# Wythoff' array and Wechsler's Sequence

#### ramsey2879

Wythoff's square array is Sloane's reference A035513 in the online encyclopedia of sequences (click the "table" button to see the sequence as a table) and Allan Wechsler's sequence is A022344. To my knowledge the following connection has not been noted before.

Let T(i,j) be defined from the table as follows T(1,1) = 1, T(1,2) = 2 and T(2,1) = 4 and let A(i) be the Wechsler's sequence starting with A(1) = 1

Then x(i,j) are integers defined by the following relation:

If T(i,j) is even then

$$\frac{5*T_{(i,j)}^{2}}{4} - A_{i}*(-1)^{i} = x_{(i,j)}^{2}$$

If T(i,j) is odd then

$$\frac{5*T_{(i,j)}^{2} - 1}{4} -A_{i}*(-1)^{i} = x_{(i,j)}^{2} + x_{(i,j)}$$

for j>2 and $$T_{(i,j)}$$ is odd

$$x_{(i,j)} = x_{(i,j-1)} + x_{(i,j-2)} [\tex] for j>2 and [tex]T_{(i,j)}$$ is even

$$x_{(i,j)} = x_{(i,j-1)} + x_{(i,j-2)} + 1[\tex] Related Linear and Abstract Algebra News on Phys.org #### ramsey2879 Wythoff's square array is Sloane's reference A035513 in the online encyclopedia of sequences (click the "table" button to see the sequence as a table) and Allan Wechsler's sequence is A022344. To my knowledge the following connection has not been noted before. Let T(i,j) be defined from the table as follows T(1,1) = 1, T(1,2) = 2 and T(2,1) = 4 and let A(i) be the Wechsler's sequence starting with A(1) = 1 Then x(i,j) are integers defined by the following relation: If T(i,j) is even then [tex]\frac{5*T_{(i,j)}^{2}}{4} - A_{i}*(-1)^{i} = x_{(i,j)}^{2}$$

If T(i,j) is odd then

$$\frac{5*T_{(i,j)}^{2} - 1}{4} -A_{i}*(-1)^{i} = x_{(i,j)}^{2} + x_{(i,j)}$$

for j>2 and $$T_{(i,j)}$$ is odd

$$x_{(i,j)} = x_{(i,j-1)} + x_{(i,j-2)}$$

for j>2 and $$T_{(i,j)}[\tex] is even [tex]x_{(i,j)} = x_{(i,j-1)} + x_{(i,j-2)} + 1$$
On the other hand since $$A_{i} = |T_{i,j}^{2}-T_{i,j-1}*T_{i,j+1}|$$, and since an odd square -1 divided by 4 is an oblong number, it occured to me that the above could be simplified to:

$$T_(n,k)^2 + 4*T_(n,k-1)*T_(n,k+1) = g(i,k)^2$$

The formula for T_(n,k)

$$T_{n,k} = F_{k+1}*floor(n*tau) + (n-1)*F_{k}$$

F(n) are the fibonacci numbers with F(0) = 0 and F(1) = 1 and tau is the golden ratio.

Also I found that g(i,k) is another row of the Wythoff array

The mapping is as follows:

g(1,k) = T(2,k-1)
g(2,k) = T(16,k-3)
g(3,k) = T(9,k-1)
g(4,k) = T(13,k-1)
g(5,k) = T(45,k-3)
g(6,k) = T(20,k-1)
g(7,k) = T(63,k-3)
g(8,k) = T(27,k-1)
g(9,k) = T(31,k-1)
g(10,k) =T(92,k-3)

I find it interesting that the 3's and 1's follow the Fibonacci rabbit sequence with 3's substituted for the zeros; and there appears to be two separate rabbit patterns associated with the series of differences of the row numbers with the differences 7 and 4 associated with the 1's and the differences 29 and 18 associated with the 3's. Does anyone care to attempt a proof or to verify my findings?

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#### ramsey2879

On the other hand since $$A_{i} = |T_{i,j}^{2}-T_{i,j-1}*T_{i,j+1}|$$, and since an odd square -1 divided by 4 is an oblong number, it occured to me that the above could be simplified to:

$$T_(n,k)^2 + 4*T_(n,k-1)*T_(n,k+1) = g(i,k)^2$$
Hey, it turns out that the above Fibonacci relationship can be used to derive the formula for Pythagorean triples and vice versa. Great for introducing math to young students.

if a,b,c are three consecutive numbers in a Fibonacci sequence then b^2 + 4ac is a square. How to make this the square hypotenuse of a primative Pythagorean triple? Simple make a and c coprime squares. Why havn't I seen this posted before?

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