A248188 - OEIS

%I #5 Oct 06 2014 23:01:08

%S 4,5,7,8,10,11,12,14,15,17,18,20,21,23,24,25,27,28,30,31,33,34,36,37,

%T 38,40,41,43,44,46,47,49,50,51,53,54,56,57,59,60,62,63,64,66,67,69,70,

%U 72,73,74,76,77,79,80,82,83,85,86,87,89,90,92,93,95,96,98

%N Numbers k such that A248186(k+1) = A248186(k) + 1.

%C a(n) = A059539(n+2) = [3^(1/3)*(n+2)] for n = 1..655, but a(656) = 948 = A059539(658)-1.

%H Clark Kimberling, <a href="/A248188/b248188.txt">Table of n, a(n) for n = 1..1000</a>

%e The difference sequence of A248186 is (0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, ...), so that A248187 = (1, 2, 3, 6, 9, 13, 16, 19, 22,...) and A248188 = (4, 5, 7, 8, 10, 11, 12, 14, 15, 17,...), the complement of A248186.

%t $MaxExtraPrecision = Infinity;

%t z = 800; p[k_] := p[k] = Sum[1/(h*(h + 1)*(h + 2)*(h + 3)), {h, 1, k}];

%t N[Table[1/18 - p[n], {n, 1, z/10}]]

%t f[n_] := f[n] = Select[Range[z], 1/18 - p[#] < 1/n^3 &, 1]

%t u = Flatten[Table[f[n], {n, 1, z}]]   (* A248186 *)

%t Flatten[Position[Differences[u], 0]]  (* A248187 *)

%t Flatten[Position[Differences[u], 1]]  (* A248188 *)

%Y Cf. A248186, A248187, A248183.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Oct 04 2014