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a(n) = least k such that (3n Pi/4)^(2k+1)/(2k+1)! < 1.
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%I #4 Oct 19 2024 22:13:32

%S 1,2,5,8,11,15,18,21,24,27,30,34,37,40,43,46,50,53,56,59,62,66,69,72,

%T 75,78,82,85,88,91,94,97,101,104,107,110,113,117,120,123,126,129,133,

%U 136,139,142,145,149,152,155,158,161,165,168,171,174,177,181,184

%N a(n) = least k such that (3n Pi/4)^(2k+1)/(2k+1)! < 1.

%C The numbers (3n Pi/4)^(2k+1)/(2k+1)! are the coefficients in the Maclaurin series for sin x when x = 3Pi/4. If m>a(n), then (n 3Pi/4)^(2k+1)/(2k+1)! < 1.

%t a[n_] := Select[Range[z], (3n Pi/4)^(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, A376959.

%K nonn

%O 0,2

%A _Clark Kimberling_, Oct 17 2024