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a(n) = least k such that (3n*Pi/4)^(2k)/(2 k)! < 1.
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%I #8 Oct 16 2024 21:37:59

%S 1,3,6,9,12,15,18,21,25,28,31,34,37,41,44,47,50,53,56,60,63,66,69,72,

%T 76,79,82,85,88,92,95,98,101,104,108,111,114,117,120,124,127,130,133,

%U 136,140,143,146,149,152,156,159,162,165,168,172,175,178,181,184

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

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

%F a(n) ~ 3*Pi*exp(1)*n/8 - log(n)/4. - _Vaclav Kotesovec_, Oct 13 2024

%t a[n_] := Select[Range[200], (3n Pi/4)^(2 #)/(2 #)! < 1 &, 1];

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

%Y Cf. A370507, A376952, A376953, A376954, A376956, A376957, A376958, A376959, A376960.

%K nonn

%O 0,2

%A _Clark Kimberling_, Oct 12 2024