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8.5   Pan Magic Squares, based on Franklin Properties

8.5.1 Analytic Solution, Pan Magic Squares, Franklin Properties 2 and 5

If the linear equations for Pan Magic Squares shown in section 8.1.1 are combined with property 2 and 5 of the Franklin Squares, as mentioned in section 8.4.1, this will result in following set of linear equations:

Every half-row and half-column sums to half the Magic Constant:

a( 1) + a( 2) + a( 3) + a( 4) = s1/2
a( 9) + a(10) + a(11) + a(12) = s1/2
a(17) + a(18) + a(19) + a(20) = s1/2
a(25) + a(26) + a(27) + a(28) = s1/2

a( 1) + a( 9) + a(17) + a(25) = s1/2
a( 2) + a(10) + a(18) + a(26) = s1/2
a( 3) + a(11) + a(19) + a(27) = s1/2
a( 4) + a(12) + a(20) + a(28) = s1/2

a( 5) + a( 6) + a( 7) + a( 8) = s1/2
a(13) + a(14) + a(15) + a(16) = s1/2
a(21) + a(22) + a(23) + a(24) = s1/2
a(29) + a(30) + a(31) + a(32) = s1/2

a( 5) + a(13) + a(21) + a(29) = s1/2
a( 6) + a(14) + a(22) + a(30) = s1/2
a( 7) + a(15) + a(23) + a(31) = s1/2
a( 8) + a(16) + a(24) + a(32) = s1/2

a(33) + a(34) + a(35) + a(36) = s1/2
a(41) + a(42) + a(43) + a(44) = s1/2
a(49) + a(50) + a(51) + a(52) = s1/2
a(57) + a(58) + a(59) + a(60) = s1/2

a(33) + a(41) + a(49) + a(57) = s1/2
a(34) + a(42) + a(50) + a(58) = s1/2
a(35) + a(43) + a(51) + a(59) = s1/2
a(36) + a(44) + a(52) + a(60) = s1/2

a(37) + a(38) + a(39) + a(40) = s1/2
a(45) + a(46) + a(47) + a(48) = s1/2
a(53) + a(54) + a(55) + a(56) = s1/2
a(61) + a(62) + a(63) + a(64) = s1/2

a(37) + a(45) + a(53) + a(61) = s1/2
a(38) + a(46) + a(54) + a(62) = s1/2
a(39) + a(47) + a(55) + a(63) = s1/2
a(40) + a(48) + a(56) + a(64) = s1/2

The numbers of the long diagonals and all the broken diagonals parallel to it sum to the Magic Constant:

a( 1) + a(10) + a(19) + a(28) + a(37) + a(46) + a(55) + a(64) = s1
a( 2) + a(11) + a(20) + a(29) + a(38) + a(47) + a(56) + a(57) = s1
a( 3) + a(12) + a(21) + a(30) + a(39) + a(48) + a(49) + a(58) = s1
a( 4) + a(13) + a(22) + a(31) + a(40) + a(41) + a(50) + a(59) = s1
a( 5) + a(14) + a(23) + a(32) + a(33) + a(42) + a(51) + a(60) = s1
a( 6) + a(15) + a(24) + a(25) + a(34) + a(43) + a(52) + a(61) = s1
a( 7) + a(16) + a(17) + a(26) + a(35) + a(44) + a(53) + a(62) = s1
a( 8) + a( 9) + a(18) + a(27) + a(36) + a(45) + a(54) + a(63) = s1

a( 8) + a(15) + a(22) + a(29) + a(36) + a(43) + a(50) + a(57) = s1
a( 7) + a(14) + a(21) + a(28) + a(35) + a(42) + a(49) + a(64) = s1
a( 6) + a(13) + a(20) + a(27) + a(34) + a(41) + a(56) + a(63) = s1
a( 5) + a(12) + a(19) + a(26) + a(33) + a(48) + a(55) + a(62) = s1
a( 4) + a(11) + a(18) + a(25) + a(40) + a(47) + a(54) + a(61) = s1
a( 3) + a(10) + a(17) + a(32) + a(39) + a(46) + a(53) + a(60) = s1
a( 2) + a( 9) + a(24) + a(31) + a(38) + a(45) + a(52) + a(59) = s1
a( 1) + a(16) + a(23) + a(30) + a(37) + a(44) + a(51) + a(58) = s1

Every 2 × 2 sub square sums to half the Magic Constant:

a(i) + a(i+1) + a(i+ 8) + a(i+ 9) = s1/2 with 1 =< i < 56 and i ≠ 8*n for n = 1, 2 ... 7

a(i) + a(i+1) + a(i+ 8) + a(i- 7) = s1/2 with i = 8*n for n = 1, 2 ... 7

a(i) + a(i+1) + a(i+56) + a(i+57) = s1/2 with i = 1, 2 ... 7

a(1) + a(8)   + a(57)   + a(64)   = s1/2

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

          →   →
     A4 * a = s

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

a(61) =  0.5 * s1 - a(62) - a(63) - a(64)
a(57) =  0.5 * s1 - a(58) - a(59) - a(60)
a(55) =  0.5 * s1 - a(56) - a(63) - a(64)
a(54) =             a(56) - a(62) + a(64)
a(53) =           - a(56) + a(62) + a(63)
a(52) =             a(56) - a(60) + a(64)
a(51) =  0.5 * s1 - a(56) - a(59) - a(64)
a(50) =             a(56) - a(58) + a(64)
a(49) =           - a(56) + a(58) + a(59) + a(60) - a(64)
a(47) =           - a(48) + a(63) + a(64)
a(46) =             a(48) + a(62) - a(64)
a(45) =  0.5 * s1 - a(48) - a(62) - a(63)
a(44) =             a(48) + a(60) - a(64)
a(43) =           - a(48) + a(59) + a(64)
a(42) =             a(48) + a(58) - a(64)
a(41) =  0.5 * s1 - a(48) - a(58) - a(59) - a(60) + a(64)
a(40) =  0.5 * s1 - a(48) - a(56) - a(64)
a(39) =             a(48) + a(56) - a(63)
a(38) =  0.5 * s1 - a(48) - a(56) - a(62)
a(37) = -0.5 * s1 + a(48) + a(56) + a(62) + a(63) + a(64)
a(36) =  0.5 * s1 - a(48) - a(56) - a(60)
a(35) =             a(48) + a(56) - a(59)
a(34) =  0.5 * s1 - a(48) - a(56) - a(58)
a(33) = -0.5 * s1 + a(48) + a(56) + a(58) + a(59) + a(60)
a(31) =           - a(32) + a(63) + a(64)
a(30) =             a(32) + a(62) - a(64)
a(29) =  0.5 * s1 - a(32) - a(62) - a(63)
a(28) =             a(32) + a(60) - a(64)
a(27) =           - a(32) + a(59) + a(64)
a(26) =             a(32) + a(58) - a(64)
a(25) =  0.5 * s1 - a(32) - a(58) - a(59) - a(60) + a(64)
a(23) =  0.5 * s1 - a(24) - a(63) - a(64)
a(22) =             a(24) - a(62) + a(64)
a(21) =           - a(24) + a(62) + a(63)
a(20) =             a(24) - a(60) + a(64)
a(19) =  0.5 * s1 - a(24) - a(59) - a(64)
a(18) =             a(24) - a(58) + a(64)
a(17) =           - a(24) + a(58) + a(59) + a(60) - a(64)
a(16) =  0.5 * s1 - a(32) - a(48) - a(58) - a(60) - a(62) + 2 * a(64)
a(15) = -0.5 * s1 + a(32) + a(48) + a(58) + a(60) + a(62) + a(63) - a(64)
a(14) =  0.5 * s1 - a(32) - a(48) - a(58) - a(60) + a(64)
a(13) =             a(32) + a(48) + a(58) + a(60) - a(63) - 2 * a(64)
a(12) =  0.5 * s1 - a(32) - a(48) - a(58) - a(62) + a(64)
a(11) = -0.5 * s1 + a(32) + a(48) + a(58) + a(59) + a(60) + a(62) - a(64)
a(10) =  0.5 * s1 - a(32) - a(48) - a(60) - a(62) + a(64)
a( 9) =             a(32) + a(48) - a(59) + a(62) - a(64)
a( 8) =           - a(24) + a(48) + a(58) + a(60) + a(62) - 2 * a(64)
a( 7) =  0.5 * s1 + a(24) - a(48) - a(58) - a(60) - a(62) - a(63) + a(64)
a( 6) =           - a(24) + a(48) + a(58) + a(60) - a(64)
a( 5) =             a(24) - a(48) - a(58) - a(60) + a(63) + 2 * a(64)
a( 4) =           - a(24) + a(48) + a(58) + a(62) - a(64)
a( 3) =  0.5 * s1 + a(24) - a(48) - a(58) - a(59) - a(60) - a(62) + a(64)
a( 2) =           - a(24) + a(48) + a(60) + a(62) - a(64)
a( 1) =             a(24) - a(48) + a(59) - a(62) + a(64)

The solutions can be obtained by guessing 24, 32, 48, 56, 58 ... 60, 62 ... 64 and filling out these guesses in the abovementioned equations.

For distinct integers also following inequalities should be applied:

0 < a(i) =< 64        for i = 1, 2, ... 23, 25, ... 31, 33, ... 47, 49, ... 55, 57 and 61
a(i) ≠ a(j)           for i ≠ j

With a(64), a(63) and a(62) constant, an optimized guessing routine (MgcSqr8g), produced 1536 Pan Magic Squares within 393 seconds, which are shown in Attachment 8.5.1.

The Pan Magic Squares composed out of Pan Magic Sub Squares, as discussed in next section, are highlighted in red.

8.5.2 Analytic Solution, Pan Magic Squares composed of Pan Magic Sub Squares,
      Franklin Property 5

If we combine the equations for Pan Magic Squares composed of Pan Magic Sub Squares as shown in section 8.2.1 with property 5 of the Franklin Squares, as mentioned in section 8.4.1, this will result in following set of linear equations:

The four sub squares are Pan Magic:

a(25) =  0.5  * s1 - a(26) - a(27) - a(28)
a(19) =  0.5  * s1 - a(20) - a(27) - a(28)
a(18) =              a(20) - a(26) + a(28)
a(17) =            - a(20) + a(26) + a(27)
a(12) =  0.25 * s1 - a(26)
a(11) = -0.25 * s1 + a(26) + a(27) + a(28)
a(10) =  0.25 * s1 - a(28)
a( 9) =  0.25 * s1 - a(27)
a( 4) =  0.25 * s1 - a(20) + a(26) - a(28)
a( 3) =  0.25 * s1 + a(20) - a(26) - a(27)
a( 2) =  0.25 * s1 - a(20)
a( 1) = -0.25 * s1 + a(20) + a(27) + a(28)

a(29) =  0.5  * s1 - a(30) - a(31) - a(32)
a(23) =  0.5  * s1 - a(24) - a(31) - a(32)
a(22) =              a(24) - a(30) + a(32)
a(21) =            - a(24) + a(30) + a(31)
a(16) =  0.25 * s1 - a(30)
a(15) = -0.25 * s1 + a(30) + a(31) + a(32)
a(14) =  0.25 * s1 - a(32)
a(13) =  0.25 * s1 - a(31)
a( 8) =  0.25 * s1 - a(24) + a(30) - a(32)
a( 7) =  0.25 * s1 + a(24) - a(30) - a(31)
a( 6) =  0.25 * s1 - a(24)
a( 5) = -0.25 * s1 + a(24) + a(31) + a(32)

a(57) =  0.5  * s1 - a(58) - a(59) - a(60)
a(51) =  0.5  * s1 - a(52) - a(59) - a(60)
a(50) =              a(52) - a(58) + a(60)
a(49) =            - a(52) + a(58) + a(59)
a(44) =  0.25 * s1 - a(58)
a(43) = -0.25 * s1 + a(58) + a(59) + a(60)
a(42) =  0.25 * s1 - a(60)
a(41) =  0.25 * s1 - a(59)
a(36) =  0.25 * s1 - a(52) + a(58) - a(60)
a(35) =  0.25 * s1 + a(52) - a(58) - a(59)
a(34) =  0.25 * s1 - a(52)
a(33) = -0.25 * s1 + a(52) + a(59) + a(60)

a(61) =  0.5  * s1 - a(62) - a(63) - a(64)
a(55) =  0.5  * s1 - a(56) - a(63) - a(64)
a(54) =              a(56) - a(62) + a(64)
a(53) =            - a(56) + a(62) + a(63)
a(48) =  0.25 * s1 - a(62)
a(47) = -0.25 * s1 + a(62) + a(63) + a(64)
a(46) =  0.25 * s1 - a(64)
a(45) =  0.25 * s1 - a(63)
a(40) =  0.25 * s1 - a(56) + a(62) - a(64)
a(39) =  0.25 * s1 + a(56) - a(62) - a(63)
a(38) =  0.25 * s1 - a(56)
a(37) = -0.25 * s1 + a(56) + a(63) + a(64)

This includes that the numbers of main diagonals sum to the Magic Constant. Following equations ensure that also the numbers of all the broken diagonals parallel to it sum to the Magic Constant as well:

a(2) + a(11) + a(20) + a(29) + a(38) + a(47) + a(56) + a(57) = s1
a(3) + a(12) + a(21) + a(30) + a(39) + a(48) + a(49) + a(58) = s1
a(4) + a(13) + a(22) + a(31) + a(40) + a(41) + a(50) + a(59) = s1
a(5) + a(14) + a(23) + a(32) + a(33) + a(42) + a(51) + a(60) = s1
a(6) + a(15) + a(24) + a(25) + a(34) + a(43) + a(52) + a(61) = s1
a(7) + a(16) + a(17) + a(26) + a(35) + a(44) + a(53) + a(62) = s1
a(8) + a( 9) + a(18) + a(27) + a(36) + a(45) + a(54) + a(63) = s1

a(7) + a(14) + a(21) + a(28) + a(35) + a(42) + a(49) + a(64) = s1
a(6) + a(13) + a(20) + a(27) + a(34) + a(41) + a(56) + a(63) = s1
a(5) + a(12) + a(19) + a(26) + a(33) + a(48) + a(55) + a(62) = s1
a(4) + a(11) + a(18) + a(25) + a(40) + a(47) + a(54) + a(61) = s1
a(3) + a(10) + a(17) + a(32) + a(39) + a(46) + a(53) + a(60) = s1
a(2) + a( 9) + a(24) + a(31) + a(38) + a(45) + a(52) + a(59) = s1
a(1) + a(16) + a(23) + a(30) + a(37) + a(44) + a(51) + a(58) = s1

Every 2 × 2 sub square sums to half the Magic Constant:

a(i) + a(i+1) + a(i+ 8) + a(i+ 9) = s1/2 with 1 =< i < 56 and i ≠ 8*n for n = 1, 2 ... 7

a(i) + a(i+1) + a(i+ 8) + a(i- 7) = s1/2 with i = 8*n for n = 1, 2 ... 7

a(i) + a(i+1) + a(i+56) + a(i+57) = s1/2 with i = 1, 2 ... 7

a(1) + a(8)   + a(57)   + a(64)   = s1/2

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

          →   →
     A5 * a = s

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

a(61) =  0.50 * s1 - a(62) - a(63) - a(64)
a(58) =            - a(60) + a(62) + a(64)
a(57) =  0.50 * s1 - a(59) - a(62) - a(64)
a(55) =  0.50 * s1 - a(56) - a(63) - a(64)
a(54) =              a(56) - a(62) + a(64)
a(53) =            - a(56) + a(62) + a(63)
a(52) =              a(56) - a(60) + a(64)
a(51) =  0.50 * s1 - a(56) - a(59) - a(64)
a(50) =              a(56) + a(60) - a(62)
a(49) =            - a(56) + a(59) + a(62)
a(48) =  0.25 * s1 - a(62)
a(47) = -0.25 * s1 + a(62) + a(63) + a(64)
a(46) =  0.25 * s1 - a(64)
a(45) =  0.25 * s1 - a(63)
a(44) =  0.25 * s1 + a(60) - a(62) - a(64)
a(43) = -0.25 * s1 + a(59) + a(62) + a(64)
a(42) =  0.25 * s1 - a(60)
a(41) =  0.25 * s1 - a(59)
a(40) =  0.25 * s1 - a(56) + a(62) - a(64)
a(39) =  0.25 * s1 + a(56) - a(62) - a(63)
a(38) =  0.25 * s1 - a(56)
a(37) = -0.25 * s1 + a(56) + a(63) + a(64)
a(36) =  0.25 * s1 - a(56) - a(60) + a(62)
a(35) =  0.25 * s1 + a(56) - a(59) - a(62)
a(34) =  0.25 * s1 - a(56) + a(60) - a(64)
a(33) = -0.25 * s1 + a(56) + a(59) + a(64)
a(31) =            - a(32) + a(63) + a(64)
a(30) =              a(32) + a(62) - a(64)
a(29) =  0.50 * s1 - a(32) - a(62) - a(63)
a(28) =              a(32) + a(60) - a(64)
a(27) =            - a(32) + a(59) + a(64)
a(26) =              a(32) - a(60) + a(62)
a(25) =  0.50 * s1 - a(32) - a(59) - a(62)
a(23) =  0.50 * s1 - a(24) - a(63) - a(64)
a(22) =              a(24) - a(62) + a(64)
a(21) =            - a(24) + a(62) + a(63)
a(20) =              a(24) - a(60) + a(64)
a(19) =  0.50 * s1 - a(24) - a(59) - a(64)
a(18) =              a(24) + a(60) - a(62)
a(17) =            - a(24) + a(59) + a(62)
a(16) =  0.25 * s1 - a(32) - a(62) + a(64)
a(15) = -0.25 * s1 + a(32) + a(62) + a(63)
a(14) =  0.25 * s1 - a(32)
a(13) =  0.25 * s1 + a(32) - a(63) - a(64)
a(12) =  0.25 * s1 - a(32) + a(60) - a(62)
a(11) = -0.25 * s1 + a(32) + a(59) + a(62)
a(10) =  0.25 * s1 - a(32) - a(60) + a(64)
a( 9) =  0.25 * s1 + a(32) - a(59) - a(64)
a( 8) =  0.25 * s1 - a(24) + a(62) - a(64)
a( 7) =  0.25 * s1 + a(24) - a(62) - a(63)
a( 6) =  0.25 * s1 - a(24)
a( 5) = -0.25 * s1 + a(24) + a(63) + a(64)
a( 4) =  0.25 * s1 - a(24) - a(60) + a(62)
a( 3) =  0.25 * s1 + a(24) - a(59) - a(62)
a( 2) =  0.25 * s1 - a(24) + a(60) - a(64)
a( 1) = -0.25 * s1 + a(24) + a(59) + a(64)

The solutions can be obtained by guessing 24, 32, 56, 59, 60, 62 ... 64 and filling out these guesses in the abovementioned equations.

For distinct integers also following inequalities should be applied:

0 < a(i) =< 64        for i = 1, 2, ... 23, 25, ... 31, 33, ... 55, 57, 58 and 61
a(i) ≠ a(j)           for i ≠ j

With a(64), a(63) and a(62) constant, an optimized guessing routine (MgcSqr8h), produced 384 Pan Magic Squares within 37,9 seconds (ref. Attachment 8.5.2).

8.5.3 Analytic Solution, Pan Magic Squares composed of Pan Magic Sub Squares,
      Franklin Properties 3 and 5

When the equations for the Franklin Bent Diagonals, as described in section 8.4.2, are combined with the equations
of section 8.5.2 above , the resulting equations can be written in matrix representation as:

          →   →
     A6 * a = s

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

a(61) =  0.50 * s1 - a(62) - a(63) - a(64)
a(58) =            - a(60) + a(62) + a(64)
a(57) =  0.50 * s1 - a(59) - a(62) - a(64)
a(55) =  0.50 * s1 - a(56) - a(63) - a(64)
a(54) =              a(56) - a(62) + a(64)
a(53) =            - a(56) + a(62) + a(63)
a(52) =              a(56) - a(60) + a(64)
a(51) =  0.50 * s1 - a(56) - a(59) - a(64)
a(50) =              a(56) + a(60) - a(62)
a(49) =            - a(56) + a(59) + a(62)
a(48) =  0.25 * s1 - a(62)
a(47) = -0.25 * s1 + a(62) + a(63) + a(64)
a(46) =  0.25 * s1 - a(64)
a(45) =  0.25 * s1 - a(63)
a(44) =  0.25 * s1 + a(60) - a(62) - a(64)
a(43) = -0.25 * s1 + a(59) + a(62) + a(64)
a(42) =  0.25 * s1 - a(60)
a(41) =  0.25 * s1 - a(59)
a(40) =  0.25 * s1 - a(56) + a(62) - a(64)
a(39) =  0.25 * s1 + a(56) - a(62) - a(63)
a(38) =  0.25 * s1 - a(56)
a(37) = -0.25 * s1 + a(56) + a(63) + a(64)
a(36) =  0.25 * s1 - a(56) - a(60) + a(62)
a(35) =  0.25 * s1 + a(56) - a(59) - a(62)
a(34) =  0.25 * s1 - a(56) + a(60) - a(64)
a(33) = -0.25 * s1 + a(56) + a(59) + a(64)
a(31) =            - a(32) + a(63) + a(64)
a(30) =              a(32) + a(62) - a(64)
a(29) =  0.50 * s1 - a(32) - a(62) - a(63)
a(28) =              a(32) + a(60) - a(64)
a(27) =            - a(32) + a(59) + a(64)
a(26) =              a(32) - a(60) + a(62)
a(25) =  0.50 * s1 - a(32) - a(59) - a(62)
a(23) =  0.50 * s1 - a(24) - a(63) - a(64)
a(22) =              a(24) - a(62) + a(64)
a(21) =            - a(24) + a(62) + a(63)
a(20) =              a(24) - a(60) + a(64)
a(19) =  0.50 * s1 - a(24) - a(59) - a(64)
a(18) =              a(24) + a(60) - a(62)
a(17) =            - a(24) + a(59) + a(62)
a(16) =  0.25 * s1 - a(32) - a(62) + a(64)
a(15) = -0.25 * s1 + a(32) + a(62) + a(63)
a(14) =  0.25 * s1 - a(32)
a(13) =  0.25 * s1 + a(32) - a(63) - a(64)
a(12) =  0.25 * s1 - a(32) + a(60) - a(62)
a(11) = -0.25 * s1 + a(32) + a(59) + a(62)
a(10) =  0.25 * s1 - a(32) - a(60) + a(64)
a( 9) =  0.25 * s1 + a(32) - a(59) - a(64)
a( 8) =  0.25 * s1 - a(24) + a(62) - a(64)
a( 7) =  0.25 * s1 + a(24) - a(62) - a(63)
a( 6) =  0.25 * s1 - a(24)
a( 5) = -0.25 * s1 + a(24) + a(63) + a(64)
a( 4) =  0.25 * s1 - a(24) - a(60) + a(62)
a( 3) =  0.25 * s1 + a(24) - a(59) - a(62)
a( 2) =  0.25 * s1 - a(24) + a(60) - a(64)
a( 1) = -0.25 * s1 + a(24) + a(59) + a(64)

which define Pan Magic Squares composed of Pan Magic Sub Squares, with Franklin properties 3 and 5. It should be noted that these equations are identical to the equations found in section 8.5.2 above.

Combining the equations of the original Franklin squares (section 8.4.2) with the equations for the long and broken diagonals (section 8.5.1), ensuring the Pan Magic property, will also result in the same set of linear equations.

With a(64) = 24, the corresponding guessing routine (MgcSqr8h) counted 46080 solutions within 1 hour (printing routine disabled).

The Pan Magic Squares analysed in this section are also known as the Most Perfect Franklin Pan Magic Squares. The total number of Most Perfect Franklin Pan Magic Squares is 64 * 46080 = 2949120.

8.5.4 Analytic Solution, Pan Magic Squares composed of Pan Magic Sub Squares,
      Franklin Properties 3 and 5, Barink Restrictions

When the equations found in section 8.5.3 above, are combined with the Barink Restrictions as discussed in section 8.4.4 (any 4 consecutive numbers starting on any odd place in a row or column sum to one half of the Magic Constant), the resulting equations can be written in the matrix representation as:

          →   →
     A7 * a = s

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

a(61) =  0.50 * s1 - a(62) - a(63) - a(64)
a(59) =            - a(60) + a(63) + a(64)
a(58) =            - a(60) + a(62) + a(64)
a(57) =  0.50 * s1 + a(60) - a(62) - a(63) - 2 * a(64)
a(55) =  0.50 * s1 - a(56) - a(63) - a(64)
a(54) =              a(56) - a(62) + a(64)
a(53) =            - a(56) + a(62) + a(63)
a(52) =              a(56) - a(60) + a(64)
a(51) =  0.50 * s1 - a(56) + a(60) - a(63) - 2 * a(64)
a(50) =              a(56) + a(60) - a(62)
a(49) =            - a(56) - a(60) + a(62) + a(63) + a(64)
a(48) =  0.25 * s1 - a(62)
a(47) = -0.25 * s1 + a(62) + a(63) + a(64)
a(46) =  0.25 * s1 - a(64)
a(45) =  0.25 * s1 - a(63)
a(44) =  0.25 * s1 + a(60) - a(62) - a(64)
a(43) = -0.25 * s1 - a(60) + a(62) + a(63) + 2 * a(64)
a(42) =  0.25 * s1 - a(60)
a(41) =  0.25 * s1 + a(60) - a(63) - a(64)
a(40) =  0.25 * s1 - a(56) + a(62) - a(64)
a(39) =  0.25 * s1 + a(56) - a(62) - a(63)
a(38) =  0.25 * s1 - a(56)
a(37) = -0.25 * s1 + a(56) + a(63) + a(64)
a(36) =  0.25 * s1 - a(56) - a(60) + a(62)
a(35) =  0.25 * s1 + a(56) + a(60) - a(62) - a(63) - a(64)
a(34) =  0.25 * s1 - a(56) + a(60) - a(64)
a(33) = -0.25 * s1 + a(56) - a(60) + a(63) + 2 * a(64)
a(31) =            - a(32) + a(63) + a(64)
a(30) =              a(32) + a(62) - a(64)
a(29) =  0.50 * s1 - a(32) - a(62) - a(63)
a(28) =              a(32) + a(60) - a(64)
a(27) =            - a(32) - a(60) + a(63) + 2 * a(64)
a(26) =              a(32) - a(60) + a(62)
a(25) =  0.50 * s1 - a(32) + a(60) - a(62) - a(63) - a(64)
a(24) =            - a(32) + a(56) + a(64)
a(23) =  0.50 * s1 + a(32) - a(56) - a(63) - 2 * a(64)
a(22) =            - a(32) + a(56) - a(62) + 2 * a(64)
a(21) =              a(32) - a(56) + a(62) + a(63) - a(64)
a(20) =            - a(32) + a(56) - a(60) + 2 * a(64)
a(19) =  0.50 * s1 + a(32) - a(56) + a(60) - a(63) - 3 * a(64)
a(18) =            - a(32) + a(56) + a(60) - a(62) + a(64)
a(17) =              a(32) - a(56) - a(60) + a(62) + a(63)
a(16) =  0.25 * s1 - a(32) - a(62) + a(64)
a(15) = -0.25 * s1 + a(32) + a(62) + a(63)
a(14) =  0.25 * s1 - a(32)
a(13) =  0.25 * s1 + a(32) - a(63) - a(64)
a(12) =  0.25 * s1 - a(32) + a(60) - a(62)
a(11) = -0.25 * s1 + a(32) - a(60) + a(62) + a(63) + a(64)
a(10) =  0.25 * s1 - a(32) - a(60) + a(64)
a( 9) =  0.25 * s1 + a(32) + a(60) - a(63) - 2 * a(64)
a( 8) =  0.25 * s1 + a(32) - a(56) + a(62) - 2 * a(64)
a( 7) =  0.25 * s1 - a(32) + a(56) - a(62) - a(63) + a(64)
a( 6) =  0.25 * s1 + a(32) - a(56) - a(64)
a( 5) = -0.25 * s1 - a(32) + a(56) + a(63) + 2 * a(64)
a( 4) =  0.25 * s1 + a(32) - a(56) - a(60) + a(62) - a(64)
a( 3) =  0.25 * s1 - a(32) + a(56) + a(60) - a(62) - a(63)
a( 2) =  0.25 * s1 + a(32) - a(56) + a(60) - 2 * a(64)
a( 1) = -0.25 * s1 - a(32) + a(56) - a(60) + a(63) + 3 * a(64)

The solutions can be obtained by guessing 32, 56, 60, 62 ... 64 and filling out these guesses in the abovementioned equations.

For distinct integers also following inequalities should be applied:

0 < a(i) =< 64        for i = 1, 2, ... 31, 33, ... 55, 57, 58, 59 and 61
a(i) ≠ a(j)           for i ≠ j

With a(64) constant, an optimized guessing routine (MgcSqr8g4), produced 720 Pan Magic Squares within 96,8 seconds, which are shown in Attachment 8.5.4.

The total number of Pan Magic Squares, composed of Pan Magic Sub Squares with Franklin properties 3 and 5 under the Barink Restrictions is 46080.

8.5.5 Most Perfect Pan Magic Squares (Compact and Complete)

A Most Perfect Magic Square of order 8 is defined as a Pan Magic Square containing the numbers 1 to 64 with two additional properties:

1. Compact : Each 2 × 2 sub square sums to 130 (Franklin Property 5);
2. Complete: All pairs of integers distant n/2 along a (main) diagonal sum to 65.

The defining properties mentioned above result, after deduction, in following set of linear equations: 

a(59) =  0.50 * s1 - a(60) - a(63) - a(64)
a(58) =              a(60) - a(62) + a(64)
a(57) =  0.50 * s1 - a(60) - a(61) - a(64)
a(55) =  0.50 * s1 - a(56) - a(63) - a(64)
a(54) =              a(56) - a(62) + a(64)
a(53) =  0.50 * s1 - a(56) - a(61) - a(64)
a(52) =              a(56) - a(60) + a(64)
a(51) =            - a(56) + a(60) + a(63)
a(50) =              a(56) - a(60) + a(62)
a(49) =            - a(56) + a(60) + a(61)
a(47) =            - a(48) + a(63) + a(64)
a(46) =              a(48) + a(62) - a(64)
a(45) =            - a(48) + a(61) + a(64)
a(44) =              a(48) + a(60) - a(64)
a(43) =  0.50 * s1 - a(48) - a(60) - a(63)
a(42) =              a(48) + a(60) - a(62)
a(41) =  0.50 * s1 - a(48) - a(60) - a(61)
a(39) =  0.50 * s1 - a(40) - a(63) - a(64)
a(38) =              a(40) - a(62) + a(64)
a(37) =  0.50 * s1 - a(40) - a(61) - a(64)
a(36) =              a(40) - a(60) + a(64)
a(35) =            - a(40) + a(60) + a(63)
a(34) =              a(40) - a(60) + a(62)
a(33) =            - a(40) + a(60) + a(61)
a(32) =  0.25 * s1 - a(60)
a(31) = -0.25 * s1 + a(60) + a(63) + a(64)
a(30) =  0.25 * s1 - a(60) + a(62) - a(64)
a(29) = -0.25 * s1 + a(60) + a(61) + a(64)
a(28) =  0.25 * s1 - a(64)
a(27) =  0.25 * s1 - a(63)
a(26) =  0.25 * s1 - a(62)
a(25) =  0.25 * s1 - a(61)
a(24) =  0.25 * s1 - a(56) + a(60) - a(64)
a(23) =  0.25 * s1 + a(56) - a(60) - a(63)
a(22) =  0.25 * s1 - a(56) + a(60) - a(62)
a(21) =  0.25 * s1 + a(56) - a(60) - a(61)
a(20) =  0.25 * s1 - a(56)
a(19) = -0.25 * s1 + a(56) + a(63) + a(64)
a(18) =  0.25 * s1 - a(56) + a(62) - a(64)
a(17) = -0.25 * s1 + a(56) + a(61) + a(64)
a(16) =  0.25 * s1 - a(48) - a(60) + a(64)
a(15) = -0.25 * s1 + a(48) + a(60) + a(63)
a(14) =  0.25 * s1 - a(48) - a(60) + a(62)
a(13) = -0.25 * s1 + a(48) + a(60) + a(61)
a(12) =  0.25 * s1 - a(48)
a(11) =  0.25 * s1 + a(48) - a(63) - a(64)
a(10) =  0.25 * s1 - a(48) - a(62) + a(64)
a( 9) =  0.25 * s1 + a(48) - a(61) - a(64)
a( 8) =  0.25 * s1 - a(40) + a(60) - a(64)
a( 7) =  0.25 * s1 + a(40) - a(60) - a(63)
a( 6) =  0.25 * s1 - a(40) + a(60) - a(62)
a( 5) =  0.25 * s1 + a(40) - a(60) - a(61)
a( 4) =  0.25 * s1 - a(40)
a( 3) = -0.25 * s1 + a(40) + a(63) + a(64)
a( 2) =  0.25 * s1 - a(40) + a(62) - a(64)
a( 1) = -0.25 * s1 + a(40) + a(61) + a(64)

With a(64) = 28, an optimized guessing routine (MgcSqr8g5), produced 46080 Most Perfect Pan Magic Squares within 1 hour, of which 384 are shown in Attachment 8.5.5.

The total number of Most Perfect Pan Magic Squares is 64 * 46080 = 2949120.

8.5.6 Additional Properties

For Most Perfect Pan Magic Squares as defined in section 8.5.5 above the following additional properties can be recognized:

  1. The corner points of all 4 x 4 sub squares (25 ea) sum to 130;
  2. The corner points of all 5 x 5 sub squares (16 ea) sum to 130;
  3. The corner points of all 6 x 6 sub squares ( 9 ea) sum to 130.

As mentioned in section 8.4.3 a variety of patterns will sum to the Magic Constant as a consequence of the fifth property (every 2 × 2 sub square sums to half the Magic Constant). More samples of these patterns are shown in Attachment 8.4.3.

8.5.7 Spreadsheet Solutions

The linear equations deducted above, have been applied in following Excel Spread Sheets:

  • CnstrSngl8h,  Pan Magic Squares composed of Pan Magic Sub Squares, Franklin Property 5

  • CnstrSngl8g2, Pan Magic Squares composed of Pan Magic Sub Squares, Franklin Properties 3 and 5

  • CnstrSngl8g4, Pan Magic Squares composed of Pan Magic Sub Squares, Franklin Properties 3 and 5,
                  Barink Restrictions

  • CnstrSngl8g5, Most Perfect Pan Magic Squares, Compact and Complete

Only the red figures have to be “guessed” to construct one of the applicable Pan Magic Squares of the 8th order (wrong solutions are obvious).

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