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%I A248560 #4 Oct 15 2014 20:56:45
%S A248560 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,
%T A248560 50,52,54,56,59,61,63,65,68,70,72,75,77,79,82,84,86,88,91,93,95,98,
%U A248560 100,102,105,107,109,112,114,116,119,121,123,126,128,130
%N A248560 Numbers k such that A248559(k+1) = A248559(k) + 1.
%H A248560 Clark Kimberling, <a href="/A248560/b248560.txt">Table of n, a(n) for n = 1..500</a>
%e A248560 (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, ...).
%t A248560 z = 200; p[k_] := p[k] = Sum[1/(h*2^h), {h, 1, k}]
%t A248560 N[Table[Log[2] - p[n], {n, 1, z/5}]]
%t A248560 f[n_] := f[n] = Select[Range[z], Log[2] - p[#] < 1/3^n &, 1]
%t A248560 u = Flatten[Table[f[n], {n, 1, z}]]    (* A248559 *)
%t A248560 Flatten[Position[Differences[u], 1]]   (* A248560 *)
%t A248560 Flatten[Position[Differences[u], 2]]   (* A248561 *)
%Y A248560 Cf. A248559, A248561.
%K A248560 nonn,easy
%O A248560 1,2
%A A248560 _Clark Kimberling_, Oct 09 2014

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