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%I #4 Oct 15 2014 20:57:43
%S 1,2,4,6,8,10,12,15,17,19,22,24,27,29,32,34,37,39,42,44,47,49,52,55,
%T 57,60,62,65,68,70,73,75,78,81,83,86,89,91,94,96,99,102,104,107,110,
%U 112,115,118,120,123,126,128,131,134,136,139,142,144,147,150,152
%N Numbers k such that A248562(k+1) = A248562(k) + 1.
%H Clark Kimberling, <a href="/A248563/b248563.txt">Table of n, a(n) for n = 1..400</a>
%e (A248562(k+1) - A248562(k)) = (1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2,...), so that A248563 = (1, 2, 4, 6, 8, 10, 12, 15, 17, ..) and A248564 = (3, 5, 7, 9, 11, 13, 14, 16, ...).
%t z = 300; p[k_] := p[k] = Sum[1/(h*3^h), {h, 1, k}];
%t N[Table[Log[3/2] - p[n], {n, 1, z/5}]]
%t f[n_] := f[n] = Select[Range[z], Log[3/2] - p[#] < 1/6^n &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248562 *)
%t Flatten[Position[Differences[u], 1]] (* A248563 *)
%t Flatten[Position[Differences[u], 2]] (* A248564 *)
%Y Cf. A248562, A248564, A248560, A248566.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Oct 09 2014
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