%I #4 Oct 08 2014 16:55:27
%S 1,2,3,4,6,7,9,11,13,16,19,21,24,28,31,34,38,42,46,51,55,60,65,70,75,
%T 81,87,93,99,105,111,118,125,132,139,147,154,162,170,178,187,195
%N Numbers k such that A248355(k+1) = A248355(k) + 1.
%C The difference sequence of A248355 is (1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1,...), so that A248356 = (5, 8, 10, 12, 14, 15, ...) and A248357 = (1, 2, 3, 4, 6, 7, 9, 11, 13, 16, 19,...); A248356 and A248357 are a complementary pair.
%H Clark Kimberling, <a href="/A248357/b248357.txt">Table of n, a(n) for n = 1..300</a>
%t z = 200; p[k_] := p[k] = k*Sin[Pi/k]; N[Table[Pi - p[n], {n, 1, z/10}]]
%t f[n_] := f[n] = Select[Range[z], Pi - p[#] < 1/(2 n) &, 1]
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248355 *)
%t v = Flatten[Position[Differences[u], 0]] (* A248356 *)
%t w = Flatten[Position[Differences[u], 1]] (* A248357 *)
%t f = Table[Floor[1/(Pi - p[n])], {n, 1, z}] (* A248358 *)
%Y Cf. A248355, A248356, A248358, A248347.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Oct 05 2014