A248232 Numbers k such that A248231(k+1) = A248231(k).

1, 4, 8, 11, 15, 18, 22, 25, 28, 32, 35, 39, 42, 46, 49, 52, 56, 59, 63, 66, 69, 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, 168, 172, 175, 179, 182, 186, 189, 192, 196

Offset

1,2

Comments

Since A248231(k+1) - A248232(k) is in {0,1} for k >= 1, A248232 and A248233 are complementary.

Links

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

Example

The difference sequence of A248231 is (0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, ...), so that A248232 = (1, 4, 8, 11, 15, 18, 22, 25, 28,...) and A248233 = (2, 3, 5, 6, 7, 9, 10, 12, 13, 14, 16, 17,...), the complement of A248232.

Mathematica

z = 400; p[k_] := p[k] = Sum[1/h^5, {h, 1, k}]; N[Table[Zeta[5] - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], Zeta[5] - p[#] < 1/n^4 &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248231 *)
Flatten[Position[Differences[u], 0]]  (* A248232 *)
Flatten[Position[Differences[u], 1]]  (* A248233 *)
Table[Floor[1/(Zeta[5] - p[n])], {n, 1, z}]  (* A248234 *)

Crossrefs

Cf. A248231, A248233, A248234, A248227.

Sequence in context: A311049 A073302 A191105 * A047346 A198270 A214971

Adjacent sequences: A248229 A248230 A248231 * A248233 A248234 A248235

Keyword

nonn,easy

Author

Clark Kimberling, Oct 05 2014

Status

approved