A250488 a(n) = Fibonacci(9^n).
1, 34, 37889062373143906
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..3
Programs
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Maple
a:= n-> (<<0|1>, <1|1>>^(9^n))[1, 2]: seq(a(n), n=0..4);
Formula
a(n) = A000045(9^n).
From Peter Bala, Nov 25 2022: (Start)
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)