OFFSET
1,1
COMMENTS
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
REFERENCES
M. N. Huxley, Area, Lattice Points and Exponential Sums, Oxford, 1996, p. 7.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Steven Edwards and William Griffiths, Generalizations of Delannoy and cross polytope numbers, Fib. Q., Vol. 55, No. 4 (2017), pp. 356-366.
Steven Edwards and William Griffiths, On Generalized Delannoy Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.3.6.
Kival Ngaokrajang, Illustration of the Pappus chain of the symmetric Arbelos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 4*(n-1) + a(n-1) for n > 1, a(1)=3. - Vincenzo Librandi, Nov 16 2010
a(n) = A046092(n) + 3. - Reinhard Zumkeller, Dec 15 2013
G.f.: x*(3 - 2*x + 3*x^2)/(1 - x)^3. - Vincenzo Librandi, Aug 03 2014
a(n) = A027575(n-2)/2. - Michel Marcus, Nov 11 2015
Sum_{n>=1} 1/a(n) = Pi*tanh(sqrt(5)*Pi/2)/(2*sqrt(5)). - Amiram Eldar, Dec 23 2022
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: exp(x)*(2*x^2 + 3) - 3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)
MATHEMATICA
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 *)
PROG
(PARI) a(n)=2*n^2-2*n+3 \\ Charles R Greathouse IV, Jun 13 2012
(PARI) Vec(x*(3-2*x+3*x^2)/(1-x)^3 + O(x^50)) \\ Altug Alkan, Nov 11 2015
(Haskell)
a097080 n = 2 * n * (n - 1) + 3 -- Reinhard Zumkeller, Dec 15 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Nov 02 2008
STATUS
approved