orders, this article adresses some of the intricacies involved to produce those. The latin component square is a new concept currently only for square orders
NOTE: I lifted the investigative status, future date investigation might be persued further)

panmagic squares of squared order (orders: m = p²)
Latin component square
generating formula
LSC(a,b)
factor: p(p-1) Latin component squares obtained by formula
LSC(a,b): LC_i,j = [a (j % p) + b (j div p) + i] % p ;
i,j ε [0,..,m-1]; a = 0 .. p - 1 ; b = 1 .. p - 1

Latin square
factor: [p(p-1)]² Latin squares obtained by combining two component squares
LS(a,b,c,d): L_i,j = p LSC(a,b)^t + LSC(c,d)

The latin components combine into latin squares, the transposition is necessairy to
avoid double number on each line. These latin squares inherit the panmagicness from
the components, digit changing however need to be done on the components in order to
keep the panmagic feature on these latin squares.

most basic panmagic squares
factor:
[p(p-1)]² *
2 * [(p(p-1)-1)(p(p-1)-2)-1]
preliminairy S = m LS(a,b,c,d) + LS(e,f,g,h) or
S = m LS(a,b,c,d) + LS^t(e,f,g,h)
Experimentations with an order 9 spreadsheet suggest 19 low components LS's with
each high component LS suggesting [p(p-1)]²[(p-1)(p-2)-1] panmagic squares
obtainable by this theory. (note: formula based on order 9 only, so might be faulty)

In order to derive the panmagic squares in normalized position from the above the same priciples of course apply
as in the prime order case, complicated by the fact that the digit changing now apply on each of the four concidered
components, since we want the upperleft corner to remain 0 each component contributes a possible (p-1) due to digit
changing, the "[p,q,r,s]-condition" (see prime orders) give rise to a factor (p-2) in stead of (p-1) in one of the 3
higher components (??), so thus argued (p-1)³(p-2) panmagic squares are thus obtained.

panmagic squares
factor:
2 * [p(p-1)]² *
[(p(p-1)-1)(p(p-1)-2)-1](p!)⁴
status: preliminairy

note the given formulae need to
be verified (generalising from
one orders samples is tricky,
prone to be faulty) S = m [p LSC(a,b)^t_p1 + LSC(c,d)_p2] + [p LSC(e,f)^t_p3 + LSC(g,h)_p4] or
S = m [p LSC(a,b)^t_p1 + LSC(c,d)_p2] + [p LSC(e,f)^t_p3 + LSC(g,h)_p4]^t
The various factors here denoted as _p1 .. _p4 are allowed here to change full range,
hence the factor is (p!)⁴ allowing this full range digit changing one also obtains
reflectional variants, also all m² panrelocations are obtained this way, thus an extra
deviding factor of 4m² is needed to obtain the amount of different panmagic squares.

2 * [p(p-1)]²[(p(p-1)-1)(p(p-1)-2)-1](p!)⁴

9 (p=3)
25 (p=5) 2 * 886.464 (2 * 2.736)
2 * 28.283.904.000.000 (2 * 11.313.561.600)

The number of LSC's is currently under investigation the Latin Component Hypercube article
suggest LSC's are formed by adding "order 4 add-onns" to the trivial hypercube. This scheme
also produces more irregularly shaped LSC's than form by the discussed 2-parameter method

LSC's by means of augmented patterns
Pattern a p by p square with digits radix p
The pattern form the basic content of each p by p subsquare

Augmentator a p by p square with digits radix p
Each augmentator cell represents a p by p block of cells with the digit as element
to each block a pattern is added (mod p) to obtain a LSC.

Colorisor a p by p square with in each cell a permutation of the digits [0..p]
Each colorisor cell contains a digit changing permutation, to be applied to the pattern
which is pasted onto the p by p block each cell represents

for order 9 the the agmentation and colorisation scheme are equivalent, currently I know of 8
augmentators, each have at least 12 working patterns as patterns which form LSC components of
order 9 magic squares. This is under investigation since recenty I found 19 working patterns
for one of the augmentators. The DataBase holds a full listing of the 124 panmagic LSC's found.
A spreadsheet is provided to combine LSC's and test the resulting squares for regularity.

Order p² panmagic squares
Given the set of panmagic LSC's a panmagic square is obtained by combining those LSC's
LSC₁ * p³ + LSC₂ * p² + LSC₃ * p + LSC₄
Currently it is mere trial and error to find LSC quadruplets which combine into regular squares
Thus for now it isn't possible to predict the total amount of squares, nor do I see possibility
to predict how many LSC's there are for other orders. The reader is invited to investigate!

A most general spreadsheet for order 9 is downloadable from the encyclopedia database utalising augmented patterns

panmagic squares of squared order (orders: m = p²)
Latin component square generating formula LSC(a,b) factor: p(p-1)	Latin component squares obtained by formula
	LSC(a,b): LC_i,j = [a (j % p) + b (j div p) + i] % p ; i,j ε [0,..,m-1]; a = 0 .. p - 1 ; b = 1 .. p - 1
Latin square factor: [p(p-1)]²	Latin squares obtained by combining two component squares
	LS(a,b,c,d): L_i,j = p LSC(a,b)^t + LSC(c,d)
	The latin components combine into latin squares, the transposition is necessairy to avoid double number on each line. These latin squares inherit the panmagicness from the components, digit changing however need to be done on the components in order to keep the panmagic feature on these latin squares.
most basic panmagic squares factor: [p(p-1)]² * 2 * [(p(p-1)-1)(p(p-1)-2)-1] preliminairy	S = m LS(a,b,c,d) + LS(e,f,g,h) or S = m LS(a,b,c,d) + LS^t(e,f,g,h)
	Experimentations with an order 9 spreadsheet suggest 19 low components LS's with each high component LS suggesting [p(p-1)]²[(p-1)(p-2)-1] panmagic squares obtainable by this theory. (note: formula based on order 9 only, so might be faulty)
In order to derive the panmagic squares in normalized position from the above the same priciples of course apply as in the prime order case, complicated by the fact that the digit changing now apply on each of the four concidered components, since we want the upperleft corner to remain 0 each component contributes a possible (p-1) due to digit changing, the "[p,q,r,s]-condition" (see prime orders) give rise to a factor (p-2) in stead of (p-1) in one of the 3 higher components (??), so thus argued (p-1)³(p-2) panmagic squares are thus obtained.
panmagic squares factor: 2 * [p(p-1)]² * [(p(p-1)-1)(p(p-1)-2)-1](p!)⁴ status: preliminairy note the given formulae need to be verified (generalising from one orders samples is tricky, prone to be faulty)	S = m [p LSC(a,b)^t_p1 + LSC(c,d)_p2] + [p LSC(e,f)^t_p3 + LSC(g,h)_p4] or S = m [p LSC(a,b)^t_p1 + LSC(c,d)_p2] + [p LSC(e,f)^t_p3 + LSC(g,h)_p4]^t
	The various factors here denoted as _p1 .. _p4 are allowed here to change full range, hence the factor is (p!)⁴ allowing this full range digit changing one also obtains reflectional variants, also all m² panrelocations are obtained this way, thus an extra deviding factor of 4m² is needed to obtain the amount of different panmagic squares.
	2 * [p(p-1)]²[(p(p-1)-1)(p(p-1)-2)-1](p!)⁴
	9 (p=3) 25 (p=5)	2 * 886.464 (2 * 2.736) 2 * 28.283.904.000.000 (2 * 11.313.561.600)
The number of LSC's is currently under investigation the Latin Component Hypercube article suggest LSC's are formed by adding "order 4 add-onns" to the trivial hypercube. This scheme also produces more irregularly shaped LSC's than form by the discussed 2-parameter method
LSC's by means of augmented patterns
Pattern	a p by p square with digits radix p
Pattern	The pattern form the basic content of each p by p subsquare
Augmentator	a p by p square with digits radix p
Augmentator	Each augmentator cell represents a p by p block of cells with the digit as element to each block a pattern is added (mod p) to obtain a LSC.
Colorisor	a p by p square with in each cell a permutation of the digits [0..p]
Colorisor	Each colorisor cell contains a digit changing permutation, to be applied to the pattern which is pasted onto the p by p block each cell represents
for order 9 the the agmentation and colorisation scheme are equivalent, currently I know of 8 augmentators, each have at least 12 working patterns as patterns which form LSC components of order 9 magic squares. This is under investigation since recenty I found 19 working patterns for one of the augmentators. The DataBase holds a full listing of the 124 panmagic LSC's found. A spreadsheet is provided to combine LSC's and test the resulting squares for regularity.
Order p² panmagic squares
Given the set of panmagic LSC's a panmagic square is obtained by combining those LSC's LSC₁ * p³ + LSC₂ * p² + LSC₃ * p + LSC₄ Currently it is mere trial and error to find LSC quadruplets which combine into regular squares Thus for now it isn't possible to predict the total amount of squares, nor do I see possibility to predict how many LSC's there are for other orders. The reader is invited to investigate!