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Fibonacci sequence (
Posted on 20231228
)
Let a
_{n}
be a sequence given recursively such that a
_{1}
=1 and
a
_{n+1}
=(7a
_{n}
+√(45a
_{n}
^{2}
36))/2
for n>0, Show that a
_{n}
* a
_{n+1}
1 is a square of an integer.
No Solution Yet
Submitted by
Danish Ahmed Khan
No Rating
Comments: (
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)
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 20231228 11:31:16
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