A248611 Numbers k such that A248610(k+1) = A248610(k).

1, 3, 5, 8, 11, 14, 18, 22, 26, 30, 34, 38, 42, 47, 51, 55, 60, 64, 69, 73, 78, 82, 87, 91, 96, 100, 105, 110, 114, 119, 123, 128, 133, 137, 142, 147, 151, 156, 161, 165, 170, 175, 179, 184, 189, 193, 198, 203, 207, 212, 217, 221, 226, 231, 236, 240, 245

Offset

1,2

Links

Clark Kimberling, Table of n, a(n) for n = 1..400

Example

(A248610(k+1) - A248610(k)) = (0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, ...), so that A248611 = (1, 3, 5, 8, 11, 14, 18, 22, 26, ..) and A248612 = (2, 4, 6, 7, 9, 10, 12, 13, ...).

Mathematica

z = 300; p[k_] := p[k] = Sum[1/((h^2)*Binomial[2 h, h]), {h, 1, k}]
d = N[Table[Pi^2/18 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], Pi^2/18 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248610 *)
d = Differences[u]
v = Flatten[Position[d, 0]]  (* A248611 *)
w = Flatten[Position[d, 1]]  (* A248612 *)

Crossrefs

Cf. A248610, A248612, A248607.

Sequence in context: A287414 A052488 A076372 * A005356 A060432 A156023

Adjacent sequences: A248608 A248609 A248610 * A248612 A248613 A248614

Keyword

nonn,easy

Author

Clark Kimberling, Oct 10 2014

Status

approved