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a(n) = least k such that (n Pi)^(2 k + 1)/(2 k + 1)! < 1.
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%I #4 Oct 02 2024 14:28:48

%S 1,3,7,11,16,20,24,28,33,37,41,45,50,54,58,62,67,71,75,79,84,88,92,96,

%T 101,105,109,113,118,122,126,131,135,139,143,148,152,156,160,165,169,

%U 173,177,182,186,190,194,199,203,207,212,216,220,224,229,233,237

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

%C The numbers (n Pi)^(2 k + 1)/(2 k + 1)! are the coefficients in the Maclaurin series for sin x when x = n*Pi.

%C (n Pi)^(2 k + 1)/(2 k + 1)! < 1 for every k >= a(n).

%t z = 300; r = Pi;

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

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

%Y Cf. A374987, A376456, A376457, A375054, A375057.

%K nonn

%O 0,2

%A _Clark Kimberling_, Oct 01 2024