A248560 Numbers k such that A248559(k+1) = A248559(k) + 1.

1, 2, 3, 5, 6, 8, 10, 11, 13, 15, 17, 19, 21, 23, 25, 28, 30, 32, 34, 36, 38, 41, 43, 45, 47, 50, 52, 54, 56, 59, 61, 63, 65, 68, 70, 72, 75, 77, 79, 82, 84, 86, 88, 91, 93, 95, 98, 100, 102, 105, 107, 109, 112, 114, 116, 119, 121, 123, 126, 128, 130

Offset

1,2

Links

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

Example

(A248559(k+1) - A248559(k)) = (1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2,...), so that A248561 = (1, 2, 3, 5, 6, 8, 10, 11, 13, ...) and A248562 = (4, 7, 9, 12, 14, 16, 18, 20, ...).

Mathematica

z = 200; p[k_] := p[k] = Sum[1/(h*2^h), {h, 1, k}]
N[Table[Log[2] - p[n], {n, 1, z/5}]]
f[n_] := f[n] = Select[Range[z], Log[2] - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]    (* A248559 *)
Flatten[Position[Differences[u], 1]]   (* A248560 *)
Flatten[Position[Differences[u], 2]]   (* A248561 *)

Crossrefs

Cf. A248559, A248561.

Sequence in context: A186322 A092979 A135260 * A179180 A085921 A005243

Adjacent sequences: A248557 A248558 A248559 * A248561 A248562 A248563

Keyword

nonn,easy

Author

Clark Kimberling, Oct 09 2014

Status

approved