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a(n) = least k such that (n*Pi/2)^(2k)/(2 k)! < 1.
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%I #9 Oct 16 2024 21:37:24

%S 1,2,4,6,8,10,12,14,16,18,20,23,25,27,29,31,33,35,37,39,42,44,46,48,

%T 50,52,54,56,59,61,63,65,67,69,71,73,76,78,80,82,84,86,88,91,93,95,97,

%U 99,101,103,105,108,110,112,114,116,118,120,122,125,127,129

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

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

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

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

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

%Y Cf. A370507, A376953, A376954, A376955, A376956, A376957, A376958, A376959, A376960.

%K nonn

%O 0,2

%A _Clark Kimberling_, Oct 12 2024