The Magic Encyclopedia ™

Panmagic squares of orders m = 4p
{note: investigative article}
(by Aale de Winkel)

The work of Kathleen Ollerenshaw and David Brée onto "Most-Perfect squares" inspired many to investigate the doubly even orders. This Author couldn't quite figure what they where doing in (if I remember correcly) their second step, however I found a much simpler two step transformation from the "Normal order m square: Nm : Ni,j = i * m + j" onto the magic squares which Ollerenshaw qualified as "most-perfect". Also I noticed that when applied to basic multiplications of N2 other sets of most perfect squares. Below a renewed study is presented summarising studies presented elsewhere on this site as well as counting arguments resulting from the given arguments. Independently George Chen came up with the same "pandiagonal transform" as I did.
(working through this I noticed that factors are currently unknown to me, "pan anti-semimagic" squares I never studied so I haven't the foggiest how many of those are around)

panmagic squares of doubly even order (orders: m = 4p)
Pandiagonal transform
P(..)
The pandiagonal transform is defined as a 2 step process on hypercubes of order 4p
[j(2p+k)] <-> [j(4p-1-k)] (j = 0..n-1; k = 0..p-1
[jk] <-> [j(2p+k)] (j = 0..n-1; k = odd; k < 2p)
The first step of the process is a mirror-swap of the third and fourth quater of each 1-agonal.
The second step is a swap of alternate 1-agonal numbers with a number 2p apart on the same 1-agonal
In overall the first step compensate the 1-agonal off sums of the numbers modulo (2p)2
while the second swap compensate the part of the off sums due to the high component of the numbers
note: the transform does not compensate diagonal sums, so the diagonals need sum already to the
magic sum, for the result to be diagonal thus the origin need to be "pan anti-semimagic" or
"pan block-magic" (must take a closer look at these qualifications)
These conditions are met by the normal square, and left multiplications with N2 or N2t of either
order 2p normal square or order 2p panmagic square.
panmagic squares of order 4
The series starts with order 4: since (N2N2t)t = (N2tN2) the three below are the only obtained
it is known that there are only 48 panmagic order 4 square also counting pan-relocations
the three below represent all possible order 3 panmagic squares
P(N4)

01 15 04 14
12 06 09 07
13 03 16 02
08 10 05 11
P(N2N2)

01 15 06 12
14 04 09 07
11 05 16 02
08 10 03 13
P(N2N2t)

01 14 07 12
15 04 09 06
10 05 16 03
08 11 02 13
panmagic squares of order m = 2p 4
Starting from the oder 4 the peculiar series of 2p 4 holds the full richness of the forementioned
besides P(Nm) we have P(N2Sm/2), P(N2tSm/2) as well as P(i=0p N2[t])
P(Nm)
factor: unknown
This singular factor is currently under investigation Nm must be seen as only a representative
of the class of order m "pan anti-semimagic" squares, this author is not familiar with this class of
squares nor its members, let alone the amount of squares thus qualified are around
P(i=0p N2[t])
factor: 2p
The full product of N2[t] ([t] denotes transpositioned or not) contributes
2p pandiagonal squares, in order to dispense with transpositional variants one needs
only consider N2 as the outer most left factor as is consistent with order 4
P(N2Sm/2)
factor: unknown
With this peculiar inductive series Sm/2 stands for all order m/2 "pan anti-semimagic" or
"pan block-magic" squares, currently I can't figure what factor since the first types of squares are
new to me. All these squares need to be put into all 8(m/2)2 pan and aspectial variants
and afterwards devided by the 4m2 pan reflectional variants of the target order. Multiplication
with only N2 ought to ensure non-occurence of transpositional variants of order m squares.
panmagic squares of order m = 2p 6
Starting from the oder 6 the peculiar series of 2p 6 is hampered by the non-existence of order 6
panmagic squares (or are there) of course N6 is a member of the not yet studied class of "pan anti-semimagic"
square. So part of the above mentioned still remains true. (more on this so needs to be left till future upload

Below I put the three panmagic squares of order 3 in normalized position and pulled out the binariy representation of those numbers, this shows that the square are based on a single binairy component square and one panrelocation, further only some transpositions of these two squares

binairy decomposition order 4 panmagic squares
P(N4)
01 08 13 12
14 11 02 07
04 05 16 09
15 10 03 06
0 0 1 1
1 1 0 0
0 0 1 1
1 1 0 0
0 1 1 0
1 0 0 1
0 1 1 0
1 0 0 1
0 1 0 1
0 1 0 1
1 0 1 0
1 0 1 0
0 1 0 1
1 0 1 0
1 0 1 0
0 1 0 1
P(N2N2)
01 08 11 14
12 13 02 07
06 03 16 09
15 10 05 04
0 0 1 1
1 1 0 0
0 0 1 1
1 1 0 0
0 1 0 1
0 1 0 1
1 0 1 0
1 0 1 0
0 1 1 0
1 0 0 1
0 1 1 0
1 0 0 1
0 1 0 1
1 0 1 0
1 0 1 0
0 1 0 1
P(N2N2t)
01 08 10 15
12 13 03 06
07 02 16 09
14 11 05 04
0 0 1 1
1 1 0 0
0 0 1 1
1 1 0 0
0 1 0 1
0 1 0 1
1 0 1 0
1 0 1 0
0 1 0 1
1 0 1 0
1 0 1 0
0 1 0 1
0 1 1 0
1 0 0 1
0 1 1 0
1 0 0 1


NOTE: The below is shown as is, a full database of order 8 binaiy panmagic squares I need to establish some squares shown below need major pan-relocation for this use. (probably I do a full iteration to establish a complete set!)

below some order 8 binairy panmagic squares are listed the first column by a systematic listing of squares I decomposed the other columns where formed by systematic application of an order 4 add on.

binairy order 8 panmagic squares
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
0 0 0 1 1 1 0 1
1 1 0 1 0 0 0 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
0 0 1 0 1 1 1 0
1 1 1 0 0 0 1 0
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
0 0 1 1 1 0 0 1
1 0 0 1 0 0 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
0 1 1 0 1 1 0 0
1 1 0 0 0 1 1 0
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
0 1 1 1 0 0 0 1
0 0 0 1 0 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 0 1 0 0 0
1 0 0 0 1 1 1 0
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 1 0 1 1 1
1 1 0 1 0 1 0 0
0 0 1 0 1 0 1 1
1 1 1 0 1 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 1 0 0 1 1 1
1 0 0 1 1 1 0 0
0 1 1 0 0 0 1 1
1 1 0 1 1 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 0 1 1 0 1
0 0 0 1 1 1 1 0
1 1 1 0 0 0 0 1
1 1 0 1 0 0 1 0
0 0 1 0 1 1 0 1
0 0 0 1 1 1 1 0
1 1 1 0 0 0 0 1
1 1 0 1 0 0 1 0
0 0 1 0 1 1 0 1
1 1 0 1 0 0 1 0
0 0 1 0 1 1 0 1
1 1 0 1 0 0 1 0
0 0 1 0 1 1 0 1
1 1 0 1 0 0 1 0
0 0 1 0 1 1 0 1
1 1 0 1 0 0 1 0


finding regular order panmagic squares from binairy magic square is finding six order 8 binairy squares, when no doubly numbers appear in a combined sextet one has found a representative of 720 differnt panmagic squares. Below just 4 of these squares I found by a trial and error run just to get a feel of the method. The above table is merely an onsett to a complete database of an order 8 panmagic binairy square database which I'll intent to complete, once this is established an identifying system can be set up, merely some transposition and panrelocations are required to obtain the sextets used below, displayed also is the square obtained with the binaries in there shown order and one in reverse order. Aside from the 720 panmagic squares found by simply permutation of the six binary squares, I think many such families will be related by mere transformation of just a few of their binairy components.
Rules needs to be established to select the "main-representative" of the 720 by permutation, ordering by the greater amount of 0's at the upper left corner. Rules also to be used by a full listing of the binairy order 8 squares, to establish a numbering system. (NOTE: the above listing does not abide these rules yet, it merely focusses my thoughts abot the idea. The first square below is I think a main representative, the third won't be I think the fifth binairy might need to move upward, probably it's origin need to be panrelecated to move the sequal of 4 0's to the top left position.)
Current use of this idea is merely to pull out the binairy squares of an existing order 8 panmagic square, reorder those (according to not yet clear rules) to find the main representative.

panmagic squares by binairy matrix aproach
(normal and reverse ordering)
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
1 1 1 1 0 0 0 0
0 0 1 1 0 0 1 1
0 0 1 1 0 0 1 1
0 0 1 1 0 0 1 1
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
1 1 0 0 1 1 0 0
1 1 0 0 1 1 0 0
1 1 0 0 1 1 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
00 05 18 23 40 45 58 63
11 14 25 28 35 38 49 52
33 36 51 54 09 12 27 30
42 47 56 61 02 07 16 21
20 17 06 03 60 57 46 43
31 26 13 08 55 50 37 32
53 48 39 34 29 24 15 10
62 59 44 41 22 19 04 01
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
00 40 18 58 05 45 23 63
52 28 38 14 49 25 35 11
33 09 51 27 36 12 54 30
21 61 07 47 16 56 02 42
10 34 24 48 15 39 29 53
62 22 44 04 59 19 41 01
43 03 57 17 46 06 60 20
31 55 13 37 26 50 08 32
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 0 0 0 1 1 1 1
1 1 1 1 0 0 0 0
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
0 0 1 1 0 0 1 1
1 1 0 0 1 1 0 0
00 21 10 29 34 55 40 63
41 60 33 54 11 30 03 20
06 19 14 25 36 49 44 59
47 58 37 50 13 24 07 16
17 04 27 12 51 38 57 46
56 45 48 39 26 15 18 05
23 02 31 08 53 32 61 42
62 43 52 35 28 09 22 01
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
0 0 1 0 1 1 0 1
0 0 0 1 1 1 1 0
1 1 1 0 0 0 0 1
1 1 0 1 0 0 1 0
0 0 1 0 1 1 0 1
0 0 0 1 1 1 1 0
1 1 1 0 0 0 0 1
1 1 0 1 0 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
00 42 20 46 17 59 05 63
37 15 33 27 52 30 48 10
24 50 28 38 09 35 13 55
61 23 41 19 44 06 56 02
34 08 54 12 51 25 39 29
07 45 03 57 22 60 18 40
58 16 62 04 43 01 47 21
31 53 11 49 14 36 26 32