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Numbers k such that A247908(k+1) = A247908(k) + 1.
3

%I #4 Sep 27 2014 19:02:44

%S 1,2,4,6,7,9,10,12,14,15,17,19,20,22,23,25,27,28,30,32,33,35,37,38,40,

%T 42,43,45,47,49,50,52,54,55,57,59,60,62,64,65,67,69,71,72,74,76,77,79,

%U 81,82,84,86,88,89,91,93,94,96,98,100,101,103,105,106,108

%N Numbers k such that A247908(k+1) = A247908(k) + 1.

%C Complement of A247909.

%H Clark Kimberling, <a href="/A247910/b247910.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 1.

%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, A247909.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 27 2014