%I #30 Nov 11 2024 20:31:30
%S 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,
%T 13,14,14,14,15,15,16,16,16,17,17,18,18,18,19,19,20,20,21,21,21,22,22,
%U 23,23,23,24,24,25,25,25,26,26,27,27,27,28,28,29,29
%N Least number k such that 1 - k*sin(1/k) < 1/n^2.
%C 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.
%C a(n+1) - a(n) is in {0,1} for n >= 1, so that the position sequences A138235 and A022840 partition the positive integers.
%C a(n) = A194986(n). - _Clark Kimberling_, Jan 15 2015
%H Clark Kimberling, <a href="/A248515/b248515.txt">Table of n, a(n) for n = 1..3000</a>
%F a(n) = ceiling (n/sqrt(6)) for n >= 1.
%e Approximations:
%e n 1-k*sin(1/k) 1/n^2
%e 1 0.158529 1
%e 2 0.041148 0.25
%e 3 0.018415 0.11111
%e 4 0.010384 0.0625
%e 5 0.006653 0.04
%e a(5) = 3 because 1 - 3*sin(1/3) < 1/25 < 1 - 2*sin(1/2).
%t z = 120; p[k_] := p[k] = k*Sin[1/k]; N[Table[1 - p[n], {n, 1, z/5}]]
%t f[n_] := f[n] = Select[Range[z], 1 - p[#] < 1/n^2 &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248515 *)
%t v = Flatten[Position[Differences[u], 0]] (* A138235 *)
%t w = Flatten[Position[Differences[u], 1]] (* A022840 *)
%t Table[Ceiling[n / Sqrt[6]], {n, 70}] (* _Vincenzo Librandi_, Jun 17 2015 *)
%o (Magma) [Ceiling(n/Sqrt(6)): n in [1..70]]; // _Vincenzo Librandi_, Jun 17 2015
%Y Cf. A194986, A033581, A248470, A138235, A022840.
%K nonn,easy
%O 1,3
%A _Clark Kimberling_, Oct 08 2014