A247973 - OEIS

%I #8 Oct 19 2014 15:41:13

%S 1,1,2,3,4,5,5,6,7,8,9,9,10,11,12,12,13,14,15,16,16,17,18,19,20,20,21,

%T 22,23,23,24,25,26,27,27,28,29,30,31,31,32,33,34,34,35,36,37,38,38,39,

%U 40,41,41,42,43,44,45,45,46,47,48,49,49,50,51,52,52,53

%N Least k such that Pi - (4*k+2)/v(2*k+2)^2 < 1/n, where the sequence v is defined in Comments.

%C The sequence v is defined as follows:  v(1) = 0, v(2) = 1, v(n) = v(n-1)/(n-2) + v(n-2).  It appears that a(n+1) - a(n) is in {0,1} for n >= 2.

%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 21.

%e Approximations for the first few terms w(n) = Pi - (4*n+2)/v(2*n+2)^2 and 1/n:

%e n ... Pi-(4*n+2)/v(2*n+2)^2 ... 1/n

%e 1 ... 0.474926 ................ 1

%e 2 ... 0.297148 ...............  0.5

%e 3 ... 0.215878 ................ 0.333333

%e 4 ... 0.169438 ................ 0.25

%e 5 ... 0.139417 ................ 0.2

%e 6 ... 0.118422 ................ 0.166666

%e a(3) = 2 because w(2) < 1/3 < w(1).

%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, A247974.

%K nonn,easy

%O 1,3

%A _Clark Kimberling_, Sep 28 2014