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%I #9 Jan 06 2019 03:58:36
%S 3,10,12,23,23,28,30,35,40,45,51,54,59,64,70,74,80,83,91,99,99
%N The smallest argument m for which an approximating sequence B_n(m) differs from Fibonacci(m).
%C Given n, an auxiliary sequence B_n(m) is defined by B_n(m) = A000045(m), 0 <= m < 3 and B_n(m) = round(x_n*B_n(m-1)), m >= 3, where x_n is a truncated approximation of the golden ratio A001622 = 1.61803398..., namely, x_n = floor(A001622*10^n)/10^n = 1, 1.6, 1.61, 1.618, ... If one were to replace x_n with the exact value of golden ratio, the B_n(m) would reproduce the Fibonacci sequence. The sequence shows the index where B_n(m) diverges first from Fibonacci(m): B_n(m) = Fibonacci(m) for 0 <= m < a(n) and B_n(m) < Fibonacci(m) for m=a(n).
%e For n=1 and m>=3, we have B_1(m) = round(1.6*B_(m-1)).By this formula with the initial conditions, B_1(3)=2, B_1(4)=3, B_1(5)=5, B_1(6)=8, B_1(7)=13, B_1(8)=21, B_1(9)=34 and B_1(10)=54. Since F(10)=55, then B_1(m) gives the first 10 Fibonacci numbers: F(0),...,F(9). Thus a(1)=10.
%p A179203 := proc(n)local a001622,x,B ; a001622 := (1+sqrt(5))/2 ; x := floor( a001622*10^n)/10^n ; B := combinat[fibonacci](2) ;
%p for m from 3 do B := round(x*B) ; if B <> combinat[fibonacci](m) then return m; end if; end do:
%p end proc:
%p seq(A179203(n),n=0..20) ; # _R. J. Mathar_, Jan 04 2011
%Y Cf. A000045, A001622, A179057.
%K nonn,base,less
%O 0,1
%A _Vladimir Shevelev_, Jul 02 2010
%E a(8), a(9) corrected, sequence extended by _R. J. Mathar_, Jan 04 2011
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