A248229 Numbers k such that A248227(k+1) = A248227(k) + 1.

2, 3, 5, 6, 7, 9, 10, 12, 13, 15, 16, 18, 19, 20, 22, 23, 25, 26, 28, 29, 31, 32, 33, 35, 36, 38, 39, 41, 42, 43, 45, 46, 48, 49, 51, 52, 54, 55, 56, 58, 59, 61, 62, 64, 65, 67, 68, 69, 71, 72, 74, 75, 77, 78, 80, 81, 82, 84, 85, 87, 88, 90, 91, 93, 94, 95

Offset

1,1

Comments

Since A248227(k+1) - A248227(k) is in {0,1} for k >= 1, A248228 and A248229 are complementary.

Links

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

Example

The difference sequence of A248227 is (0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, ...), so that A248228 = (1, 4, 8, 11, 14, 17, 2,...) and A248229 = (2, 3, 5, 6, 7, 9, 10, 12, 13, 15, 16, 18,...), the complement of A248228.

Mathematica

$MaxExtraPrecision = Infinity; z = 400; p[k_] := p[k] = Sum[1/h^4, {h, 1, k}];
N[Table[Zeta[4] - p[n], {n, 1, z/10}]]
f[n_] := f[n] = Select[Range[z], Zeta[4] - p[#] < 1/n^3 &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]   (* A248227 *)
Flatten[Position[Differences[u], 0]]  (* A248228 *)
Flatten[Position[Differences[u], 1]]  (* A248229 *)
f = Table[Floor[1/(Zeta[4] - p[n])], {n, 1, z}]  (* A248230 *)

Crossrefs

Cf. A248227, A248228, A248230.

Sequence in context: A364153 A230391 A284931 * A207672 A038161 A158746

Adjacent sequences: A248226 A248227 A248228 * A248230 A248231 A248232

Keyword

nonn,easy

Author

Clark Kimberling, Oct 05 2014

Status

approved