A248188 Numbers k such that A248186(k+1) = A248186(k) + 1.

4, 5, 7, 8, 10, 11, 12, 14, 15, 17, 18, 20, 21, 23, 24, 25, 27, 28, 30, 31, 33, 34, 36, 37, 38, 40, 41, 43, 44, 46, 47, 49, 50, 51, 53, 54, 56, 57, 59, 60, 62, 63, 64, 66, 67, 69, 70, 72, 73, 74, 76, 77, 79, 80, 82, 83, 85, 86, 87, 89, 90, 92, 93, 95, 96, 98

Offset

1,1

Comments

a(n) = A059539(n+2) = [3^(1/3)*(n+2)] for n = 1..655, but a(656) = 948 = A059539(658)-1.

Links

Clark Kimberling, Table of n, a(n) for n = 1..1000

Example

The difference sequence of A248186 is (0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, ...), so that A248187 = (1, 2, 3, 6, 9, 13, 16, 19, 22,...) and A248188 = (4, 5, 7, 8, 10, 11, 12, 14, 15, 17,...), the complement of A248186.

Mathematica

$MaxExtraPrecision = Infinity;
z = 800; p[k_] := p[k] = Sum[1/(h*(h + 1)*(h + 2)*(h + 3)), {h, 1, k}];
N[Table[1/18 - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], 1/18 - p[#] < 1/n^3 &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248186 *)
Flatten[Position[Differences[u], 0]]  (* A248187 *)
Flatten[Position[Differences[u], 1]]  (* A248188 *)

Crossrefs

Cf. A248186, A248187, A248183.

Sequence in context: A119642 A276705 A168044 * A174047 A285085 A231507

Adjacent sequences: A248185 A248186 A248187 * A248189 A248190 A248191

Keyword

nonn,easy

Author

Clark Kimberling, Oct 04 2014

Status

approved