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a(n) = least k such that (n*Pi/3)^(2k)/(2 k)! < 1.
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%I #8 Oct 16 2024 21:37:36

%S 1,1,2,4,5,6,8,9,11,12,13,15,16,18,19,20,22,23,25,26,27,29,30,32,33,

%T 35,36,37,39,40,42,43,44,46,47,49,50,52,53,54,56,57,59,60,61,63,64,66,

%U 67,69,70,71,73,74,76,77,78,80,81,83,84,86,87,88,90,91,93

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

%C The numbers (n*Pi/3)^(2k)/(2 k)! are the coefficients in the Maclaurin series for cos x when x = n*Pi/3. If m>a(n), then (m*Pi/3)^(2k)/(2 k)! < 1. A375057 is a trisection of this sequence.

%F a(n) ~ Pi*exp(1)*n/6 - log(n)/4. - _Vaclav Kotesovec_, Oct 13 2024

%t a[n_] := Select[Range[300], (n Pi/3)^(2 #)/(2 #)! < 1 &, 1];

%t Flatten[Table[a[n], {n, 0, 200}]]

%Y Cf. A370507, A376952, A376954, A376955, A376956, A376957, A376958, A376959, A376960.

%K nonn

%O 0,3

%A _Clark Kimberling_, Oct 12 2024