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a(n) = least k such that (2n*Pi/3)^(2k)/(2 k)! < 1.
10

%I #10 Oct 16 2024 21:37:46

%S 1,2,5,8,11,13,16,19,22,25,27,30,33,36,39,42,44,47,50,53,56,59,61,64,

%T 67,70,73,76,78,81,84,87,90,93,95,98,101,104,107,110,113,115,118,121,

%U 124,127,130,132,135,138,141,144,147,149,152,155,158,161,164,167

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

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

%F a(n) ~ Pi*exp(1)*n/3 - log(n)/4. - _Vaclav Kotesovec_, Oct 13 2024

%t a[n_] := Select[Range[200], (2n Pi/3)^(2 #)/(2 #)! < 1 &, 1];

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

%Y Cf. A370507, A376952, A376953, A376955, A376956, A376957, A376958, A376959, A376960.

%K nonn

%O 0,2

%A _Clark Kimberling_, Oct 12 2024