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 A247912 Numbers k such that A247911(k+1) = A247911(k). 3

%I #4 Sep 27 2014 19:02:56

%S 2,4,7,9,12,15,17,20,22,25,27,30,32,35,37,40,42,45,47,50,52,55,57,60,

%T 62,64,67,69,72,74,77,79,81,84,86,89,91,93,96,98,101,103,105,108,110,

%U 113,115,117,120,122,125,127,129,132,134,137,139,141,144,146

%N Numbers k such that A247911(k+1) = A247911(k).

%C Complement of A247913.

%H Clark Kimberling, <a href="/A247912/b247912.txt">Table of n, a(n) for n = 1..500</a>

%e A247911(n+1) - A247911(n) = (1,0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,0,...), 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], (2 # + 1)/u[2 # + 1] - E < n^-n &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A247911 *)

%t w = Differences[u]

%t Flatten[Position[w, 0]] (* A247912 *)

%t Flatten[Position[w, 1]] (* A247913 *)

%Y Cf. A247911, A247913, A247908, A247914.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Sep 27 2014

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Last modified March 2 21:56 EST 2024. Contains 370498 sequences. (Running on oeis4.)