Let T(i,j) be the jth element in the ith row of the square Wythoff array Sloane's sequence A035513 in the online encyclopedia of integer sequences https://www.research.att.com/~njas/sequences/table?a=35513&fmt=312 and A(i) be the ith element of Allan Wechsler's "J determinant" sequence A0022344 https://www.research.att.com/~njas/sequences/A022344. Let = the floor of x.

Then
If j is odd then
<3*T(i,j)^2/2>-^2 -A(i) equals a square (x*x) if T(i,j) is even or an oblong, x(x-1), if T(i,j) is odd.
If j is even then
<3*T(i,j)^2/2>-^2 (plus) A(i) equals a square (x*x) if T(i,j) is even or an oblong, x(x-1), if T(i,j) is odd.

The Table is at https://www.research.att.com/~njas/sequences/A035513
The sequence is https://www.research.att.com/~njas/sequences/A022344

If T(i,j) is even then
(5(T(i,j))^2)/4 - A(i)*(-1)^i = x(i,j)*x(i,j) where x(i,j) is an integer

If T(i,j) is odd then
(5*T(i,j)^2 - 1)/4 - A(i)*(-1)^i = x(i,j)*x(i,j) x(i,j) where x(i,j) is an integer

x(i,j) = x(i,j-1) plus x(i,j-2) if T(i,j) is odd
x(i,j) = x(i,j-1) plus x(i,j-2) plus 1 if T(i,j) is even

I don't have enough time to tabulate the x(i,j) table

Prove this identity and you have my solution.

I had forgotten ( did not know? ) the above identity, but it dropped out
pretty quickly as what was really in need of being proved.

Above is proved by induction.

Look my last post again. I think that those who try to come up with a relationship between primes should check it out. I mean the A(i) terms seem to be prime as long as T(i,j) and T(i,j+1) are coprime and are otherwise are a prime multiplied by the square of the greatest common denominator.

>Is it not what you are looking for?

It was but I found an even more impressive identity after reviewing your post.

F(n+1)*F(n-1) + F(n)*F(n+2) =
F(n+1)*{F(n+1) - F(n)} + F(n)*{F(n) + F(n+1)} =
F(n+1)^2 + F(n)^2,
which rearranges into the formula you want for your recursion.

Cheers,

Stuart Anderson

more directly since 4* a square is an even square and since 4* an oblong number + 1 is an odd square my recursion simplifies to

F(n)^2 + 4*F(n-1)*F(n+1) = a square

This too appears to not be well known if it is known at all

Note that it works for all Fibonacci like sequences in Wythoff's table

This result can be used to find Pythagorean Triangles even to derive the formula for Pythagorean triples. How is this so? Simple just make F(n-1) and F(n+1) coprime squares!