A New Procedure for Magic Squares (Part II)

Loubère Block Modified Squares

A stairs

Continuation using 7x7 Squares

As shown in the previous page a new Loubère method Modified Loubère Block Modified 5x5 Squares was introduced where numbers are added consecutively starting with 1 and ending with n2. The final square is transformed into a modified Loubère, having numbers greater than n2 by modifying a block or grid of numbers. The initial number is added either at the normal Loubère sites, i.e., (the center of the last row or last column) or on the main diagonal.

This is done by taking the non-magic squares and converting them into magic ones using a variety of means. These squares are depicted below in methods I and II. The generation of these magic squares involves an almost normal approach:
Numbers are added consecutively starting out with 1 and ending with numbers greater than n2 after modification. A break involves translational moves (up, down or sideways).

Moreover, it must be stated here that the magic sum has been modified from the known equation S = ½(n3 + n) to the general equation:

>
S = ½(n3 ± an)

which takes into account these new squares. The variable a, an odd number, is equal to 1,3,5,7 or ... and may take on + or - values. For example when a = 1 the normal magic sum S is implied. When a takes on different odd values S gives the magic sum of a modified magic square. It will be shown that the addition or subtraction of n2 to some of the cells in the square gives rise to a new magic square.

Construction of 7x7 Block Modified Loubère Squares

Method I: Start at lower left hand corner (1, 2, 3 ⇒ break)
  1. Place a 1 into the lower left hand corner.
  2. Add consecutive numbers up to the middle cell.
  3. Move, i.e. break, one cell down.
  4. Repeat the process until the square is filled as shown below in squares 1-2.
  5. As shown below this square is not magic because the three columns and three rows (in grey) sum to 147 instead of 196.
  6. Where the sums 147 cross corresponds to the small square in blue.
  7. Since the numbers 1, 2 and 3 sit on a diagonal which sums to 196, these numbers remain unchanged.
  8. Adding n2 = 49 to (34, 10 and 4) or to (42, 43, 12) (color change to light green) gives the two magic squares 3 and 4 where S = ½(n3 + 7n).
1
 
 
 
 
3
2 4
1
2
196
9 11 2029 384049 196
17 19 2537 39488196
18 27 3645 47716196
26 35 4446 61524196
34 43 35 142325147
42 2 413 223133147
1101221 30 3241147
147147147 196 196196 196 196
3
196
9 11 2029 384049 196
17 19 2537 39488196
18 27 3645 47716196
26 35 4446 61524196
83 43 35 142325196
42 2 5313 223133196
1591221 30 3241196
196196196 196 196196 196 196
+
4
196
9 11 2029 384049 196
17 19 2537 39488196
18 27 3645 47716196
26 35 4446 61524196
34 92 35 142325196
91 2 413 223133196
1106121 30 3241196
196196196 196 196196 196 196
Method II: Start at lower left hand corner (1, 2, 3, 4 ⇒ break)
  1. Place a 1 into the lower left hand corner.
  2. Add 4 consecutive numbers to past the center cell.
  3. Move, i.e. break, one cell down.
  4. Repeat the process until the square is filled as shown below in squares 1-2.
  5. As shown below this square is not magic because the two columns and two rows (in grey) sum to 154 instead of 203.
  6. Where the sums 154 cross corresponds to the small square in blue.
  7. Since the numbers 1, 2, 3 and 4 sit on a diagonal which sums to 154, at least one of these numbers must be changed and since the left diagonal also sums to 154 the number to change is 4.
  8. Adding n2 = 49 to either (4, 10, 11, 34) or to (4, 12, 42, 43) gives the magic squares 3 and 4 where S = ½(n3 + 9n).
1
 
 
 
4
35
2
1
2
154
9 18 2029 384049 203
17 19 2837 46488203
25 27 3645 47716203
26 35 44 461524154
34 43 3 5142332154
42 2 11 13223133154
11012 2130 3941154
154154154 154 203203 203 154
3
203
9 18 2029 384049 203
17 19 2837 46488203
25 27 3645 47716203
26 35 44 5361524203
83 43 3 5142332203
42 2 60 13223133203
15912 2130 3941203
203203203 203 203203 203 203
+
4
203
9 18 2029 384049 203
17 19 2837 46488203
25 27 3645 47716203
26 35 44 5361524203
34 92 3 5142332203
91 2 11 13223133203
11061 2130 3941203
203203203 203 203203 203 203
Method III: Start at lower left hand corner (1, 2, 3, 4, 5, 6 ⇒ break)
  1. Place a 1 into the lower left hand corner.
  2. Add 5 consecutive numbers to past the center cell.
  3. Move, i.e. break, one cell down.
  4. Repeat the process until the square is filled as shown below in squares 1-2.
  5. As shown below this square is not magic because the two columns and two rows (in grey) sum to 70 or 168 instead of 217.
  6. Where the sums 168 cross corresponds to the small square in blue.
  7. Since the numbers 1, 2, 3, 4, 5 and 6 sit on a diagonal which sums to 70, at least two of these numbers must be changed and since the left diagonal also sums to 167 the number to change is 4.
  8. Adding n2 = 49 to either (4, 6, 11, 25,30, 43) or to (1, 4, 7, 19, 43, 46) gives the magic squares 3 and 4 where S = ½(n3 + 13n).
1
 
6
57
4
3
2
1
2
70
9 18 2729 384749 217
17 26 28 37 4668168
25 34 36 45 5716168
33 35 44 4131524 168
41 43 3 12142332 168
42 2 11 20223140 168
11019 2130 3948 168
168168168 168 168168 217 168
3
217
9 18 2729 384749 217
17 26 28 37 46558217
74 34 36 45 5716217
33 35 44 53131524 217
41 92 3 12142332 217
42 2 60 20223140 217
11019 2179 3948 217
217217217 217 217217 217 217
+
4
217
9 18 2729 384749 217
17 26 28 37 9568217
25 34 36 45 55616217
33 35 44 53131524 217
41 92 3 12142332 217
42 2 60 20223140 217
501019 2130 3948 217
217217217 217 217217 217 217
Method IV: Start at the center of the last column (1, 2, 3, 4 ⇒ break))
  1. Place a 1 into the lower left hand corner.
  2. Add consecutive numbers 2,3,4 up to the top cell.
  3. Move, i.e. break, one cell down.
  4. Repeat the process until the square is filled as shown below in squares 1-2.
  5. As shown below this square is not magic because the two columns and two rows (in grey) sum to 154 instead of 203.
  6. Where the sums 154 cross corresponds to the small square in blue.
  7. Since the numbers 35,3 and 13 on the left diagonal which sums to 154 at least one of these numbers must be changed.
  8. Two examples are obtained by adding n2 = 49 to either (1, 5, 11, 35) or to (1, 2, 3, 4) giving the magic squares 3 and 4 where S = ½(n3 + 9n).
1
4
3 5
2
1
 
 
 
2
203
35 44 46 152426 154
43 3 5 14233234154
2 11 13 22313342154
10 12 21 3039411154
18 20 29 3840499 203
19 28 37 4648817 203
273645 477 1625 203
154154154 203 203203 154 154
3
203
84 44 46 152426 203
43 3 54 14233234203
2 60 13 22313342203
10 12 21 30394150203
18 20 29 3840499 203
19 28 37 4648817 203
273645 477 1625 203
203203203 203 203203 203 203
+
4
203
35 44 536 152426 203
43 52 5 14233234203
51 11 13 22313342203
10 12 21 30394150203
18 20 29 3840499 203
19 28 37 4648817 203
273645 477 1625 203
203203203 203 203203 203 203

This completes this section on the new block Loubère Method (Part I). The next section deals with A New Procedure for Méziriac type magic squares (Part III). To return to Part I. To return to homepage.


Copyright © 2009 by Eddie N Gutierrez. E-Mail: Fiboguti89@Yahoo.com