A250488 - OEIS

%I #12 Nov 27 2022 06:39:57

%S 1,34,37889062373143906

%N a(n) = Fibonacci(9^n).

%H Alois P. Heinz, <a href="/A250488/b250488.txt">Table of n, a(n) for n = 0..3</a>

%F a(n) = A000045(9^n).

%F From _Peter Bala_, Nov 25 2022: (Start)

%F 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.

%F a(n) == 7 (mod 9) for n >= 1.

%F a(n+1) == a(n) mod (9^n).

%F 5*a(n)^2 == 2 (mod 9^n).

%F 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)

%p a:= n-> (<<0|1>, <1|1>>^(9^n))[1, 2]:

%p seq(a(n), n=0..4);

%Y Row n = 9 of A250486. Bisection of A045529.

%Y Cf. A000045.

%K nonn,easy

%O 0,2

%A _Alois P. Heinz_, Nov 24 2014