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Numbers k such that A248559(k+1) = A248559(k) + 1.
5

%I #4 Oct 15 2014 20:56:45

%S 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 50,52,54,56,59,61,63,65,68,70,72,75,77,79,82,84,86,88,91,93,95,98,

%U 100,102,105,107,109,112,114,116,119,121,123,126,128,130

%N Numbers k such that A248559(k+1) = A248559(k) + 1.

%H Clark Kimberling, <a href="/A248560/b248560.txt">Table of n, a(n) for n = 1..500</a>

%e (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 z = 200; p[k_] := p[k] = Sum[1/(h*2^h), {h, 1, k}]

%t N[Table[Log[2] - p[n], {n, 1, z/5}]]

%t f[n_] := f[n] = Select[Range[z], Log[2] - p[#] < 1/3^n &, 1]

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248559 *)

%t Flatten[Position[Differences[u], 1]] (* A248560 *)

%t Flatten[Position[Differences[u], 2]] (* A248561 *)

%Y Cf. A248559, A248561.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Oct 09 2014