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%I #4 Oct 06 2014 23:00:45
%S 3,4,6,7,9,10,11,13,14,16,17,19,20,21,23,24,26,27,28,30,31,33,34,36,
%T 37,38,40,41,43,44,45,47,48,50,51,53,54,55,57,58,60,61,62,64,65,67,68,
%U 69,71,72,74,75,77,78,79,81,82,84,85,86,88,89,91,92,94,95
%N Numbers k such that A248183(k+1) = A248183(k) + 1.
%H Clark Kimberling, <a href="/A248185/b248185.txt">Table of n, a(n) for n = 1..1000</a>
%e The difference sequence of A248183 is (0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1,...), so that A248184 = (1, 2, 5, 8, 12, 15, 18, 22,...) and A248185 = (3, 4, 6, 7, 9, 10, 11, 13, 14, 16, 17,...), the complement of A248184.
%t z = 200; p[k_] := p[k] = Sum[1/(h*(h + 1)*(h + 2)), {h, 1, k}] ;
%t N[Table[1/4 - p[n], {n, 1, z/10}]]
%t f[n_] := f[n] = Select[Range[z], 1/4 - p[#] < 1/n^2 &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248183 *)
%t Flatten[Position[Differences[u], 0]] (* A248184 *)
%t Flatten[Position[Differences[u], 1]] (* A248185 *)
%Y Cf. A248183, A248184.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Oct 04 2014
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