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A248566 Numbers k such that A248565(k+1) = A248565(k) + 1.

%I A248566 #4 Oct 15 2014 20:58:31
%S A248566 1,2,4,5,7,8,10,12,14,15,17,19,21,23,25,27,29,30,32,34,36,38,40,42,44,
%T A248566 46,48,50,52,54,56,58,59,61,63,65,67,69,71,73,75,77,79,81,83,85,87,89,
%U A248566 91,93,95,97,99,101,103,105,107,109,111,113,115,116,118
%N A248566 Numbers k such that A248565(k+1) = A248565(k) + 1.
%H A248566 Clark Kimberling, <a href="/A248566/b248566.txt">Table of n, a(n) for n = 1..500</a>
%e A248566 (A248565(k+1) - A248565(k)) = (1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2,...), so that A248566 = (1, 2, 4, 5, 7, 8, 10, 12, 14, ..) and A248567 = (3, 6, 9, 11, 13, 16, 18, 20, ...).
%t A248566 z = 2500; p[k_] := p[k] = Sum[1/(h*4^h), {h, 1, k}];
%t A248566 N[Table[p[k], {k, 1, z/5}], 12];
%t A248566 N[Table[Log[4/3] - p[n], {n, 1, z/5}]];
%t A248566 f[n_] := f[n] = Select[Range[z], Log[4/3] - p[#] < 1/8^n &, 1];
%t A248566 u = Flatten[Table[f[n], {n, 1, z}]] ;   (* A248565 *)
%t A248566 Flatten[Position[Differences[u], 1]];   (* A248566 *)
%t A248566 Flatten[Position[Differences[u], 2]];   (* A248567 *)
%Y A248566 Cf. A248565, A248567, A248560, A248563.
%K A248566 nonn,easy
%O A248566 1,2
%A A248566 _Clark Kimberling_, Oct 09 2014