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

Author

Clark Kimberling, Oct 05 2014

Keywords

• nonn
• easy

Comments

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

Examples

			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.
		

Links

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

Crossrefs

Cf. A248227, A248228, A248230.

Programs

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 *)