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A248196 Numbers k such that A248180(k+1) = A248180(k) + 1.

%I A248196 #4 Oct 06 2014 23:01:22
%S A248196 3,5,7,9,11,13,15,17,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,
%T A248196 50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,87,89,91,93,
%U A248196 95,97,99,101,103,105,107,109,111,113,115,117,119,121,123
%N A248196 Numbers k such that A248180(k+1) = A248180(k) + 1.
%C A248196 Complement of A248195.
%H A248196 Clark Kimberling, <a href="/A248196/b248196.txt">Table of n, a(n) for n = 0..1500</a>
%e A248196 The difference sequence of A248180 is (0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1,...), so that A248195 = (1,2,4,6,8,10,12,14,16,19,...) and A248196 = (3,5,7,9,11,13,15,17,18,...)
%t A248196 $MaxExtraPrecision = Infinity;
%t A248196 z = 300; p[k_] := p[k] = Sum[1/Binomial[2 h + 1, h], {h, 0, k}] ;
%t A248196 r = Sum[1/Binomial[2 h + 1, h], {h, 0, Infinity}]  (* A248179 *)
%t A248196 r = 2/27 (9 + 2 Sqrt[3] \[Pi]); N[r, 20]
%t A248196 N[Table[r - p[n], {n, 0, z/10}]]
%t A248196 f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^n &, 1]
%t A248196 u = Flatten[Table[f[n], {n, 0, z}]]  (* A248180 *)
%t A248196 Flatten[Position[Differences[u], 0]] (* A248195 *)
%t A248196 Flatten[Position[Differences[u], 1]] (* A248196 *)
%Y A248196 Cf. A248180, A248195.
%K A248196 nonn,easy
%O A248196 0,1
%A A248196 _Clark Kimberling_, Oct 03 2014