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

Author

Clark Kimberling, Oct 05 2014

Keywords

• nonn
• easy

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A248231, A248233, A248234, A248227.

Programs

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 *)