a(n+1) = 625*a(n)^9 - 1125*a(n)^7 + 675*a(n)^5 - 150*a(n)^3 + 9*a(n) with a(0) = 1.
a(n) == 7 (mod 9) for n >= 1.
a(n+1) == a(n) mod (9^n).
5*a(n)^2 == 2 (mod 9^n).
In the ring of 9-adic integers, the sequence {a(n)} is a Cauchy sequence. It converges to a 9-adic root of the quadratic equation 5*x^2 - 2 = 0 (the 9-adic Cauchy sequence {Fibonacci(3*9^n)} converges to the other root). (End)
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