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10.2   Concentric and Eccentric Magic Squares

10.2.1 Concentric Magic Squares

In general an even Concentric Magic Square consists of 2 x 2 cells, around which borders can be constructed again and again.

A 10th order Concentric Magic Square consists of an embedded Magic Square of the 8th order with an embedded Magic Square of the 6th order with an embedded (Pan) Magic Square of the 4th order.

 a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a13 a14 a15 a16 a17 a18 a19 a20 a21 a22 a23 a24 a25 a26 a27 a28 a29 a30 a31 a32 a33 a34 a35 a36 a37 a38 a39 a40 a41 a42 a43 a44 a45 a46 a47 a48 a49 a50 a51 a52 a53 a54 a55 a56 a57 a58 a59 a60 a61 a62 a63 a64 a65 a66 a67 a68 a69 a70 a71 a72 a73 a74 a75 a76 a77 a78 a79 a80 a81 a82 a83 a84 a85 a86 a87 a88 a89 a90 a91 a92 a93 a94 a95 a96 a97 a98 a99 a100

With the inner 4 x 4 square pan magic, the embedded Magic Squares can be described by following linear equations:

a(82) =  404 - a(83) - a(84) - a(85) - a(86) - a(87) - a(88) - a(89)
a(73) =  303 - a(74) - a(75) - a(76) - a(77) - a(78)
a(72) =  101 - a(79)
a(64) =  202 - a(65) - a(66) - a(67)
a(63) =  101 - a(68)
a(62) =  101 - a(69)
a(56) =  202 - a(57) - a(66) - a(67)
a(55) =        a(57) - a(65) + a(67)
a(54) =      - a(57) + a(65) + a(66)
a(53) =  101 - a(58)
a(52) =  101 - a(59)
a(47) =  101 - a(65)
a(46) = -101 + a(65) + a(66) + a(67)
a(45) =  101 - a(67)
a(44) =  101 - a(66)
a(43) =  101 - a(48)
a(42) =  101 - a(49)
a(38) =  202 - a(48) - a(58) - a(68) + a(73) - a(78)
a(37) =  101 - a(57) + a(65) - a(67)
a(36) =  101 + a(57) - a(65) - a(66)
a(35) =  101 - a(57)
a(34) = -101 + a(57) + a(66) + a(67)
a(33) = -404 + a(48) + a(58) + a(68) + a(74) + a(75) + a(76) + a(77) + 2*a(78)
a(32) =  101 - a(39)
a(29) =  303 - a(39) - a(49) - a(59) - a(69) - a(79) + a(82) - a(89)
a(28) = -202 + a(74) + a(75) + a(76) + a(77) + a(78)
a(27) =  101 - a(77)
a(26) =  101 - a(76)
a(25) =  101 - a(75)
a(24) =  101 - a(74)
a(23) =  101 - a(78)
a(22) =  101 - a(29)
a(19) =  101 - a(82)
a(18) =  101 - a(88)
a(17) =  101 - a(87)
a(16) =  101 - a(86)
a(15) =  101 - a(85)
a(14) =  101 - a(84)
a(13) =  101 - a(83)
a(12) =  101 - a(89)

which can be completed with the equations describing the outer border, which results in following linear equations:

a(91) =  505 - a(92) - a(93) - a(94) - a(95) - a(96) - a(97) - a(98) - a(99) - a(100)
a(82) =  404 - a(83) - a(84) - a(85) - a(86) - a(87) - a(88) - a(89)
a(81) =  101 - a(90)
a(73) =  303 - a(74) - a(75) - a(76) - a(77) - a(78)
a(72) =  101 - a(79)
a(71) =  101 - a(80)
a(64) =  202 - a(65) - a(66) - a(67)
a(63) =  101 - a(68)
a(62) =  101 - a(69)
a(61) =  101 - a(70)
a(56) =  202 - a(57) - a(66) - a(67)
a(55) =        a(57) - a(65) + a(67)
a(54) =      - a(57) + a(65) + a(66)
a(53) =  101 - a(58)
a(52) =  101 - a(59)
a(51) =  101 - a(60)
a(47) =  101 - a(65)
a(46) = -101 + a(65) + a(66) + a(67)
a(45) =  101 - a(67)
a(44) =  101 - a(66)
a(43) =  101 - a(48)
a(42) =  101 - a(49)
a(41) =  101 - a(50)
a(38) =  202 - a(48) - a(58) - a(68) + a(73) - a(78)
a(37) =  101 - a(57) + a(65) - a(67)
a(36) =  101 + a(57) - a(65) - a(66)
a(35) =  101 - a(57)
a(34) = -101 + a(57) + a(66) + a(67)
a(33) = -404 + a(48) + a(58) + a(68) + a(74) + a(75) + a(76) + a(77) + 2*a(78)
a(32) =  101 - a(39)
a(31) =  101 - a(40)
a(29) =  303 - a(39) - a(49) - a(59) - a(69) - a(79) + a(82) - a(89)
a(28) = -202 + a(74) + a(75) + a(76) + a(77) + a(78)
a(27) =  101 - a(77)
a(26) =  101 - a(76)
a(25) =  101 - a(75)
a(24) =  101 - a(74)
a(23) =  101 - a(78)
a(22) =  101 - a(29)
a(21) =  101 - a(30)
a(20) =  404 - a(30) - a(40) - a(50) - a(60) - a(70) - a(80) - a(90) + a(91) - a(100)
a(19) =  101 - a(82)
a(18) =  101 - a(88)
a(17) =  101 - a(87)
a(16) =  101 - a(86)
a(15) =  101 - a(85)
a(14) =  101 - a(84)
a(13) =  101 - a(83)
a(12) =  101 - a(89)
a(11) =  101 - a(20)
a(10) =  101 - a(91)
a( 9) =  101 - a(99)
a( 8) =  101 - a(98)
a( 7) =  101 - a(97)
a( 6) =  101 - a(96)
a( 5) =  101 - a(95)
a( 4) =  101 - a(94)
a( 3) =  101 - a(93)
a( 2) =  101 - a(92)
a( 1) =  101 - a(100)

which can be applied in an Excel spreadsheet (Ref. CnstrSngl10b).

Note: The Embedded Magic Square is based on the consecutive integers 19, 20, ... 82.

With a(74), a(78), the variables of the two exterior borders and the 4th order most inner Pan Magic Square constant, an optimized guessing routine (MgcSqr10b), produced 1440 Magic Squares in 242 seconds which are shown in Attachment 10.2.1.

With the 8th order embedded Magic Square constant, while varying a(92), a(93), a(94), a(30), a(40) and a(50), the same routine produced 1728 Magic Squares within 161 seconds, which are shown in Attachment 10.2.2.

10.2.2 Bordered Magic Squares

Also other Magic Squares of the 8th order as described and constructed in Section 8.1 thru Section 8.5, can be used as Center Squares for 10th order Bordered Magic Squares.

The Embedded Magic Squares will have a Magic Sum s8 = 404 and might be based on the consecutive integers 19, 20, ... 82.

Attachment 10.2.6 contains - based on some of the described Magic Squares of order 8 - examples of Bordered Magic Squares for the first occurring border.

10.2.3 Eccentric Magic Squares

An Eccentric Magic Square can be defined as a Magic Corner Square of order n, supplemented with two or more (i) rows and columns to a Magic Square of order (n + i).

A 10th order Eccentric Magic Square consists of one Magic Corner Square of the 8th order, supplemented with two rows and two columns.

 a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 a12 a13 a14 a15 a16 a17 a18 a19 a20 a21 a22 a23 a24 a25 a26 a27 a28 a29 a30 a31 a32 a33 a34 a35 a36 a37 a38 a39 a40 a41 a42 a43 a44 a45 a46 a47 a48 a49 a50 a51 a52 a53 a54 a55 a56 a57 a58 a59 a60 a61 a62 a63 a64 a65 a66 a67 a68 a69 a70 a71 a72 a73 a74 a75 a76 a77 a78 a79 a80 a81 a82 a83 a84 a85 a86 a87 a88 a89 a90 a91 a92 a93 a94 a95 a96 a97 a98 a99 a100

Rather than starting with the equations of the Magic Corner Square, the equations of the supplementary rows and columns can be used as a starting point for the generation of Eccentric Magic Squares.

The supplementary rows and columns can be described by following linear equations:

a( 1) + a( 2) + a( 3) + a( 4) + a( 5) + a( 6) + a( 7) + a( 8) + a( 9) + a(10) = 505
a(11) + a(12) + a(13) + a(14) + a(15) + a(16) + a(17) + a(18) + a(19) + a(20) = 505
a( 1) + a(11) + a(21) + a(31) + a(41) + a(51) + a(61) + a(71) + a(81) + a(91) = 505
a( 2) + a(12) + a(22) + a(32) + a(42) + a(52) + a(62) + a(72) + a(82) + a(92) = 505
a(10) + a(19) + a(28) + a(37) + a(46) + a(55) + a(64) + a(73) + a(82) + a(91) = 505
a( 1) + a(12) = 101
a(21) + a(22) = 101
a(31) + a(32) = 101
a(41) + a(42) = 101
a(51) + a(52) = 101
a(61) + a(62) = 101
a(71) + a(72) = 101
a(81) + a(82) = 101
a(91) + a(92) = 101
a( 3) + a(13) = 101
a( 4) + a(14) = 101
a( 5) + a(15) = 101
a( 6) + a(16) = 101
a( 7) + a(17) = 101
a( 8) + a(18) = 101
a( 9) + a(19) = 101
a(10) + a(20) = 101

Which can be reduced, by means of row and column manipulations, to:

a(91) = 101 - a(92)
a(81) = 101 - a(82)
a(71) = 101 - a(72)
a(61) = 101 - a(62)
a(51) = 101 - a(52)
a(41) = 101 - a(42)
a(31) = 101 - a(32)
a(21) = 101 - a(22)
a(19) = 404 + a(20) - a(28) - a(37) - a(46) - a(55) - a(64) - a(73) - a(82) - a(91)
a(12) = 50.5 + 0.5*(- a(13) - a(14) - a(15) - a(16) - a(17) - a(18) - a(19) - a(20) +
+ a(21) + a(31) + a(41) + a(51) + a(61) + a(71) + a(81) + a(91))
a(11) = 505 - a(12) - a(13) - a(14) - a(15) - a(16) - a(17) - a(18) - a(19) - a(20)
a(10) = 101 - a(20)
a( 9) = 101 - a(19)
a( 8) = 101 - a(18)
a( 7) = 101 - a(17)
a( 6) = 101 - a(16)
a( 5) = 101 - a(15)
a( 4) = 101 - a(14)
a( 3) = 101 - a(13)
a( 2) = 101 - a(11)
a( 1) = 101 - a(12)

The linear equations, deducted above, can be applied in an Excel spreadsheet. Following typical cases have been considered:

- The 8th order Magic Corner Square is an Eccentric Magic Square itself (ref. CnstrSngl10c);
- The 8th order Magic Corner Square is a  Pan Magic Square (ref. CnstrSngl10d).

Note: The Magic Corner Square is based on the consecutive integers 19, 20, ... 82.

In both cases it is obvious that the number of Eccentric Magic Squares is determined by the sum s2 of the values of the key variables a(28), a(37), a(46), a(55), a(64) and a(73).

An optimized guessing routine (MgcSqr10c) produced, with the Eccentric Magic Corner Square of the 8th order as shown in the first Spreadsheet Solution above constant, while varying the variables a(13), a(14), a(15), a(22), a(32) and a(42), 1800 Eccentric Magic Squares within 150 seconds, which are shown in Attachment 10.2.3.

The same routine produced, based on óne Pan Magic Corner Square of the eighth order constant and while varying the same variables 504 Eccentric Magic Squares within 52 seconds, which are shown in Attachment 10.2.4.

10.2.4 Order 3 Square Inlays (4 ea)

The 10th order Inlaid Magic Square shown below, contains four each 3th order Simple Magic Square Inlays with Magic Sums s(1) = 111, s(2) = 201, s(3) = 102 and s(4) = 192.

 86 5 96 19 40 20 88 9 70 72 1 36 28 47 2 90 66 78 57 100 99 48 37 26 87 4 58 67 76 3 22 27 46 38 59 81 77 56 68 31 21 92 95 89 83 71 10 11 15 18 82 97 94 91 30 17 8 12 14 60 62 35 23 44 39 41 65 73 54 69 42 43 34 25 32 52 53 64 75 85 61 24 45 33 49 50 74 55 63 51 29 98 7 93 84 79 6 80 13 16
 111 201 102 192

The relation between the Magic Sums s(1), s(2), s(3) and s(4) is:

```s(1) + s(2) + s(3) + s(4) = 6 * s10 / 5
```

With s10 = 505 the Magic Sum of the 10th order Inlaid Magic Square.

The 3th order Simple Magic Square Inlays can be constructed by means of suitable selected Latin Squares based on the two 3th order magic series {2, 3, 4} and {5, 6, 7} as illustrated below:

A
 5 7 6 5 7 6 7 6 5 7 6 5 6 5 7 6 5 7 4 2 3 4 2 3 2 3 4 2 3 4 3 4 2 3 4 2
B = R(A)
 3 2 4 6 7 5 4 3 2 5 6 7 2 4 3 7 5 6 3 2 4 6 7 5 4 3 2 5 6 7 2 4 3 7 5 6
M = A + 10 * B + 1
 36 28 47 66 78 57 48 37 26 58 67 76 27 46 38 77 56 68 35 23 44 65 73 54 43 34 25 53 64 75 24 45 33 74 55 63

The remainder of the 10th order Inlaid Magic Square ('Window') can be completed based on the defining equations as incorporated in procedure Inlaid103.

The square shown above corresponds with 84 * (3!)4 = 5.308.416 solutions, which can be obtained by selecting other aspects of the four inlays and variation of the window.

Attachment 10.2.7 shows a few more order 10 Inlaid Magic Squares for miscellaneous order 3 Inlays, which can be constructed based on order 8 Inlaid Magic Squares as discussed in Section 8.8.6.

10.2.5 Order 4 Pan Magic Square Inlays (4 ea)

The 10th order Inlaid Magic Square shown below, contains four each 4th order Pan Magic Square Inlays with Magic Sums s(1) = 220, s(2) = 180, s(3) = 224 and s(4) = 184.

 100 10 20 30 40 60 71 80 91 3 6 84 22 68 46 74 12 58 36 99 8 66 48 82 24 56 38 72 14 97 9 42 64 26 88 32 54 16 78 96 94 28 86 44 62 18 76 34 52 11 2 85 23 69 47 75 13 59 37 95 5 67 49 83 25 57 39 73 15 92 90 43 65 27 89 33 55 17 79 7 93 29 87 45 63 19 77 35 53 4 98 51 41 31 21 81 70 61 50 1
 220 180 224 184

The relation between the Magic Sums s(1), s(2), s(3) and s(4) is:

```s(1) + s(2) + s(3) + s(4) = 8 * s10 / 5
```

With s10 = 505 the Magic Sum of the 10th order Inlaid Magic Square.

The 4th order Pan Magic Square Inlays can be constructed by means of suitable selected Latin Diagonal Squares based on the two 4th order magic series {1, 3, 5, 7} and {2, 4, 6, 8} as illustrated below:

A
 3 1 7 5 3 1 7 5 5 7 1 3 5 7 1 3 1 3 5 7 1 3 5 7 7 5 3 1 7 5 3 1 4 2 8 6 4 2 8 6 6 8 2 4 6 8 2 4 2 4 6 8 2 4 6 8 8 6 4 2 8 6 4 2
B = R(A)
 8 2 6 4 7 1 5 3 6 4 8 2 5 3 7 1 4 6 2 8 3 5 1 7 2 8 4 6 1 7 3 5 8 2 6 4 7 1 5 3 6 4 8 2 5 3 7 1 4 6 2 8 3 5 1 7 2 8 4 6 1 7 3 5
M = A + 10 * B + 1
 84 22 68 46 74 12 58 36 66 48 82 24 56 38 72 14 42 64 26 88 32 54 16 78 28 86 44 62 18 76 34 52 85 23 69 47 75 13 59 37 67 49 83 25 57 39 73 15 43 65 27 89 33 55 17 79 29 87 45 63 19 77 35 53

The remainder of the 10th order Inlaid Magic Square ('Border') can be completed based on the defining equations as incorporated in procedure Inlaid104.

The square shown above corresponds with 3844 * (4!)4 = 7,214 1015 solutions, which can be obtained by selecting other aspects of the four inlays and variation of the border.