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%I #4 Mar 17 2017 22:04:09
%S 2,1,3,16,8,21,55,288,144,377,1974,987,2584,6765,35422,17711,46368
%N a(n) = least m such that (2^n-1)/2^n < f(m) < (2*2^n-1)/2^(n+1), where f(m) = fractional part of m*(golden ratio).
%C This is column 1 of A283741; |a(n+1)-a(n)| is a Fibonacci number for n>=1.
%t g = GoldenRatio; z = 100000; t = Table[N[FractionalPart[n*g]], {n, 1, z}];
%t r[k_] := Select[Range[z], (2^k - 1)/2^k < t[[#]] < (2*2^k - 1)/2^(k + 1) &, 1];
%t Flatten[Table[r[k], {k, 0, 50}]] (* A283748 *)
%Y Cf. A000045, A001622, A283740, A283748.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Mar 17 2017
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