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A248195 Numbers k such that A248180(k+1) = A248180(k).

%I A248195 #5 Oct 06 2014 23:01:15
%S A248195 1,2,4,6,8,10,12,14,16,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,
%T A248195 49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83,85,88,90,92,94,
%U A248195 96,98,100,102,104,106,108,110,112,114,116,118,120,122,124
%N A248195 Numbers k such that A248180(k+1) = A248180(k).
%H A248195 Clark Kimberling, <a href="/A248195/b248195.txt">Table of n, a(n) for n = 0..1500</a>
%e A248195 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 A248195 $MaxExtraPrecision = Infinity;
%t A248195 z = 300; p[k_] := p[k] = Sum[1/Binomial[2 h + 1, h], {h, 0, k}] ;
%t A248195 r = Sum[1/Binomial[2 h + 1, h], {h, 0, Infinity}]  (* A248179 *)
%t A248195 r = 2/27 (9 + 2 Sqrt[3] \[Pi]); N[r, 20]
%t A248195 N[Table[r - p[n], {n, 0, z/10}]]
%t A248195 f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^n &, 1]
%t A248195 u = Flatten[Table[f[n], {n, 0, z}]]  (* A248180 *)
%t A248195 Flatten[Position[Differences[u], 0]] (* A248195 *)
%t A248195 Flatten[Position[Differences[u], 1]] (* A248196 *)
%Y A248195 Cf. A248180, A248196.
%K A248195 nonn,easy
%O A248195 0,2
%A A248195 _Clark Kimberling_, Oct 03 2014