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Numbers k such that A247973(k+1) = A247973(k).
4

%I #4 Sep 30 2014 15:56:20

%S 1,6,11,15,20,25,29,34,39,43,48,52,57,62,66,71,76,80,85,90,94,99,104,

%T 108,113,118,122,127,132,136,141,146,150,155,160,164,169,174,178,183,

%U 188,192,197,202,206,211,216,220,225,230,234,239,244,248,253,258

%N Numbers k such that A247973(k+1) = A247973(k).

%t $RecursionLimit = Infinity; z = 400; v[1] = 0; v[2] = 1;

%t v[n_] := v[n] = v[n - 1]/(n - 2) + v[n - 2];

%t TableForm[Table[{n, N[Pi - (4 n + 2)/(v[2 (n + 1)]^2)], N[1/n]}, {n, 1, 10}]]

%t g[n_] := g[n] = Select[Range[z], Pi - (4 # + 2)/(v[2 (# + 1)]^2) < 1/n &, 1];

%t u = Flatten[Table[g[n], {n, 1, z}]] (* A247973 *)

%t d = Differences[u]

%t Flatten[Position[d, 0]] (* A247974 *)

%Y Cf. A247971, A247972, A247973.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 28 2014