A248357 Numbers k such that A248355(k+1) = A248355(k) + 1.

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, 81, 87, 93, 99, 105, 111, 118, 125, 132, 139, 147, 154, 162, 170, 178, 187, 195

Offset

1,2

Comments

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.

Links

Clark Kimberling, Table of n, a(n) for n = 1..300

Mathematica

z = 200; p[k_] := p[k] = k*Sin[Pi/k]; N[Table[Pi - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], Pi - p[#] < 1/(2 n) &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]        (* A248355 *)
v = Flatten[Position[Differences[u], 0]]   (* A248356 *)
w = Flatten[Position[Differences[u], 1]]   (* A248357 *)
f = Table[Floor[1/(Pi - p[n])], {n, 1, z}] (* A248358 *)

Crossrefs

Cf. A248355, A248356, A248358, A248347.

Sequence in context: A321195 A249020 A258470 * A047872 A173254 A015851

Adjacent sequences: A248354 A248355 A248356 * A248358 A248359 A248360

Keyword

nonn,easy

Author

Clark Kimberling, Oct 05 2014

Status

approved