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A247971
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Least k such that 4*k/v(2*k)^2 - Pi < 1/n, where the sequence v is defined in Comments.
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4
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1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 11, 12, 12, 13, 14, 15, 16, 16, 17, 18, 19, 19, 20, 21, 22, 23, 23, 24, 25, 26, 27, 27, 28, 29, 30, 30, 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
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OFFSET
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1,2
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COMMENTS
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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.
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REFERENCES
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Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19.
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LINKS
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EXAMPLE
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Approximations for the first few terms w(n) = 4*n/v(2*n)^2 - Pi and 1/n:
n ... 4*n/v(2*n)^2-Pi ... 1/n
1 ... 0.858407 .......... 1
2 ... 0.413963 ......... 0.5
3 ... 0.271741 .......... 0.333333
4 ... 0.202081 .......... 0.25
5 ... 0.160801 .......... 0.2
6 ... 0.133508 .......... 0.166666
a(2) = 2 because w(2) < 1/2 < w(1).
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MATHEMATICA
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$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[4 n/(v[2 n]^2) - Pi], N[1/n]}, {n, 1, 10}]]
f[n_] := f[n] = Select[Range[z], 4 #/(v[2 #]^2) - Pi < 1/n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]] (* A247971 *)
d = Differences[u]
v = Flatten[Position[d, 0]] (* A247972 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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