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a(n) = least k such that (n Pi/2)^(2k+1)/(2k+1)! < 1.
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%I #4 Oct 19 2024 22:13:05

%S 1,1,3,5,7,9,11,14,16,18,20,22,24,26,28,30,33,35,37,39,41,43,45,47,50,

%T 52,54,56,58,60,62,64,67,69,71,73,75,77,79,82,84,86,88,90,92,94,96,99,

%U 101,103,105,107,109,111,113,116,118,120,122,124,126,128

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

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

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

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

%Y Cf. A370507, A376284, A376952, A376953, A376954, A376955, A376956, A376958, A376959, A376960.

%K nonn

%O 0,3

%A _Clark Kimberling_, Oct 17 2024