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A248187 Numbers k such that A248186(k+1) = A248186(k).

%I A248187 #4 Oct 06 2014 23:01:00
%S A248187 1,2,3,6,9,13,16,19,22,26,29,32,35,39,42,45,48,52,55,58,61,65,68,71,
%T A248187 75,78,81,84,88,91,94,97,101,104,107,110,114,117,120,123,127,130,133,
%U A248187 136,140,143,146,150,153,156,159,163,166,169,172,176,179,182,185
%N A248187 Numbers k such that A248186(k+1) = A248186(k).
%H A248187 Clark Kimberling, <a href="/A248187/b248187.txt">Table of n, a(n) for n = 1..500</a>
%e A248187 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 A248187 $MaxExtraPrecision = Infinity;
%t A248187 z = 800; p[k_] := p[k] = Sum[1/(h*(h + 1)*(h + 2)*(h + 3)), {h, 1, k}];
%t A248187 N[Table[1/18 - p[n], {n, 1, z/10}]]
%t A248187 f[n_] := f[n] = Select[Range[z], 1/18 - p[#] < 1/n^3 &, 1]
%t A248187 u = Flatten[Table[f[n], {n, 1, z}]]   (* A248186 *)
%t A248187 Flatten[Position[Differences[u], 0]]  (* A248187 *)
%t A248187 Flatten[Position[Differences[u], 1]]  (* A248188 *)
%Y A248187 Cf. A248186, A248189, A248183.
%K A248187 nonn,easy
%O A248187 1,2
%A A248187 _Clark Kimberling_, Oct 04 2014