(consecutively concentric squares)

The bones **4(n-1)** border numbers form **2(n-1)** ±
pairs. The + numbers are in the range:

**((n-2)²+1)/2** to **(n²-1)/2**

The sum of each bones row, column, and main diagonal is 0.

Therefore, the border constraints are:

- the sum of row 1, (row n), is 0
- the sum of column 1, (column n), is 0
- the other rows and columns, and the main diagonals have matching ±numbers at each end

The square is divided into the following regions:

- center cell
- NW,SE main diagonal
- NE,SW main diagonal
- W,E center row
- N,S center column
- NNW sector, SSW sector
- NNE sector, SSE sector
- WNW sector, ENE sector
- WSW sector, ESE sector

Define rrel = row - (n+1)/2, crel = column - (n+1)/2.

Then, the basic formula used here to compute a cell number is 2x² + 2y + c,
where, for each region:

- x is rrel, or crel
- y is crel, or rrel, or 0
- c is a constant

The algorithm is:

- assign x: rrel
- assign y: crel
- determine the region of the square
- reset y to 0 in NW, SE, NE, SW
- swap x and y in WNW, ENE, WSW, ESE, W, E
- assign c:
- -1 in NNW, SSW, NE, SW
- 0 in WNW, ENE, W, center, E
- 1 in NNE, SSE, NW, SE
- 2 in WSW, ESE, N, S

- compute 2x² + 2y + c
- change sign in NNE, ESE, SSW, WNW, SE, SW, N, W
- optionally, add (n²+1)/2

Summary:

region x y c sign -------- ----------- ----------- ------- ---------- center rrel crel 0 0 NW rrel 0 1 + SE rrel 0 1 - NNW rrel crel -1 + SSW rrel crel -1 - N rrel crel 2 - S rrel crel 2 + NNE rrel crel 1 - SSE rrel crel 1 + NE rrel 0 -1 + SW rrel 0 -1 - ENE crel rrel 0 + WNW crel rrel 0 - E crel rrel 0 + W crel rrel 0 - ESE crel rrel 2 - WSW crel rrel 2 +

Is the algorithm correct for all odd order squares?

It is sufficient to check the border, since the test applies recursively through the nested squares down to order 3.

Let B be the bones, m = (n+1)/2, and X = 2(1-m)² = 2(n-m)².

In row 1, rrel = 1-m, crel ranges from 1-m to n-m.

B(1,1) = X + 1 B(1,m) = -(X + 2) B(1,n) = X - 1 The sum of these is X + 1 - X - 2 + X - 1 =X - 2. NNW numbers are X + 2crel - 1. NNE numbers are -(X + 2crel + 1). Summing NNW,NNE number pairs from left to right, we have: B(1,2) + B(1,m+1) = X + 2(2-m) - 1 - X - 2(1) - 1 = -(n+1) … B(1,m-1) + B(1,n-1) = X + 2(-1) - 1 - X - 2(n-1-m) - 1 = -(n+1) The sum of these (n-3)/2 pairs is -(n+1)(n-3)/2 =-(X - 2).

Row 1 total is **(X - 2) - (X - 2)** = 0.

Rows 2 to n-1 have ± the same numbers at each end.

Row n contains ± the same numbers as row 1.

In column 1, rrel ranges from 1-m to n-m, crel = 1-m.

B(1,1) = X + 1 B(m,1) = -X B(n,1) = -(X - 1) The sum of these is X + 1 - X - X + 1 =-(X - 2). WNW numbers are -(X + 2rrel). WSW numbers are X + 2rrel + 2. Summing the WNW,WSW number pairs from top to bottom, we have: B(2,1) + B(m+1,1) = -X - 2(2-m) + X + 2(1) + 2 = n+1 … B(m-1,1) + B(n-1,1) = -X - 2(-1) + X + 2(n - 1 - m) + 2 = n+1 The sum of these (n-3)/2 pairs is (n+1)(n-3)/2 =X - 2.

Column 1 total is **-(X - 2) + (X - 2)** = 0.

Columns 2 to n-1 have ± the same numbers at each end.

Column n contains ± the same numbers as column 1.

Are the border numbers correct, that is, in the right range with no duplicates?

There are 4(n-1) border numbers or 2(n-1) ± pairs.

There is 1 pair in each of the main diagonals,

1 in the center row, and 1 in the center column.

The remaining 2(n-1) - 4 pairs are in the (NNW,SSW), (NNE,SSE), (WNW,ENE),

and (WSW,ESE) sectors.

The placing of the 2(n-1) positive numbers is:

----- n-3 numbers alternating NNW, ENE NNW: B(1,2) = X + 2(2-m) - 1 = X - n + 2 ENE: B(2,n) = X + 2(2-m) = X - n + 3 … NNW: B(1,m-1) = X + 2(-1) - 1 = X - 3 ENE: B(m-1,n) = X + 2(-1) = X - 2 NE: B(1,n) = X - 1 E: B(m,n) = X NW: B(1,1) = X + 1 S: B(n,m) = X + 2 ----- n-3 numbers alternating SSE, WSW SSE: B(n,m+1) = X + 2(1) + 1 = X + 3 WSW: B(m+1,1) = X + 2(1) + 2 = X + 4 … SSE: B(n,n-1) = X + 2(n-1-m) + 1 = X + n - 2 WSW: B(n-1,1) = X + 2(n-1-m) + 2 = X + n - 1

The next smaller odd order square is order k = n-2.

Its biggest border number is 2(1-(k+1)/2)² + k - 1 = **X - n + 1**.

The smallest border number of the order n is **X - n + 2**.

That is, the smallest border number of the order n is 1 greater

than the biggest border number of the order n-2.