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A376957
a(n) = least k such that (n Pi/2)^(2k+1)/(2k+1)! < 1.
10
1, 1, 3, 5, 7, 9, 11, 14, 16, 18, 20, 22, 24, 26, 28, 30, 33, 35, 37, 39, 41, 43, 45, 47, 50, 52, 54, 56, 58, 60, 62, 64, 67, 69, 71, 73, 75, 77, 79, 82, 84, 86, 88, 90, 92, 94, 96, 99, 101, 103, 105, 107, 109, 111, 113, 116, 118, 120, 122, 124, 126, 128
OFFSET
0,3
COMMENTS
The numbers (n Pi/2)^(2k+1)/(2k+1)! are the coefficients in the Maclaurin series for sin x when x = Pi/2. If m>a(n), then (n Pi/2)^(2k+1)/(2k+1)! < 1.
MATHEMATICA
a[n_] := Select[Range[z], (n Pi/2)^(2 # + 1)/(2 # + 1)! < 1 &, 1]
Flatten[Table[a[n], {n, 0, 100}]]
KEYWORD
nonn
AUTHOR
Clark Kimberling, Oct 17 2024
STATUS
approved