A132209 a(0) = 0 and a(n) = 2*n^2 + 2*n - 1, for n>=1.

0, 3, 11, 23, 39, 59, 83, 111, 143, 179, 219, 263, 311, 363, 419, 479, 543, 611, 683, 759, 839, 923, 1011, 1103, 1199, 1299, 1403, 1511, 1623, 1739, 1859, 1983, 2111, 2243, 2379, 2519, 2663, 2811, 2963, 3119, 3279, 3443, 3611, 3783, 3959, 4139, 4323, 4511

Offset: 0

Author

Mohamed Bouhamida, Nov 06 2007

Keywords

• nonn

Comments

Previous name was: Sequence gives X values that satisfy the integer equation 2*X^3 + 3*X^2 = Y^2.

To find Y values: b(n) = (2*n^2 + 2*n - 1)*(2*n - 1).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Crossrefs

Cf. A005563, A046092, A001082, A002378, A036666, A062717, A028347, A087475, A000217.

Programs

Magma

[0] cat [2*n^2+2*n-1: n in [1..50]]; // Vincenzo Librandi, Sep 22 2015

Mathematica

Join[{0}, LinearRecurrence[{3, -3, 1}, {3, 11, 23}, 40]] (* Vincenzo Librandi, Sep 22 2015 *)

PARI

for(n=0,50, print1(if(n==0, 0, 2*n^2 + 2*n -1), ", ")) \\ G. C. Greubel, Jul 13 2017

Formula

a(n) = 2*n^2 + 2*n - 1 for n>=1.

G.f.: x*(1+x)*(3-x)/(1-x)^3. - R. J. Mathar, Nov 14 2007

E.g.f.: 1 + (2*x^2 + 4*x -1)*exp(x). - G. C. Greubel, Jul 13 2017

From Amiram Eldar, Mar 07 2021: (Start)

Sum_{n>=1} 1/a(n) = 1 + sqrt(3)*Pi*tan(sqrt(3)*Pi/2)/6.

Product_{n>=1} (1 + 1/a(n)) = -Pi*sec(sqrt(3)*Pi/2)/2.

Product_{n>=1} (1 - 1/a(n)) = cos(sqrt(5)*Pi/2)*sec(sqrt(3)*Pi/2)/2. (End)

Extensions

Edited by the Associate Editors of the OEIS, Nov 15 2009

More terms from Vincenzo Librandi, Sep 22 2015

Shorter name (using formula given) from Joerg Arndt, Sep 27 2015