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A097080
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a(n) = 2*n^2 - 2*n + 3.
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22
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3, 7, 15, 27, 43, 63, 87, 115, 147, 183, 223, 267, 315, 367, 423, 483, 547, 615, 687, 763, 843, 927, 1015, 1107, 1203, 1303, 1407, 1515, 1627, 1743, 1863, 1987, 2115, 2247, 2383, 2523, 2667, 2815, 2967, 3123, 3283, 3447, 3615, 3787, 3963, 4143, 4327, 4515, 4707
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OFFSET
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1,1
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COMMENTS
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The rational numbers may be totally ordered, first by height (see A002246) and then by magnitude; every rational number of height n appears in this ordering at a position <= a(n).
The old entry with this sequence number was a duplicate of A027356.
This is also the sum of the pairwise averages of five consecutive triangular numbers in A000217. Start with A000217(0): (0+1)/2 + (1+3)/2 + (3+6)/2 + (6+10)/2 = 15, which is the third term of this sequence. Simply continue to create this sequence. - J. M. Bergot, Jun 13 2012
2*a(n) is inverse radius (curvature) of the touching circle of the large semicircle (radius 1) and the n-th and (n-1)-st circles of the Pappus chain of the symmetric Arbelos. One can use Descartes three circle theorem to find the solution 4*n^2 - 4*n + 6, n >= 1. Note that the circle with curvature 6 is also the third circle of the clockwise Pappus chain, which is A059100(2) (by symmetry). See the illustration. - Wolfdieter Lang and Kival Ngaokrajang, Jul 01 2015
Numbers k such that 2*k - 5 is a square. - Bruno Berselli, Nov 08 2017
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REFERENCES
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M. N. Huxley, Area, Lattice Points and Exponential Sums, Oxford, 1996; p. 7.
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LINKS
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FORMULA
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Sum_{n>=1} 1/a(n) = Pi*tanh(sqrt(5)*Pi/2)/(2*sqrt(5)). - Amiram Eldar, Dec 23 2022
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MATHEMATICA
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Table[2n^2-2n+3, {n, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {3, 7, 15}, 50] (* Harvey P. Dale, Aug 02 2014 *)
CoefficientList[Series[(3 - 2 x + 3 x^2)/(1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 03 2014 *)
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PROG
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(PARI) Vec(x*(3-2*x+3*x^2)/(1-x)^3 + O(x^50)) \\ Altug Alkan, Nov 11 2015
(Haskell)
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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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