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%I #4 Sep 27 2014 19:02:32
%S 3,5,8,11,13,16,18,21,24,26,29,31,34,36,39,41,44,46,48,51,53,56,58,61,
%T 63,66,68,70,73,75,78,80,83,85,87,90,92,95,97,99,102,104,107,109,111,
%U 114,116,119,121,123,126,128,131,133,135,138,140,142,145,147
%N Numbers k such that A247908(k+1) = A247908(k).
%C Complement of A247910.
%H Clark Kimberling, <a href="/A247909/b247909.txt">Table of n, a(n) for n = 1..500</a>
%e A247908(n+1) - A247908(n) = (1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,1,...), and a(n) is the position of the n-th 0.
%t $RecursionLimit = 1000; $MaxExtraPrecision = 1000;
%t z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
%t f[n_] := f[n] = Select[Range[z], E - 2 #/u[2 #] < 1/n^n &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A247908 *)
%t w = Differences[u]
%t Flatten[Position[w, 0]] (* A247909 *)
%t Flatten[Position[w, 1]] (* A247910 *)
%Y Cf. A247908, A247910.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, Sep 27 2014
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