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6.09   Pan Magic Squares, Non Consecutive Integers, Part 1

6.09.1 Introduction

As proven in Section 6.1 it is not possible to construct Pan Magic Squares of the 6th order based on the consecutive distinct integers 1 ... 36 and related Magic Constant 111.

However in 1917 L.S. Pierson published a 6th order Pan Magic Square based on a series non consecutive distinct integers with the related Magic Constant 150.

1

47

6

43

5

48

35

17

30

21

31

16

36

12

41

8

40

13

7

45

2

49

3

44

29

19

34

15

33

20

42

10

37

14

38

9

According to William Symes Andrews subject square was originally constructed by Dr. C. Planck (ref. Magic Squares and Cubes (1909), Fig. 393).

It should be noted that this Pan Magic Square has also following properties:

1. Each 2 × 2 sub square sums to 100 (2/3 * Magic Constant);
2. Each 3 × 3 sub square sums to 225 (3/2 * Magic Constant);
3. All pairs of integers distant 6/2 along each diagonal sum to 50 (1/3 * Magic Constant).

Which makes the square comparable with Most Perfect Magic Squares as defined in Section 2.4.

6.09.2 Analysis (Pan Magic, Most Perfect)

The properties mentioned in section 6.09.1 above result in following set of linear equations:

All rows, columns and (pan)diagonals sum to the Magic Constant:

a( 1) + a( 2) + a( 3) + a( 4) + a( 5) + a( 6) = 150
a( 7) + a( 8) + a( 9) + a(10) + a(11) + a(12) = 150
a(13) + a(14) + a(15) + a(16) + a(17) + a(18) = 150
a(19) + a(20) + a(21) + a(22) + a(23) + a(24) = 150
a(25) + a(26) + a(27) + a(28) + a(29) + a(30) = 150
a(31) + a(32) + a(33) + a(34) + a(35) + a(36) = 150

a( 1) + a( 7) + a(13) + a(19) + a(25) + a(31) = 150
a( 2) + a( 8) + a(14) + a(20) + a(26) + a(32) = 150
a( 3) + a( 9) + a(15) + a(21) + a(27) + a(33) = 150
a( 4) + a(10) + a(16) + a(22) + a(28) + a(34) = 150
a( 5) + a(11) + a(17) + a(23) + a(29) + a(35) = 150
a( 6) + a(12) + a(18) + a(24) + a(30) + a(36) = 150

a( 1) + a( 8) + a(15) + a(22) + a(29) + a(36) = 150
a( 2) + a( 9) + a(16) + a(23) + a(30) + a(31) = 150
a( 3) + a(10) + a(17) + a(24) + a(25) + a(32) = 150
a( 4) + a(11) + a(18) + a(19) + a(26) + a(33) = 150
a( 5) + a(12) + a(13) + a(20) + a(27) + a(34) = 150
a( 6) + a( 7) + a(14) + a(21) + a(28) + a(35) = 150

a( 6) + a(11) + a(16) + a(21) + a(26) + a(31) = 150
a( 5) + a(10) + a(15) + a(20) + a(25) + a(36) = 150
a( 4) + a( 9) + a(14) + a(19) + a(30) + a(35) = 150
a( 3) + a( 8) + a(13) + a(24) + a(29) + a(34) = 150
a( 2) + a( 7) + a(18) + a(23) + a(28) + a(33) = 150
a( 1) + a(12) + a(17) + a(22) + a(27) + a(32) = 150

Each 2 × 2 sub square sums to 2/3 * Magic Constant:

a(i) + a(i+1) + a(i+ 6) + a(i+ 7) = 100 with 1 =< i < 30 and i ≠ 6*n for n = 1, 2 ... 5

a(i) + a(i+1) + a(i+ 6) + a(i- 5) = 100 with i = 6*n for n = 1, 2 ... 5

a(i) + a(i+1) + a(i+30) + a(i+31) = 100 with i = 1, 2 ... 5

a(1) + a(6)   + a(31)   + a(36)   = 100

Each 3 × 3 sub square sums to 3/2 * Magic Constant:

       Σ   a(i + j) = 225 with 1 =< i =< 22 and i ≠ 6 * n and i ≠ (6 * n - 1) for n = 1 ... 4
j = 0, 1, 2
    6, 7, 8
   12,13,14

       Σ   a(i + j) = 225 with i = (6 * n - 1) for n = 1 ... 4
j = 0, 1, 2
    6, 7, 8
   12,13,-4

       Σ   a(i + j) = 225 with i = 6 * n       for n = 1 ... 4
j = 0, 1, 2
    6, 7, 8
   12,-5,-4

       Σ   a(i + j) = 225 with i = 1 ... 4
j = 0, 1, 2
   24,25,26
   30,31,32

       Σ   a(i + j) = 225 with i = 1 ... 4
j = 0, 1, 2
    6, 7, 8
   30,31,32

a( 1) + a( 5) + a( 6) + a(25) + a(29) + a(30) + a(31) + a(35) + a(36) = 225
a( 1) + a( 2) + a( 6) + a(25) + a(26) + a(30) + a(31) + a(32) + a(36) = 225
a( 1) + a( 5) + a( 6) + a( 7) + a(11) + a(12) + a(31) + a(35) + a(36) = 225
a( 1) + a( 2) + a( 6) + a( 7) + a( 8) + a(12) + a(31) + a(32) + a(36) = 225

All pairs of integers distant 6/2 along each diagonal sum to 1/3 * Magic Constant:

a( 1) + a(22)= 50
a( 7) + a(28)= 50
a(13) + a(34)= 50
a(19) + a( 4)= 50
a(25) + a(10)= 50
a(31) + a(16)= 50

a( 2) + a(23)= 50
a( 8) + a(29)= 50
a(14) + a(35)= 50
a(20) + a( 5)= 50
a(26) + a(11)= 50
a(32) + a(17)= 50

a( 3) + a(24)= 50
a( 9) + a(30)= 50
a(15) + a(36)= 50
a(21) + a( 6)= 50
a(27) + a(12)= 50
a(33) + a(18)= 50

The resulting number of equations can be written in the matrix representation as:

             
     AB * a = s

which can be reduced, by means of row and column manipulations, and results in following set of linear equations:

a(33) =  150 - 2 * a(34) - 2 * a(35) - a(36)
a(32) = -150 + 2 * a(34) + 3 * a(35) + 2 * a(36)
a(31) =  150 - a(34) - 2 * a(35) - 2 * a(36)
a(29) =  100 - a(30) - a(35) - a(36)
a(28) =        a(30) - a(34) + a(36)
a(27) = - 50 - a(30) + 2 * a(34) + 2 * a(35)
a(26) =  150 + a(30) - 2 * a(34) - 3 * a(35) - a(36)
a(25) = - 50 - a(30) + a(34) + 2 * a(35) + a(36)
a(24) =  125 - a(30) - a(34) - a(35) - a(36)
a(23) = -125 + a(30) + a(34) + 2 * a(35) + 2 * a(36)
a(22) =  125 - a(30) - a(35) - 2 * a(36)
a(21) =   25 + a(30) - a(34) - a(35) + a(36)
a(20) = - 25 - a(30) + a(34) + 2 * a(35)
a(19) =   25 + a(30) - a(35)
a(18) = -100 + 2 * a(34) + 2 * a(35) + a(36)
a(17) =  200 - 2 * a(34) - 3 * a(35) - 2 * a(36)
a(16) = -100 + a(34) + 2 * a(35) + 2 * a(36)
a(15) =   50 - a(36)
a(14) =   50 - a(35)
a(13) =   50 - a(34)
a(12) =  100 + a(30) - 2 * a(34) - 2 * a(35)
a(11) = -100 - a(30) + 2 * a(34) + 3 * a(35) + a(36)
a(10) =  100 + a(30) - a(34) - 2 * a(35) - a(36)
a( 9) =   50 - a(30)
a( 8) = - 50 + a(30) + a(35) + a(36)
a( 7) =   50 - a(30) + a(34) - a(36)
a( 6) =   25 - a(30) + a(34) + a(35) - a(36)
a( 5) =   75 + a(30) - a(34) - 2 * a(35)
a( 4) =   25 - a(30) + a(35)
a( 3) = - 75 + a(30) + a(34) + a(35) + a(36)
a( 2) =  175 - a(30) - a(34) - 2 * a(35) - 2 * a(36)
a( 1) = - 75 + a(30) + a(35) + 2 * a(36)

The solutions can be obtained by guessing a(36), a(35), a(34) and a(30) and filling out these guesses in the abovementioned equations.

For distinct integers also following inequalities should be applied:

0 < a(i) =< 49       for i = 1, 2 ... 29, 31 ... 33
a(i) ≠ a(j)          for i ≠ j

An optimized guessing routine (MgcSqr6g1), produced 288 Most Perfect Pan Magic Squares (ref. Attachment 6.10.2) within 10.5 seconds.

This collection is identical to the class {Aijk}, which can be obtained by rotation, reflection, column and/or row shifts performed on the Most Perfect Square of L.S. Pierson shown in Section 6.09.1 above.

It should be noted that the non occurring integers of the range 1 thru 49, as shown below (shaded):

1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31 32 33 34 35
36 37 38 39 40 41 42
43 44 45 46 47 48 49

are a consequence of the defining properties of subject square.

6.09.3 Analysis (Pan Magic, 2 x 2 Compact)

If the defining properties mentioned in Section 6.09.1 above are moderated, much more Pan Magic Squares with non consecutive integers can be found.

It should be noted that the second property in Section 6.09.1 (each 3 × 3 Sub Square sums to 225) is a consequence of property 1 and 3.

So let’s concentrate on Pan Magic Squares for which only the first property (each 2 × 2 sub square sums to 100) applies.

The resulting equations for this case are:

a(31) = 150 - a(32) - a(33) - a(34) - a(35) - a(36)	
a(29) = 100 - a(30) - a(35) - a(36)	
a(28) =       a(30) - a(34) + a(36)	
a(27) = 100 - a(30) - a(33) - a(36)	
a(26) =       a(30) - a(32) + a(36)	
a(25) = -50 - a(30) + a(32) + a(33) + a(34) + a(35)	
a(23) =     - a(24) + a(35) + a(36)	
a(22) =       a(24) + a(34) - a(36)	
a(21) =     - a(24) + a(33) + a(36)	
a(20) =       a(24) + a(32) - a(36)	
a(19) = 150 - a(24) - a(32) - a(33) - a(34) - a(35)	
a(17) = 100 - a(18) - a(35) - a(36)	
a(16) =       a(18) - a(34) + a(36)	
a(15) = 100 - a(18) - a(33) - a(36)	
a(14) =       a(18) - a(32) + a(36)	
a(13) = -50 - a(18) + a(32) + a(33) + a(34) + a(35)	
a(12) =  75 - a(24) - a(32) - a(34) + a(36)	
a(11) = -75 + a(24) + a(32) + a(34) + a(35)	
a(10) =  75 - a(24) - a(32)	
a( 9) = -75 + a(24) + a(32) + a(33) + a(34)	
a( 8) =  75 - a(24) - a(34)	
a( 7) =  75 + a(24) - a(33) - a(35) - a(36)	
a( 6) =  75 - a(18) - a(30) + a(32) + a(34) - 2 * a(36)	
a( 5) =  25 + a(18) + a(30) - a(32) - a(34) - a(35) + a(36)	
a( 4) =  75 - a(18) - a(30) + a(32) - a(36)	
a( 3) =  25 + a(18) + a(30) - a(32) - a(33) - a(34) + a(36)	
a( 2) =  75 - a(18) - a(30) + a(34) - a(36)	
a( 1) =-125 + a(18) + a(30) + a(33) + a(35) + 2 * a(36)	

The solutions can be obtained by guessing a(36) ... a(32), a(30), a(24) and a(18) and filling out these guesses in the abovementioned equations.

For distinct integers also following inequalities should be applied:

0 < a(i) =< 49       for i = 1, 2 ... 17, 19 ... 23, 25 ... 29, 31
a(i) ≠ k             for i = 1 ... 36 and k = 4, 11, 18, 22 ... 28, 32, 39, 46
a(i) ≠ a(j)          for i ≠ j

An optimized guessing routine (MgcSqr6g2), produced 36 * 864 = 31104 Compact Pan Magic Squares within 1.5 hour.

This collection contains 31104/288 = 108 Classes {Aijk}, of which one Class is shown in Attachment 6.10.3.

Much more solutions can be found when the non consecutive integers are selected from the whole range 1 thru 49.

6.09.4 Analysis (Pan Magic, Associated)

Associated (Pan) Magic Squares can be defined as Center Symmetric (Pan) Magic Squares.

The sum of each pair of elements, which can be connected with a straight line through the centre is 2 * s1/n.

For a 6th order Associated Pan Magic Square, based on the non consecutive integers applied above and corresponding Magic Sum s1 = 150, these pairs sum to 50.

For Associated Pan Magic Squares, following equations should be added to the equations defining a Pan Magic Square:

a( 1) + a(36) = s1/3
a( 2) + a(35) = s1/3
a( 3) + a(34) = s1/3
a( 4) + a(33) = s1/3
a( 5) + a(32) = s1/3

a( 6) + a(31) = s1/3
a( 7) + a(30) = s1/3
a( 8) + a(29) = s1/3
a( 9) + a(28) = s1/3
a(10) + a(27) = s1/3

a(11) + a(26) = s1/3
a(12) + a(25) = s1/3
a(13) + a(24) = s1/3
a(14) + a(23) = s1/3

a(15) + a(22) = s1/3
a(16) + a(21) = s1/3
a(17) + a(20) = s1/3
a(18) + a(19) = s1/3

which results in following linear equations describing Associated Pan Magic Squares:

a(31) =     s1 - a(32) - a(33) - a(34) - a(35) - a(36)
a(27) = 2 * s1 - a(28) - 2 * a(29) - 2 * a(30) + a(32) - 2 * a(34) - 3 * a(35) - 2 * a(36)
a(26) =     s1 - 2 * a(27) - a(29) - 2 * a(30)
a(25) =              a(27) - a(28) + a(30)
a(24) = 3 * s1/2 - a(29) - 2 * a(30) - a(34) - 2 * a(35) - 2 * a(36)
a(23) =     s1/2 - a(25) - a(28) - a(29) + a(32)
a(22) = 3 * s1/2 - 2 * a(28) - a(29) - 2 * a(34) - 2 * a(35) - a(36)
a(21) =     s1 - a(24) - a(32) - a(33) - a(35) - a(36)
a(20) =          a(22) + a(28) - a(30) + a(32) - a(36)
a(19) =        - a(20) + a(28) - a(30) + a(32) + a(33)

a(18) = s1/3 - a(19)
a(17) = s1/3 - a(20)
a(16) = s1/3 - a(21)
a(15) = s1/3 - a(22)
a(14) = s1/3 - a(23)

a(13) = s1/3 - a(24)
a(12) = s1/3 - a(25)
a(11) = s1/3 - a(26)
a(10) = s1/3 - a(27)
a( 9) = s1/3 - a(28)

a(8) = s1/3 - a(29)
a(7) = s1/3 - a(30)
a(6) = s1/3 - a(31)
a(5) = s1/3 - a(32)

a(4) = s1/3 - a(33)
a(3) = s1/3 - a(34)
a(2) = s1/3 - a(35)
a(1) = s1/3 - a(36)

The solutions can be obtained by guessing a(36) ... a(32), a(30), a(29) and a(28) and filling out these guesses in the abovementioned equations.

For distinct integers also following inequalities should be applied:

0 < a(i) =< 49       for i = 1, 2 ... 27, 31
a(i) ≠ k             for i = 1 ... 36 and k = 4, 11, 18, 22 ... 28, 32, 39, 46
a(i) ≠ a(j)          for i ≠ j

An optimized guessing routine (MgcSqr6g4), produced 1664 Associated Pan Magic Squares, which are shown in Attachment 6.10.4.

6.09.5 Analysis (Pan Magic, Associated, Non Overlapping Subsquares)

For Associated Pan Magic Squares (Ultra Magic) of the sixth order, composed out of 9 Non Overlapping Sub Squares (2 x 2) following equations should be added to the equations defining an Ultra Magic Square:

a( 1) + a( 2) + a( 7) + a( 8) = 2 * s1 / 3
a( 3) + a( 4) + a( 9) + a(10) = 2 * s1 / 3
a( 5) + a( 6) + a(11) + a(12) = 2 * s1 / 3
a(13) + a(14) + a(19) + a(20) = 2 * s1 / 3
a(15) + a(16) + a(21) + a(22) = 2 * s1 / 3
a(17) + a(18) + a(23) + a(24) = 2 * s1 / 3
a(25) + a(26) + a(31) + a(32) = 2 * s1 / 3
a(27) + a(28) + a(33) + a(34) = 2 * s1 / 3
a(29) + a(30) + a(35) + a(36) = 2 * s1 / 3

which results in following linear equations describing Ultra Magic Squares composed out of Non Overlapping Sub Squares:

a(32) =             - a(33) + a(34) + a(35)
a(31) =      s1 - 2 * a(34) - 2 * a(35) - a(36)
a(29) =  2 * s1 / 3 - a(30) - a(35) - a(36)
a(27) =  2 * s1 / 3 - a(28) - a(33) - a(34)
a(26) =    - s1 + 2 * a(28) - a(30) + 2 * a(33) + 2 * a(34) + a(35) + a(36)
a(25) =  2 * s1 / 3 - 2 * a(28) + a(30) - a(33) - a(34)
a(24) =  5 * s1 / 6 - a(30) - a(34) - a(35) - a(36)
a(23) = -5 * s1 / 6 + a(28) + 2 * a(34) + 2 * a(35) + a(36)
a(22) =  5 * s1 / 6 - 2 * a(28) + a(30) - 2 * a(34) - a(35)
a(21) =      s1 / 6 + a(30) - a(35)
a(20) =  5 * s1 / 6 - a(28) - a(33) - a(34) - a(36)
a(19) = -5 * s1 / 6 + 2 * a(28) - a(30) + a(33) + 2 * a(34) + a(35) + a(36)

a(18) = s1/3 - a(19)
a(17) = s1/3 - a(20)
a(16) = s1/3 - a(21)
a(15) = s1/3 - a(22)
a(14) = s1/3 - a(23)

a(13) = s1/3 - a(24)
a(12) = s1/3 - a(25)
a(11) = s1/3 - a(26)
a(10) = s1/3 - a(27)
a( 9) = s1/3 - a(28)

a(8) = s1/3 - a(29)
a(7) = s1/3 - a(30)
a(6) = s1/3 - a(31)
a(5) = s1/3 - a(32)

a(4) = s1/3 - a(33)
a(3) = s1/3 - a(34)
a(2) = s1/3 - a(35)
a(1) = s1/3 - a(36)

The solutions can be obtained by guessing a(36) ... a(33), a(30) and a(28) and filling out these guesses in the abovementioned equations.

For distinct integers also following inequalities should be applied:

0 < a(i) =< 49       for i = 1, 2 ... 27, 29, 31, 32
a(i) ≠ k             for i = 1 ... 36 and k = 4, 11, 18, 22 ... 28, 32, 39, 46
a(i) ≠ a(j)          for i ≠ j

An optimized guessing routine (Priem6f), produced 288 of subject Ultra Magic Squares, which are shown in Attachment 6.10.6.

It should be noted that for the selected non consecutive integers all 288 Ultra Magic Squares are Compact.

6.09.6 Analysis (Pan Magic, Associated, 2 x 2 Compact)

For Compact Ultra Magic Squares of the sixth order, following equations should be added to the equations defining an Ultra Magic Square:

a(i) + a(i+1) + a(i+ 6) + a(i+ 7) = 2*s1/3 with 1 =< i < 30 and i ≠ 6*n for n = 1, 2 ... 5

a(i) + a(i+1) + a(i+ 6) + a(i- 5) = 2*s1/3 with i = 6*n for n = 1, 2 ... 5

a(i) + a(i+1) + a(i+30) + a(i+31) = 2*s1/3 with i = 1, 2 ... 5

a(1) + a(6)   + a(31)   + a(36)   = 2*s1/3

which results in following linear equations describing Compact Ultra Magic Squares:

a(33) =      s1   - a(34) - 2 * a(35) - 2 * a(36)
a(32) =    - s1   + 2 * a(34) + 3 * a(35) + 2 * a(36)
a(31) =      s1   - 2 * a(34) - 2 * a(35) - a(36)
a(29) =  2 * s1/3 - a(30) - a(35) - a(36)
a(28) =             a(30) - a(34) + a(36)
a(27) =    - s1/3 - a(30) + a(34) + 2 * a(35) + a(36)
a(26) =      s1   + a(30) - 2 * a(34) - 3 * a(35) - a(36)
a(25) =    - s1/3 - a(30) + 2 * a(34) + 2 * a(35)
a(24) =  5 * s1/6 - a(30) - a(34) - a(35) - a(36)
a(23) = -5 * s1/6 + a(30) + a(34) + 2 * a(35) + 2 * a(36)
a(22) =  5 * s1/6 - a(30) - a(35) - 2 * a(36)
a(21) =      s1/6 + a(30) - a(35)
a(20) =    - s1/6 - a(30) + a(34) + 2 * a(35)
a(19) =      s1/6 + a(30) - a(34) - a(35) + a(36)

a(18) = s1/3 - a(19)
a(17) = s1/3 - a(20)
a(16) = s1/3 - a(21)
a(15) = s1/3 - a(22)
a(14) = s1/3 - a(23)

a(13) = s1/3 - a(24)
a(12) = s1/3 - a(25)
a(11) = s1/3 - a(26)
a(10) = s1/3 - a(27)
a( 9) = s1/3 - a(28)

a(8) = s1/3 - a(29)
a(7) = s1/3 - a(30)
a(6) = s1/3 - a(31)
a(5) = s1/3 - a(32)

a(4) = s1/3 - a(33)
a(3) = s1/3 - a(34)
a(2) = s1/3 - a(35)
a(1) = s1/3 - a(36)

The solutions can be obtained by guessing a(36) ... a(34) and a(30) and filling out these guesses in the abovementioned equations.

For distinct integers also following inequalities should be applied:

0 < a(i) =< 49       for i = 1, 2 ... 29, 31 ... 33
a(i) ≠ k             for i = 1 ... 36 and k = 4, 11, 18, 22 ... 28, 32, 39, 46
a(i) ≠ a(j)          for i ≠ j

An optimized guessing routine (Priem6h), produced the same 288 Compact Ultra Magic Squares as found in Section 6.09.5, which are shown in Attachment 6.10.7.

6.09.7 Analysis (Pan Magic, Concentric, Associated Center Square)

For Concentric Pan Magic Squares with Associated Center Square, following equations should be added to the equations defining a Pan Magic Square:

a( 1) + a(36) = s1/3
a( 2) + a(32) = s1/3
a( 3) + a(33) = s1/3
a( 4) + a(34) = s1/3
a( 5) + a(35) = s1/3

a( 6) + a(31) = s1/3
a( 7) + a(12) = s1/3
a( 8) + a(29) = s1/3
a( 9) + a(28) = s1/3
a(10) + a(27) = s1/3

a(11) + a(26) = s1/3
a(13) + a(18) = s1/3
a(14) + a(23) = s1/3
a(15) + a(22) = s1/3

a(16) + a(21) = s1/3
a(17) + a(20) = s1/3
a(19) + a(24) = s1/3
a(25) + a(30) = s1/3

which results in following linear equations describing Pan Magic Concentric Squares with Associated Center Square:

a(31) =    s1   - a(32) - a(33) - a(34) - a(35) - a(36)
a(29) =  2*s1/3 - a(30) - a(35) - a(36)
a(28) = (  s1/3 + 2 * a(30) - a(32) - a(33) - a(34) + a(35))/2
a(27) =  2*s1/3 - a(28) - a(33) - a(34)
a(26) = -2*s1/3 + a(30) + a(33) + a(34) + a(35) + a(36)
a(25) =    s1/3 - a(30)
a(24) =  5*s1/6 - a(30) - a(34) - a(35) - a(36)
a(23) = -3*s1/6 - a(28) + 2 * a(30) + 2 * a(35) + a(36)
a(22) =  3*s1/6 - a(30) - a(35)
a(21) =    s1/6 + a(30) - a(32)
a(20) = (4*s1/3 - 2 * a(30) + a(32) - a(33) - a(34) - a(35) - 2 * a(36))/2
a(19) =    s1/3 - a(24)
a(18) =  5*s1/6 - a(30) - a(33) - a(35) - a(36)
a(17) =    s1/3 - a(20)
a(16) =    s1/3 - a(21)
a(15) =    s1/3 - a(22)
a(14) =    s1/3 - a(23)
a(13) =    s1/3 - a(18)
a(12) =           a(30) - a(32) + a(35)
a(11) =    s1/3 - a(26)
a(10) =    s1/3 - a(27)
a( 9) =    s1/3 - a(28)
a( 8) =    s1/3 - a(29)
a( 7) =    s1/3 - a(12)
a( 6) =    s1/3 - a(31)
a( 5) =    s1/3 - a(35)
a( 4) =    s1/3 - a(34)
a( 3) =    s1/3 - a(33)
a( 2) =    s1/3 - a(32)
a( 1) =    s1/3 - a(36)

The solutions can be obtained by guessing a(36) ... a(32) and a(30) and filling out these guesses in the abovementioned equations.

The minimum Magic Sum s1 = 120 occurs for the range {i} = {1 ... 13, 15 ... 19, 21 ... 25, 27 ... 39}.

An optimized guessing routine (MgcSqr6g5), produced 128 Concentric Pan Magic Squares with Associated Center Square, which are shown in Attachment 6.10.5.

6.09.8 Analysis (Pan Magic, Complete, Composed of Semi Magic Sub Squares)

The defining equations for Pan Magic Squares composed of Semi Magic Sub Squares (6 Magic Lines) result - after deduction - in following set of linear equations:

a(34) =      s1 / 2 - a(35) - a(36)
a(31) =      s1 / 2 - a(32) - a(33)
a(28) =      s1 / 2 - a(29) - a(30)
a(25) =      s1 / 2 - a(26) - a(27)
a(24) =      s1 / 2 - a(30) - a(36)
a(23) =      s1 / 2 - a(29) - a(35)
a(22) =               a(29) + a(30) - a(34)
a(21) =      s1 / 2 - a(27) - a(33)
a(20) =      s1 / 2 - a(26) - a(32)
a(19) =      s1 / 2 - a(25) - a(31)

a(18) = s1 / 3 - a(33)
a(17) = s1 / 3 - a(32)
a(16) = s1 / 3 - a(31)
a(15) = s1 / 3 - a(36)
a(14) = s1 / 3 - a(35)
a(13) = s1 / 3 - a(34)

a(12) = s1 / 3 - a(27)
a(11) = s1 / 3 - a(26)
a(10) = s1 / 3 - a(25)
a( 9) = s1 / 3 - a(30)
a( 8) = s1 / 3 - a(29)
a( 7) = s1 / 3 - a(28)

a(6) = s1 / 3 - a(21)
a(5) = s1 / 3 - a(20)
a(4) = s1 / 3 - a(19)
a(3) = s1 / 3 - a(24)
a(2) = s1 / 3 - a(23)
a(1) = s1 / 3 - a(22)

which illustrate the consequential symmetry (complete).

The solutions can be obtained by guessing a(26), a(27), a(29), a(30), a(32), a(33), a(35), a(36) and filling out these guesses in the abovementioned equations.

The minimum Magic Sum s1 = 120 occurs for the range {i} = {1 ... 13, 15 ... 19, 21 ... 25, 27 ... 39}.

An optimized guessing routine (Priem6e3), produced 82944 Pan Magic and Complete Magic Squares Composed of Semi Magic Sub Squares within 1,85 hour, of which the first occuring for a(36) = i are shown in Attachment 6.10.8a.

Attachment 6.10.8b shows the corresponding Simple Associated Magic Squares Composed of Semi Magic Sub Squares (Euler).

6.09.9 Spreadsheet Solutions

The linear equations deducted in Section 6.09.2 thru 6.09.7 above, have been applied in following Excel Spread Sheets:

The red figures have to be “guessed” to construct a Pan Magic Square of the 6th order (wrong solutions are obvious).


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