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A376958
a(n) = least k such that (n Pi/3)^(2k+1)/(2k+1)! < 1.
0
1, 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 16, 17, 19, 20, 21, 23, 24, 26, 27, 28, 30, 31, 33, 34, 35, 37, 38, 40, 41, 43, 44, 45, 47, 48, 50, 51, 52, 54, 55, 57, 58, 60, 61, 62, 64, 65, 67, 68, 69, 71, 72, 74, 75, 77, 78, 79, 81, 82, 84, 85, 86, 88, 89, 91, 92
OFFSET
0,3
COMMENTS
The numbers (n Pi/3)^(2k+1)/(2k+1)! are the coefficients in the Maclaurin series for sin x when x = Pi/3. If m>a(n), then (n Pi/3)^(2k+1)/(2k+1)! < 1.
MATHEMATICA
a[n_] := Select[Range[z], (n Pi/3)^(2 # + 1)/(2 # + 1)! < 1 &, 1]
Flatten[Table[a[n], {n, 0, 100}]]
KEYWORD
nonn,new
AUTHOR
Clark Kimberling, Oct 17 2024
STATUS
approved