Border Wheel Squares - Number of Squares (Part VIIIa)

Table for Calculating the Number of (a,b,c) Squares

In the previous two pages Part VI and Part VII it is shown that for each Δ value of n there is produced a certain number of squares. When we tabulate these two sets of values for n=5 and n=7 it can be seen that the differences (δ₁) among the values in the n=5 and n=7 column are 1 and 2, respectively. Since we know that the n+2 Δ values for each n is 1 (last value in the column), we can expand the table by taking as the constant of separation (δ₁) the consecutive numbers: 3 for n=7, 4 for n=9, ... up to 8 for n=19 or by use of the equation (n−3)/2.

Therefore, adding these δs to the 1s at the end of each appropriate column, i.e., the n+2 Δ value for each n, and working backwards up to the Δ=3 row we produce an updated Table I. However, an even better method has been found for generating all the terms in this table using a unlimited number of similar quadratic equations each based on the order n. In column 2, each row is constructed from a different second order equation where the coefficients of the first order nth term is a consecutive even number, while the zeroth order constant varies from one equation to the next by a difference of 6. Part XII shows how one of these equations was conceived.

Thus, we can use Table I to find all the number of squares that may be formed. For instance, the number of squares in a 15th order square with a Δ=5 is 37. This may be visualized by seeing what happens when the entire three spokes (the 3 columns) is moved two positions to the right so that the numbers 1 and 2 in Table s1 now become non spoke numbers. Table s2 is now the new wheel consisting of new spoke numbers, besides the main diagonal, for the square s2 in the series. As one continues to the right of the complement table, in increments of two, eventually one reaches the last number 105 at s37. Further movement to the right is not possible since the next number 106, part of the main diagonal, is unavailable for placement into any of the other spokes. In addition, only three spokes have been shown to keep things simple. The complements of these three spokes, each containing 7 numbers, have been left out for clarity.

The Wheel Octagon Triangle

Table I can be converted into a Pascal type triangle, the Wheel Octagon Triangle (The WOT), by tilting the table by 45^o on its side. By adding three ones to the top of the figure using equations one and two in Table I, we can complete the triangle even though these ones are not actual part of Table I since an n=1 or 3 cannot employ the Octagon algorithm. If we look at the triangle it can be seen that each entry in each row is obtained by adding a consecutive number to the initial 1 and to each sum (i.e. each term) after that until the number of terms equals the row number. This is equivalent to Table I where n (the consecutive number) is added first to 1 then to the sum created then to the next until one reaches the row labeled Δ=3. The triangle is known and has a Sloane number in the OEIS database of A162609.

where the sum (∑) of each row takes on a value from the equation m(m² −3m + 4)/2, starting at m=1, an equation which can be used to derive the sequence stored in the OEIS database under the Sloane number A060354.

This completes Part VIIIa. Part VIIIb continues with the WOT triangle and produces the first instance of interleaving ascending diagonals.

To go back to Part VII. Go back to homepage.