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a(n) = least k such that (2n Pi/3)^(2k+1)/(2k+1)! < 1.
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%I #4 Oct 19 2024 22:13:23

%S 1,2,5,7,10,13,16,19,21,24,27,30,33,35,38,41,44,47,50,52,55,58,61,64,

%T 67,69,72,75,78,81,84,86,89,92,95,98,101,104,106,109,112,115,118,121,

%U 123,126,129,132,135,138,140,143,146,149,152,155,157,160,163,166

%N a(n) = least k such that (2n Pi/3)^(2k+1)/(2k+1)! < 1.

%C The numbers (2n Pi/3)^(2k+1)/(2k+1)! are the coefficients in the Maclaurin series for sin x when x = 2Pi/3. If m>a(n), then (n 2Pi/3)^(2k+1)/(2k+1)! < 1.

%t a[n_] := Select[Range[z], (2n Pi/3)^(2 # + 1)/(2 # + 1)! < 1 &, 1]

%t Flatten[Table[a[n], {n, 0, 100}]]

%Y Cf. A370507, A376284, A376952, A376953, A376954, A376955, A376956, A376957, A376958, A376960.

%K nonn

%O 0,2

%A _Clark Kimberling_, Oct 17 2024