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a(n) = least k such that (n Pi/3)^(2k+1)/(2k+1)! < 1.
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%I #4 Oct 19 2024 22:13:15

%S 1,1,2,3,5,6,7,9,10,11,13,14,16,17,19,20,21,23,24,26,27,28,30,31,33,

%T 34,35,37,38,40,41,43,44,45,47,48,50,51,52,54,55,57,58,60,61,62,64,65,

%U 67,68,69,71,72,74,75,77,78,79,81,82,84,85,86,88,89,91,92

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

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

%t a[n_] := Select[Range[z], (n 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, A376959, A376960.

%K nonn

%O 0,3

%A _Clark Kimberling_, Oct 17 2024