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

%I #4 Oct 19 2024 22:12:58

%S 1,1,2,3,4,6,7,8,10,11,12,14,15,16,18,19,20,22,23,24,26,27,28,30,31,

%T 32,34,35,36,38,39,41,42,43,45,46,47,49,50,51,53,54,55,57,58,59,61,62,

%U 64,65,66,68,69,70,72,73,74,76,77,78,80,81,83,84,85,87,88

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

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

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

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

%Y Cf. A370507, A376284, A376952, A376953, A376954, A376955, A376956, A376957, A376958, A376959, A376960.

%K nonn

%O 0,3

%A _Clark Kimberling_, Oct 17 2024