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%I #5 Mar 17 2017 22:03:34
%S 1,4,7,2,10,31,52,18,39,94,5,60,115,26,170,81,225,369,136,280,47,191,
%T 568,335,102,479,246,623,13,390,157,1144,534,911,301,678,1665,68,445,
%U 1432,822,1809,212,2186,589,1576,2563,966,356,1343,3927,733,3317,1720
%N a(n) = least m such that 1/(n+1) < f(m) < 1/n, where f(m) = fractional part of m*(golden ratio).
%C This is column 1 of A283738; |a(n+1)-a(n)| is a Fibonacci number for n>=1.
%t g = GoldenRatio; z = 10000; t = Table[N[FractionalPart[n*g]], {n, 1, z}];
%t r[k_] := Select[Range[z], 1/(k + 1) < t[[#]] < 1/k &, 1];
%t Flatten[Table[r[k], {k, 1, 200}]] (* A283745 *)
%Y Cf. A000045, A001622, A283738, A283746.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Mar 17 2017