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 A247973 Least k such that Pi - (4*k+2)/v(2*k+2)^2 < 1/n, where the sequence v is defined in Comments. 4
 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, 22, 23, 23, 24, 25, 26, 27, 27, 28, 29, 30, 31, 31, 32, 33, 34, 34, 35, 36, 37, 38, 38, 39, 40, 41, 41, 42, 43, 44, 45, 45, 46, 47, 48, 49, 49, 50, 51, 52, 52, 53 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS 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. REFERENCES Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 21. LINKS EXAMPLE Approximations for the first few terms w(n) = Pi - (4*n+2)/v(2*n+2)^2 and 1/n: n ... Pi-(4*n+2)/v(2*n+2)^2 ... 1/n 1 ... 0.474926 ................ 1 2 ... 0.297148 ...............  0.5 3 ... 0.215878 ................ 0.333333 4 ... 0.169438 ................ 0.25 5 ... 0.139417 ................ 0.2 6 ... 0.118422 ................ 0.166666 a(3) = 2 because w(2) < 1/3 < w(1). MATHEMATICA \$RecursionLimit = Infinity; z = 400; v[1] = 0; v[2] = 1; v[n_] := v[n] = v[n - 1]/(n - 2) + v[n - 2]; TableForm[Table[{n, N[Pi - (4 n + 2)/(v[2 (n + 1)]^2)], N[1/n]}, {n, 1, 10}]] g[n_] := g[n] = Select[Range[z], Pi - (4 # + 2)/(v[2 (# + 1)]^2) < 1/n &, 1]; u = Flatten[Table[g[n], {n, 1, z}]]  (* A247973 *) d = Differences[u] Flatten[Position[d, 0]] (* A247974 *) CROSSREFS Cf. A247971, A247972, A247974. Sequence in context: A331266 A172103 A123731 * A352241 A195181 A003005 Adjacent sequences:  A247970 A247971 A247972 * A247974 A247975 A247976 KEYWORD nonn,easy AUTHOR Clark Kimberling, Sep 28 2014 STATUS approved

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Last modified August 18 04:47 EDT 2022. Contains 356204 sequences. (Running on oeis4.)