The Siamese method which includes both the Loubère and Méziriac magic squares have the property that the center cell must always contain the middle number of the series of numbers used, i.e. a number which is equal to one half the sum of the first and last numbers of the series, or
The previous page gave an account of a new method to generate 7×7 and 9×9 Méziriac type squares, a method previously unreported since Loubère and Méziriac concentrated only on those squares where the initial digit 1 were located on the periphery of the square (i.e., the cell at the center of row one) or on the cell adjacent to the central cell. There was no attempt to look into those squares where the digit 1 resided on cells other than these two positions. It was shown on that page that the 7×7 square was the first to contain three cells to the right of the central cell and that one could also place the digit 1 in the second cell to generate a new type of Méziriac square.
It was also shown that certain squares, particularly those where the size of the number n used to generate the square was divisible by three, produced less than the required number of squares. Thus, where the starting digit 1 is placed to the right of the central cell towards C determines what right hand shift to use when moving from one diagonal end to the start of the next diagonal. It was mentioned that there are four positions in which the digit 1 can be placed. An E with an arrow meant that a right cell shift of the value in the cell is required towards the periphery of the square. On the other hand, if it points inwards the movement is towards C:
C | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 | 26 | ... | E → |
In addition, the caption for each square included both a letter and the two arrows showing the shift to the right and the accompanying shift to the left, e.g.,
Examination of the 11×11 (K) and the 13×13 (L) Méziriac type squares where the initial digit 1 is place three cells to the right of the central cell are shown below. Again the position of this cell implies that a shift of 6 cells to the right is to be used when moving from one diagonal to the next. Although the right hand shifts of these squares are 6 one may also visualize them as left hand shifts of 5 for (K) and 7 for (L) where the sum of the
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So the upshot is that the full complement of m is possible for these squares when n is prime and it doesn't matter whether n is of type 4N+1 or 4N+3.
The next square, the 15×15 (M), shows (1) that filling in the right hand diagonal, (2) then placing the starting digit 1 six cells to the right of the central cell, (3) doing a stepwise construction, will eventually result in a cell number being superimposed over another. The same result is produced when placing the digit 1 thre cells to the right of the central cell (not shown). In both cases the 30 superimposes over 111 even though the former is a 12 cell right shift and the latter is a 6 cell right shift.
24 | 8 | 217 | 120 | |||||||||||
23 | 7 | 218 | 119 | |||||||||||
22 | 6 | 215 | 118 | |||||||||||
5 | 214 | 117 | 21 | |||||||||||
4 | 213 | 116 | 20 | |||||||||||
3 | 212 | 115 | 19 | |||||||||||
211 | 114 | 18 | 2 | |||||||||||
225 | 113 | 17 | 1 | |||||||||||
224 | 112 | 16 | 15 | |||||||||||
111 | 30 | 14 | 233 | |||||||||||
110 | 29 | 13 | 222 | |||||||||||
109 | 28 | 12 | 221 | |||||||||||
108 | 27 | 11 | 220 | |||||||||||
107 | 26 | 10 | 219 | |||||||||||
106 | 25 | 9 | 218 |
Since n = 3×5 then those right shifts divisible by 5 will not produce a square. Thus, placing the digit 1 five cells to the right of the central cell will, according to the shift table below, result in a right cell shift of 10 cells to the right, whereby the number 15 superimposes over the cell value of 112. The upshot is that one more square is not constructible as seen in partial square Mp.
C | 2 | 4 | 6 | 8 | 10 | 12 | 14 | E → |
115 | 3 | |||||||||||||
114 | 2 | |||||||||||||
113 | 1 | |||||||||||||
112 | 15 | |||||||||||||
111 | 14 |
Although, the square (N) with the digit 1 in position 4 cells from the center is magic (below), the square with the digit in 2 cells to the east of the central cell is not magic (square not shown here but the identical square rotated by 180o degrees along the main diagonal is shown at the end of Part IV). Thus according to the 15×15 right shift table above, only those right shifts of 2, 8 and 14 are possible while the 4 is non magic the 6, 10, and 12 being impossible.
137 | 169 | 201 | 8 | 40 | 72 | 104 | 121 | 153 | 185 | 217 | 24 | 56 | 88 | 120 |
168 | 200 | 7 | 39 | 71 | 103 | 135 | 152 | 184 | 216 | 23 | 55 | 87 | 119 | 136 |
199 | 6 | 38 | 70 | 102 | 134 | 151 | 183 | 215 | 22 | 54 | 86 | 118 | 150 | 167 |
5 | 37 | 69 | 101 | 133 | 165 | 182 | 214 | 21 | 53 | 85 | 117 | 149 | 166 | 198 |
36 | 68 | 100 | 132 | 164 | 181 | 213 | 20 | 52 | 84 | 116 | 148 | 180 | 197 | 4 |
67 | 99 | 131 | 163 | 195 | 212 | 19 | 51 | 83 | 115 | 147 | 179 | 196 | 3 | 35 |
98 | 130 | 162 | 194 | 211 | 18 | 50 | 82 | 114 | 146 | 178 | 210 | 2 | 34 | 66 |
129 | 161 | 193 | 225 | 17 | 49 | 81 | 113 | 145 | 177 | 209 | 1 | 33 | 65 | 97 |
160 | 192 | 224 | 16 | 48 | 80 | 112 | 144 | 176 | 208 | 15 | 32 | 64 | 96 | 128 |
191 | 223 | 30 | 47 | 79 | 111 | 143 | 175 | 207 | 14 | 31 | 63 | 95 | 127 | 159 |
222 | 29 | 46 | 78 | 110 | 142 | 174 | 206 | 13 | 45 | 62 | 94 | 126 | 158 | 190 |
28 | 60 | 77 | 109 | 141 | 173 | 205 | 12 | 44 | 61 | 93 | 125 | 157 | 189 | 221 |
59 | 76 | 108 | 140 | 172 | 204 | 11 | 43 | 75 | 92 | 124 | 156 | 188 | 220 | 27 |
90 | 107 | 139 | 171 | 203 | 10 | 42 | 74 | 91 | 123 | 155 | 187 | 219 | 26 | 58 |
106 | 138 | 170 | 202 | 9 | 41 | 73 | 105 | 122 | 154 | 186 | 218 | 25 | 57 | 89 |
Finally, for the larger squares, we can mention that when n = 25, i.e., p1 × p2= 25 and p1 = p2 then those cells in the shift table divisible by five will be impossible to construct. For example, placing the digit 1 into the 5th cell from center will result in a right shift of 10, superimposing the digit 51 onto number 311 of the main diagonal. Similarly, for n = 3 × 7, placing the digit 1 in cell number 7 will result in a right shift of 14 and superimpose the digit 21 onto 220 of the main diagonal. The superimpositions, thereby, reduce the number of Méziriac type magic squares.
This completes this section (Part IB). To go to next section Part II. To return to Part IA. To return to homepage.
Copyright © 2020 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com