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Fibonacci sequence (Posted on 2023-12-28) Difficulty: 3 of 5
Let an be a sequence given recursively such that a1=1 and

an+1=(7an+√(45an2-36))/2

for n>0, Show that an * an+1-1 is a square of an integer.

No Solution Yet Submitted by Danish Ahmed Khan    
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Some Thoughts A finding without a proof | Comment 1 of 2
When the formula is put into a spreadsheet, and the required equation calculated, and then the square root taken, it leads to the following sequence:
2, 13, 89, 610, 4181, 28657, 196418, 1346269, 9227465, ...
Which is oeis A033891:  a(n) = Fibonacci(4*n+3).

I have not attempted a proof that the sequence always produces a perfect square.

  Posted by Larry on 2023-12-28 11:31:16
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