
COMMENTS

Given n, an auxiliary sequence A_n(m) is defined by A_n(m)=A000045(m), 0<=m<5 and
A_n(m)=round( log_2(x_n^A_n(m1)+x_n^A_n(m2))), m>=5, where x_n is a truncated
approximation of 2^A001622 = 3.0695645076529..., namely
x_n = floor(2^A001622*10^n)/10^n = 3, 3.0, 3.06, 3.069, 3.0695,... for n = 0, 1, 2, 3,...
If one would replace x_n by the exact value of 2^(golden ratio), the A_n(m) would reproduce the Fibonacci sequence.
The sequence shows the index where A_n(m) diverges first from Fibonacci(m): A_n(m) = Fibonacci(m) for 0<=m<a(n) and A_n(m) <> Fibonacci(m) for m=a(n). More exactly, it could be proved that, for m=a(n),A_n(m)=Fibonacci(m)1.


EXAMPLE

For n=0 and m>=5, we have A_0(m) = round(log_2(3^A_0(m1)+3^A_0(m2))). By this formula with the initial conditions, A_0(5)=5, A_0(6)=8, A_0(7)=13, A_0(8)=21 and A_0(9)=33. Since F(9)=34, then A_(m) gives the first 9 Fibonacci numbers: F(0),...,F(8). Thus a(0)=9.
