A248184 Numbers k such that A248183(k+1) = A248183(k).

1, 2, 5, 8, 12, 15, 18, 22, 25, 29, 32, 35, 39, 42, 46, 49, 52, 56, 59, 63, 66, 70, 73, 76, 80, 83, 87, 90, 93, 97, 100, 104, 107, 110, 114, 117, 121, 124, 128, 131, 134, 138, 141, 145, 148, 151, 155, 158, 162, 165, 169, 172, 175, 179, 182, 186, 189, 192

Offset

1,2

Comments

The difference sequence of A248183 is (0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1,...), so that A248184 = (1, 2, 5, 8, 12, 15, 18, 22,...) and A248185 = (3, 4, 6, 7, 9, 10, 11, 13, 14, 16, 17,...), the complement of A248185.

Links

Clark Kimberling, Table of n, a(n) for n = 1..500

Mathematica

z = 200; p[k_] := p[k] = Sum[1/(h*(h + 1)*(h + 2)), {h, 1, k}] ;
N[Table[1/4 - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], 1/4 - p[#] < 1/n^2 &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248183 *)
Flatten[Position[Differences[u], 0]]  (* A248184 *)
Flatten[Position[Differences[u], 1]]  (* A248185 *)

Crossrefs

Cf. A248183, A248185.

Sequence in context: A092767 A190000 A080753 * A156745 A087347 A062468

Adjacent sequences: A248181 A248182 A248183 * A248185 A248186 A248187

Keyword

nonn,easy

Author

Clark Kimberling, Oct 04 2014

Status

approved