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

Author

Clark Kimberling, Oct 04 2014

Keywords

• nonn
• easy

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A248186, A248187, A248183.

Programs

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 *)