A248515 Least number k such that 1 - k*sin(1/k) < 1/n^2.

1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10, 11, 11, 12, 12, 12, 13, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 18, 19, 19, 20, 20, 21, 21, 21, 22, 22, 23, 23, 23, 24, 24, 25, 25, 25, 26, 26, 27, 27, 27, 28, 28, 29, 29

Offset: 1

Author

Clark Kimberling, Oct 08 2014

Keywords

• nonn
• easy

Comments

This sequences provides insight into the manner of convergence of n*sin(1/n). One may also consider: [1/(1 - n*sin(1/n))] = 6*n^2 = A033581(n) for n >= 1.

a(n+1) - a(n) is in {0,1} for n >= 1, so that the position sequences A138235 and A022840 partition the positive integers.

a(n) = A194986(n). - Clark Kimberling, Jan 15 2015

Examples

			Approximations:
n      1-k*sin(1/k)     1/n^2
1      0.158529         1
2      0.041148         0.25
3      0.018415         0.11111
4      0.010384         0.0625
5      0.006653         0.04
a(5) = 3 because 1 - 3*sin(1/3) < 1/25 < 1 - 2*sin(1/2).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..3000

Crossrefs

Cf. A194986, A033581, A248470, A138235, A022840.

Programs

Magma

[Ceiling(n/Sqrt(6)): n in [1..70]]; // Vincenzo Librandi, Jun 17 2015

Mathematica

z = 120; p[k_] := p[k] = k*Sin[1/k]; N[Table[1 - p[n], {n, 1, z/5}]]
f[n_] := f[n] = Select[Range[z], 1 - p[#] < 1/n^2 &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]        (* A248515 *)
v = Flatten[Position[Differences[u], 0]]   (* A138235 *)
w = Flatten[Position[Differences[u], 1]]   (* A022840 *)
Table[Ceiling[n / Sqrt[6]], {n, 70}] (* Vincenzo Librandi, Jun 17 2015 *)

Formula

a(n) = ceiling (n/sqrt(6)) for n >= 1.