The method of J. Travers also works from the outer border towards the center. First the outer border of the square of order n is filled, then the border of the square of order n − 2 embedded in it, etc. The entire border includes exactly
2n + (2n − 2) =4n − 2
numbers. To compensate in pairs, he summarizes the numbers 1, … , 2n-2 and their complements in pairs, as already described by Frénicle. Since n is odd, it follows with n=2k+1
2n - 2=2 · (2k+1) - 2=4k+2 - 2=4k
that a total of 4k numbers are to be distributed. Travers divides them into four groups of equal size:
1, 2, … , k
k+1, k+2, … , 2k
2k+1, 2k+2, … , 3k
3k+1, 3k+2, … , 4k
For order n=7, the number sequences (1,2,3), (4,5,6), (7,8,9) and (10,11,12) are obtained. He now distributes them using the following steps:
1 | 2 | 3 | ||||
5 | ||||||
6 | ||||||
7 | ||||||
12 | ||||||
11 | ||||||
10 | 8 | 9 | 4 |
46 | 1 | 2 | 3 | 42 | 41 | 40 |
45 | 5 | |||||
44 | 6 | |||||
7 | 43 | |||||
12 | 38 | |||||
11 | 39 | |||||
10 | 49 | 48 | 47 | 8 | 9 | 4 |
Once all the numbers have been entered, the complementary numbers are entered in the horizontally or vertically opposite cells. Only the two lower corners are an exception because they write the complement in the diagonally opposite corners.
When the border of the square of order 7 is completely filled, Travers continues with the border of the embedded inner square of order n=5. Since a total of 8 numbers with their complement must be placed and the numbers 1, 2, … , 12 have already been assigned, the numbers 13, 14, … , 20 are used now.
46 | 1 | 2 | 3 | 42 | 41 | 40 |
45 | 13 | 14 | 5 | |||
44 | 16 | 6 | ||||
7 | 17 | 43 | ||||
12 | 20 | 38 | ||||
11 | 19 | 18 | 15 | 39 | ||
10 | 49 | 48 | 47 | 8 | 9 | 4 |
46 | 1 | 2 | 3 | 42 | 41 | 40 |
45 | 35 | 13 | 14 | 32 | 31 | 5 |
44 | 34 | 16 | 6 | |||
7 | 17 | 33 | 43 | |||
12 | 20 | 30 | 38 | |||
11 | 19 | 37 | 36 | 18 | 15 | 39 |
10 | 49 | 48 | 47 | 8 | 9 | 4 |
The same procedure is followed with the embedded inner square of order n=3.
46 | 1 | 2 | 3 | 42 | 41 | 40 |
45 | 35 | 13 | 14 | 32 | 31 | 5 |
44 | 34 | 21 | 16 | 6 | ||
7 | 17 | 23 | 33 | 43 | ||
12 | 20 | 24 | 22 | 30 | 38 | |
11 | 19 | 37 | 36 | 18 | 15 | 39 |
10 | 49 | 48 | 47 | 8 | 9 | 4 |
46 | 1 | 2 | 3 | 42 | 41 | 40 |
45 | 35 | 13 | 14 | 32 | 31 | 5 |
44 | 34 | 28 | 21 | 26 | 16 | 6 |
7 | 17 | 23 | 27 | 33 | 43 | |
12 | 20 | 24 | 29 | 22 | 30 | 38 |
11 | 19 | 37 | 36 | 18 | 15 | 39 |
10 | 49 | 48 | 47 | 8 | 9 | 4 |
Then only the center cell remains, in which the median value of all the numbers involved is entered, in this case 25. The following figure shows the multi-bordered magic square of the order n=7 generated by Travers using this algorithm.
46 | 1 | 2 | 3 | 42 | 41 | 40 |
45 | 35 | 13 | 14 | 32 | 31 | 5 |
44 | 34 | 28 | 21 | 26 | 16 | 6 |
7 | 17 | 23 | 25 | 27 | 33 | 43 |
12 | 20 | 24 | 29 | 22 | 30 | 38 |
11 | 19 | 37 | 36 | 18 | 15 | 39 |
10 | 49 | 48 | 47 | 8 | 9 | 4 |
46 | 1 | 2 | 3 | 42 | 41 | 40 |
45 | 35 | 13 | 14 | 32 | 31 | 5 |
44 | 34 | 28 | 21 | 26 | 16 | 6 |
7 | 17 | 23 | 25 | 27 | 33 | 43 |
12 | 20 | 24 | 29 | 22 | 30 | 38 |
11 | 19 | 37 | 36 | 18 | 15 | 39 |
10 | 49 | 48 | 47 | 8 | 9 | 4 |
The following figure shows further magic squares of the orders n=5 and n=9 created using the method of Travers.
23 | 1 | 2 | 20 | 19 |
22 | 16 | 9 | 14 | 4 |
5 | 11 | 13 | 15 | 21 |
8 | 12 | 17 | 10 | 18 |
7 | 25 | 24 | 6 | 3 |
77 | 1 | 2 | 3 | 4 | 72 | 71 | 70 | 69 |
76 | 62 | 17 | 18 | 19 | 58 | 57 | 56 | 6 |
75 | 61 | 51 | 29 | 30 | 48 | 47 | 21 | 7 |
74 | 60 | 50 | 44 | 37 | 42 | 32 | 22 | 8 |
9 | 23 | 33 | 39 | 41 | 43 | 49 | 59 | 73 |
16 | 28 | 36 | 40 | 45 | 38 | 46 | 54 | 66 |
15 | 27 | 35 | 53 | 52 | 34 | 31 | 55 | 67 |
14 | 26 | 65 | 64 | 63 | 24 | 25 | 20 | 68 |
13 | 81 | 80 | 79 | 78 | 10 | 11 | 12 | 5 |