A248233 Numbers k such that A248231(k+1) = A248231(k) + 1.

2, 3, 5, 6, 7, 9, 10, 12, 13, 14, 16, 17, 19, 20, 21, 23, 24, 26, 27, 29, 30, 31, 33, 34, 36, 37, 38, 40, 41, 43, 44, 45, 47, 48, 50, 51, 53, 54, 55, 57, 58, 60, 61, 62, 64, 65, 67, 68, 70, 71, 72, 74, 75, 77, 78, 79, 81, 82, 84, 85, 86, 88, 89, 91, 92, 94

Offset

1,1

Comments

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

This appears to be a duplicate of A097432. - R. J. Mathar, Oct 10 2014

Links

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

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, A248232, A248234, A248227.

Sequence in context: A371188 A195121 A047332 * A097432 A364153 A230391

Adjacent sequences: A248230 A248231 A248232 * A248234 A248235 A248236

Keyword

nonn,easy

Author

Clark Kimberling, Oct 05 2014

Status

approved