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A376955
a(n) = least k such that (3n*Pi/4)^(2k)/(2 k)! < 1.
0
1, 3, 6, 9, 12, 15, 18, 21, 25, 28, 31, 34, 37, 41, 44, 47, 50, 53, 56, 60, 63, 66, 69, 72, 76, 79, 82, 85, 88, 92, 95, 98, 101, 104, 108, 111, 114, 117, 120, 124, 127, 130, 133, 136, 140, 143, 146, 149, 152, 156, 159, 162, 165, 168, 172, 175, 178, 181, 184
OFFSET
0,2
COMMENTS
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.
FORMULA
a(n) ~ 3*Pi*exp(1)*n/8 - log(n)/4. - Vaclav Kotesovec, Oct 13 2024
MATHEMATICA
a[n_] := Select[Range[200], (3n Pi/4)^(2 #)/(2 #)! < 1 &, 1];
Flatten[Table[a[n], {n, 0, 200}]]
KEYWORD
nonn,new
AUTHOR
Clark Kimberling, Oct 12 2024
STATUS
approved